// Numbas version: exam_results_page_options {"name": "Natural Sciences", "metadata": {"description": "

A portfolio of NUMBAS questions created for first year Natural Sciences students. The questions cover the topics:

Linear functions
Quadratic functions
Differentiation
Integration
Explonatial and logarithms
Further differentiation
Further Integration
Trigonometric Functions

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Interpreting line graphs depicting the decrease of temperature in a mixture over time. Estimating the temperature of the mixture at a given time point and vice versa.

", "licence": "None specified"}, "statement": "

Zinc is mixed with an acid in an insulated container. After the initial temperature rise, the temperature of the mixture, $T$ $^\\circ$C, falls over time, $t$ minutes, as shown on the graph:

{geogebra_applet('https://www.geogebra.org/m/frzwhq9u', defs)}

", "advice": "

a) To find the initial temperature of the mixture, we need to find the point on the $T$-axis where $t=0$. So, we need to find the point where the line intersects with the $T$-axis.

Here, a good estimate could be $\\var{b} ^\\circ C$.

{geogebra_applet('https://www.geogebra.org/m/evvd3nua', defadvice1)}

b) We need to find the point on $T$-axis where the temperature $t=\\var{t_1}$. So we first need to estimate the point on the $t$-axis where $t=\\var{t_1}$. From there, we draw a vertical line to find the point on the graph. From the point on the graph, we draw a horizontal line and estimate the value on the $T$-axis where the horizontal line and the axis intersect.

Here, a good estimate could be $ \\var{ans_c_estimate} ^\\circ C$.

{geogebra_applet('https://www.geogebra.org/m/evvd3nua', defadvice2)}

c) We need to find the point on $t$-axis where the temperature $T=\\var{temp_1}$. So we first need to estimate the point on the $T$-axis where $T=\\var{temp_1}$. From there, we draw a horizontal line to find the point on the graph. From the point on the graph, we draw a vertical line and estimate the value on the $T$-axis where the vertical line and the axis intersect.

Here, a good estimate could be $ \\var{ans_d_estimate}$ minutes.

{geogebra_applet('https://www.geogebra.org/m/evvd3nua', defadvice3)}

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Use the graph to estimate the initial temperature of the mixture.

The initial temperature of the mixture is [[0]] $ ^\\circ C$.

Use the graph to estimate the temperature of the mixture after $\\var{t_1}$ minutes.

The temperature after $\\var{t_1}$ minutes is [[0]] $ ^\\circ C$.

Use the graph to estimate the time taken for the mixture to reach $\\var{temp_1}^\\circ C$.

The time taken is [[0]] minutes.

Practicing skills required for sketching line graphs depicting the temperature of a mixture according to time. The question includes: a) choosing the accurate sketch from a list, and b) identifying the initial temperature of the mixture.

", "licence": "None specified"}, "statement": "

Zinc is mixed with an acid in an insulated container. After the initial temperature rise, the temperature of the mixture, $T$ $^\\circ$ C, falls over time, $t$ minutes, according to the formula:

\\[ T=\\var{b}-\\var{a}t \\]

Sketch the function $T$ for $0\\leq t \\leq \\var{t_max}$ minutes and then answer the questions below.

", "advice": "

a) To sketch the graph, one thing we need to consider is the gradient. Here the gradient is $-\\var{a}$ which is a negative number. So, the line graph will go downwords over time. We only have two options showing graphs with negative gradient. The difference between the two graphs is the last point shown on the graph. Therefore, we need to calculate the value of $T$ when $t=25$ seconds.

\\[ T=\\var{b}-\\var{a} \\times 25\\]

\\[ T=\\var{b}-\\simplify{{{a}*25}}\\]

\\[ T=\\simplify{{b}-{{a}*25}}\\]

Therefore, our sketch should look like the graph below

{geogebra_applet('https://www.geogebra.org/m/cyvdfer7', defs)}

b) To calculate the initial temperature of the mixture we need to find the value of $T$ when $t=0$ seconds.

\\[ T=\\var{b}-\\var{a}\\times 0\\]

\\[ T=\\var{b}-\\simplify{{{a}*0}}\\]

\\[ T=\\simplify{{b}-{{a}*0}}\\]

Therefore, the initial temperature of the mixture is $T=\\simplify{{b}-{{a}*0}} ^\\circ C$.

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Choose the graph that looks more like your sketch.

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What is the initial temperature of the mixture?

The initial temperature of the mixture is [[0]] $ ^\\circ$ C.

Practicing skills required for sketching line graphs depicting DNA melting temperature according to the percentage of GC content. The question includes: a) identifying the vertical intercept, b) choosing the accurate sketch from a list, and c) interpreting elements of the sketch in context.

", "licence": "None specified"}, "statement": "

The melting temperature of DNA, $T$ $^ \\circ$C, depends on the percentage of GC content of the DNA, $G~$%, according to the formula:

\\[T=\\simplify[fractionNumbers]{{a}G+{b}} \\]

Sketch the function $T$ for $0\\leq G \\leq \\var{G_max}$ and then answer the questions below.

", "advice": "

a) The vertical axis shows possible values for $T$. In order to find the intersection of the graph with the vertical axis, we need to find the value of $T$ when $G=0$.

Therefore,

\\[T=\\simplify[fractionNumbers]{{a}*0+{b}} \\]

\\[T=\\simplify[!zeroTerm]{{a*0}+{b}} \\]

\\[T=\\simplify{{a*0}+{b}} \\]

So, the intersection of the graph with the vertical axis is the point $(0,\\var{b})$.

b) We need to consider the gradient of the graph. The gradient here is $\\simplify[fractionNumbers]{{a}}$ which is a positive number. Therefore, the graph is going upwards. We also know that the intersection of the graph with the veritcal axis is $(0,\\var{b})$ from part (a).

Therefore, our graph should look like the one below:

{geogebra_applet('https://www.geogebra.org/m/a9wf5jjk', defs)}

c) The intersection point $(0,\\var{b})$ tells us that when the GC content (G) is $0$, the melting temperature ($T$) is $\\var{b}$.

So, the correct answer is \"The melting temperature of DNA with 0% GC content\".

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Find the intersection of the graph with the vertical axis.

The intersection is the point $(0,$ [[0]] $)$.

Choose the graph which looks most like your sketch.

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What does the intersection of the graph with the vertical axis represent?

", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["The melting temperature of DNA with $0$% GC content.", "The percentage of GC content when $T=0 ^\\circ C$.", "The maximum melting temperature of DNA.", "Nothing, it is just a point on the graph."], "matrix": ["1", 0, 0, 0], "distractors": ["", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Line Graphs: Interesting Graphs - DNA Melting Temperature", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Evi Papadaki", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18113/"}], "tags": [], "metadata": {"description": "

Interpreting line graphs depicting the melting temperature of DNA depending on the percentage of GC content. Estimating the melting temperature given a GC percentage and vice versa.

", "licence": "None specified"}, "statement": "

The melting temperature of DNA, $T$ $^\\circ$C, depends on the percentage of GC content of the DNA, $G~$%. The following graph depicts the relation between $T$ and $G$, for $0\\leq G \\leq \\var{G_max}$:

{geogebra_applet('https://www.geogebra.org/m/rcx6wdzu', defs)}

", "advice": "

a) We need to find the point on the $T$-axis where $G=\\var{g_1}$ . So, we first need to find the point on the $G$-axis where $G=\\var{g_1}$. From that point, we draw a line vertically to find a point on the graph. From the point on the graph, we need to draw a horizontal line, and then estimate the value on $T$.

Here, a good estimate could be $\\var{roundedA} ^\\circ C$.

{geogebra_applet('https://www.geogebra.org/m/z8gwsgjk', defsadvice1)}

b) We need to find the point on the $G$-axis where $T=\\var{t_1}$. So, we first need to estimate the point on the $T$-axis where $T=\\var{t_1}$. From here, we draw horizontally to find a point on the graph. From the point on the graph, we draw a vertical line, and then estimate the value on $G$.

Here, a good estimate could be $\\var{roundedB}$%.

{geogebra_applet('https://www.geogebra.org/m/z8gwsgjk', defsadvice2)}

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Estimate the melting temperature of DNA with $\\var{G_1}$% GC content.

The melting temperature in this case is [[0]] $ ^\\circ C$.

Estimate the percentage GC content of DNA with a melting temperature of $\\var{T_1}^\\circ C$.

The percentage GC content of DNA is [[0]] %.

The question includes a quadratic graph depicting the relationship between the frequency of an allele A at a genetic locus in a diploid population and the fitness of a population with this frequency of allele A. The aim is to estimate the maximum and minimum fitness of the population and the corresponding frequency of allele A.

", "licence": "None specified"}, "statement": "

The following graph depicts the relationship between the frequency, $x$, of an allele A at a genetic locus in a diploid population and the fitness, $w$, of a population with this frequency of allele A.

{geogebra_applet('https://www.geogebra.org/m/rtngjns4', defs)}

", "advice": "

{geogebra_applet('https://www.geogebra.org/m/xw4jjqxa',defsad1)}

Note: you can zoom in and out of the figure as needed.

a) We first need to locate the highest point on the graph. From that point, we can draw a line perpendicular to the vertical axis to estimate the maximum fitness and a line perpendicular to the horizontal axis to estimate the corresponding frequency of allele A.

Here, a good estimate could be:

maximum fitness, $w_{max}=\\var{max}$
frequency of allele A, $x=\\var{xmax}$.

b)We first need to locate the lowest point on the graph. From that point, we can draw a line perpendicular to the vertical axis to estimate the minimum fitness and a line perpendicular to the horizontal axis to estimate the corresponding frequency of allele A.

Here, a good estimate could be:

minimum fitness, $w_{min}=\\var{min}$
frequency of allele A, $x=\\var{xmin}$.

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Use the graph to find the maximum possible fitness and the value of $x$ at which this maximum is attained.

Maximum fitness, $w_{max}=$ [[0]].
Frequency of allele A, $x=$ [[1]].

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Use the graph to find the minimum possible fitness and the value of $x$ at which this maximum is attained.

Minimum fitness, $w_{min}=$ [[0]].
Frequency of allele A, $x=$ [[1]].

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Estimating the proportion of sodium carbonate in a solutionat a specific timepoint and vice versa, depicted as a quadratic graph.

", "licence": "None specified"}, "statement": "

A small volume of sodium carbonate is added to a beaker of water. The proportion of the sodium carbonate, $p$, which has dissolved by time $t$ seconds is illustrated in the graph below for $0\\leq t\\leq \\var{tmax}$.

{geogebra_applet('https://www.geogebra.org/m/uhnbdxaf', defs)}

", "advice": "

a) We first need to find the point on the time axis for $\\var{t_1}$ seconds. From that point, we draw a line vertically to find a point on the graph. From the point on the graph, we need to draw a horizontal line, and then estimate the value of $p$.

{geogebra_applet('https://www.geogebra.org/m/wnracghs', defsad)}

Here, a good estimate could be $p=\\var{ans_a_rounded} $.

b) We first need to estimate the point where the $\\var{p_1*100}$% would lay on would lie on the p-axis. That is $ \\frac{\\var{p_1*100}}{100}=\\var{p_1}$. From here, we draw a line horizontally to find a point on the graph. From the point on the graph, we draw a vertical line, and then estimate the value of $t$.

{geogebra_applet('https://www.geogebra.org/m/wnracghs', defsad1)}

Here, a good estimate could be $\\var{ans_t}$ seconds.

c) At 7 seconds, the proportion of sodium carbonate which has dessolved is $p=1$ or, if transformed to a percentage, $100$%. That means all the sodium carbonate will have dissolved completely at $t=7$ seconds and the quadratic graph cannot describe the situation beyond the 7 seconds.

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['a',a],
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Use the graph to estimate the proportion of sodium carbonate which has dissolved after $\\var{t_1}$ seconds.

$p=$ [[0]]

Give your answer as a decimal, if needed, rounded to 2 decimal places.

Use the graph to estimate the time taken for $\\simplify{{p_1}*100}$ % of the sodium carbonate to dissolve.

$t=$ [[0]] sec.

Give your answer rounded to 2 decimal places.

Why is this model not appropriate when $t\\geq \\var{tmax}$. You can choose only one of the answers.

[[0]]

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The relationship between the frequency of an allele A, $x$, at a genetic locus in a diploid population and the fitness of a population with this frequency of allele A, $w$, is described by the function $w=ax^2+x(b-x)+c(b-x)^2$ . The aims are (a) ti simplify the algebraic expression, (b) calculate the fitness of a population with a given allele A frequency, and (c) calculate the allele A frequency when the fitness of the population is given.

", "licence": "None specified"}, "statement": "

Let $x$ represent the frequency of an allele A at a genetic locus in a diploid population. The fitness, $w$, of a population with this frequency of allele A is given by:

\\[ \\simplify{w={a}x^2+x({b}-x)+{c}({b}-x)^2} \\]

", "advice": "

a) We need to open the brackets and collect like terms:

\\[ \\begin{split} w &= \\simplify{{a}x^2+x({b}-x)+{c}({b}-x)^2} \\\\ \\implies w &= \\simplify[!basic]{{a}x^2+x({b}-x)+{c}({b}-x)({b}-x)}\\\\ \\implies w &= \\simplify [unitFactor, !basic] {{a}x^2+x({b}-x)+{c}({b}^2-{b}x-{b}x+x^2)} \\\\ \\implies w &= \\simplify {{a}x^2+x({b}-x)+{c}({b}^2-{b}x-{b}x+x^2)} \\\\ \\implies w &= \\simplify[unitFactor, !collectNumbers] {{a}x^2+{b}*x-x^2+{c}*{b^2}-{c}*{2*b}x+{c}x^2} \\\\ \\implies w &=\\simplify[all, !noLeadingMinus] {{a}x^2+{b}*x-x^2+{c}*{b}^2-2*{c}*{b}x+{c}x^2} \\end{split}\\]

b) We substitute $x=\\var{x_1}$ in the simplified formula:

\\[w=\\simplify[unitfactor, !basic]{{coef_a}*{x_1}^2+{coef_b}*{x_1}+{coef_c}} \\\\ \\implies w=\\simplify [unitfactor, !basic]{{coef_a}*{x_1^2}+{coef_b}*{x_1}+{coef_c}} \\\\ \\implies w=\\var{w_ans_rounded} \\text{ (2 d.p.)}\\]

Note: we get the same result regardless of which version of the formula we use. Using the simplified formula however is quicker.

c) We need to substitute $w=\\var{w_1}$ and solve the quadratic equation:

\\[ \\simplify[!noLeadingMinus]{{w_1}={coef_a}*x^2+{coef_b}*x+{coef_c}} \\\\ \\implies \\simplify[!noLeadingMinus]{{w_1}-{coef_a}*x^2-{coef_b}*x-{coef_c}=0} \\\\ \\implies \\simplify{{w_1}-{coef_a}*x^2-{coef_b}*x-{coef_c}=0}\\]

Using the quadratic formula

\\[ x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\]

We get the solutions:

\\[ \\begin{split} x_1 &\\,= \\frac{\\simplify{-{co_b}+sqrt({co_b}^2-4*{co_a}*{co_c})}}{\\simplify{{2*co_a}}} \\qquad \\text{and} \\qquad x_2 &\\,= \\frac{\\simplify{-{co_b}-sqrt({co_b}^2-4*{co_a}*{co_c})}}{\\simplify{{2*co_a}}} \\\\\\\\ &\\,= \\var{ansx_1_rounded},\\quad &\\,=\\var{ansx_2_rounded}. \\end{split}\\]

Taking into consideration the context of the problem, we need to reject one of the two answers if it is negative. Thus, the final answer is $x=\\var{ans_c}$.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "problem posing variables", "definition": "random (0 .. 0.5 # 0.1 except 0)", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "problem posing variables", "definition": "random([1,2])", "description": "", "templateType": "anything", "can_override": false}, "x_1": {"name": "x_1", "group": "problem posing variables", "definition": "random (0..1 #0.05 except [0,1]) ", "description": "", "templateType": "anything", "can_override": false}, "w_1": {"name": "w_1", "group": "problem posing variables", "definition": "random (0..1 #0.1 except [0,1]) ", "description": "", "templateType": "anything", "can_override": false}, "coef_a": {"name": "coef_a", "group": "answer variables", "definition": "{a}+{c}-1", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "problem posing variables", "definition": "random (0 .. 0.5 # 0.1 except 0)", "description": "", "templateType": "anything", "can_override": false}, "coef_b": {"name": "coef_b", "group": "answer variables", "definition": "{b}-2*{c}*{b}", "description": "", "templateType": "anything", "can_override": false}, "coef_c": {"name": "coef_c", "group": "answer variables", "definition": "{c}*{b}^2", "description": "", "templateType": "anything", "can_override": false}, "w_ans": {"name": "w_ans", "group": "answer variables", "definition": "coef_a*x_1^2+coef_b*x_1+coef_c", "description": "", "templateType": "anything", "can_override": false}, "disc": {"name": "disc", "group": "answer variables", "definition": "(-coef_b)^2 - 4*coef_a*(coef_c-w_1)", "description": "", "templateType": "anything", "can_override": false}, "ansx_1": {"name": "ansx_1", "group": "Final part coefficients", "definition": "(-co_b+sqrt(co_b^2-4*co_a*co_c))/(2*co_a)", "description": "", "templateType": "anything", "can_override": false}, "ansx_2": {"name": "ansx_2", "group": "Final part coefficients", "definition": "(-co_b-sqrt(co_b^2-4*co_a*co_c))/(2*co_a)", "description": "", "templateType": "anything", "can_override": false}, "w_ans_rounded": {"name": "w_ans_rounded", "group": "answer variables", "definition": "precround(w_ans,2)", "description": "", "templateType": "anything", "can_override": false}, "ansx_1_rounded": {"name": "ansx_1_rounded", "group": "Final part coefficients", "definition": "precround(ansx_1,2)", "description": "", "templateType": "anything", "can_override": false}, "ansx_2_rounded": {"name": "ansx_2_rounded", "group": "Final part coefficients", "definition": "precround(ansx_2,2)", "description": "", "templateType": "anything", "can_override": false}, "ans_c": {"name": "ans_c", "group": "Final part coefficients", "definition": "if(ansx_1_rounded>0, ansx_1_rounded,ansx_2_rounded)", "description": "", "templateType": "anything", "can_override": false}, "co_a": {"name": "co_a", "group": "Final part coefficients", "definition": "-coef_a", "description": "", "templateType": "anything", "can_override": false}, "co_b": {"name": "co_b", "group": "Final part coefficients", "definition": "-coef_b", "description": "", "templateType": "anything", "can_override": false}, "co_c": {"name": "co_c", "group": "Final part coefficients", "definition": "w_1-coef_c", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "(coef_a<0)and(0<=w_ans<=1)and(disc>0)and(coef_c-w_1>0)", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [{"name": "problem posing variables", "variables": ["a", "b", "x_1", "w_1", "c"]}, {"name": "answer variables", "variables": ["coef_a", "coef_b", "coef_c", "w_ans", "w_ans_rounded", "disc"]}, {"name": "Final part coefficients", "variables": ["co_a", "co_b", "co_c", "ansx_1", "ansx_2", "ansx_1_rounded", "ansx_2_rounded", "ans_c"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Write $w$ in the form $ax^2+bx+c$.

$w=$[[0]]

Calculate the fitness of a population with allele frequency $x=\\var{x_1}$.

$w=$[[0]]

Give your answer rounded to 2 decimal places.

Find the allele frequency, $x$, of a population with fitness $\\var{w_1}$.

$x=$[[0]]

Give your answer rounded to 2 decimal places, if needed.

Calculating the gradient of a quadratic equation at a specific point and finding the stationary point (maximum) in a contextualised problem.

", "licence": "None specified"}, "statement": "

The discharge of water into a stream, $D$ dm$^3/$s, affects the growth of the fish in the stream.

The average length of the fish, $x$ cm, is modelled by the curve:

$ x=\\var{a}D^2+\\var{b}D+\\var{c} $

", "advice": "

a) We first need to find differentiate the formula of the curve in respect of D.

\\[ \\frac{dx}{dD} = 2 \\times (\\var{a}) D+ \\var{b}+0 \\\\ \\implies \\frac{dx}{dD}= \\simplify{2*{a}}D+\\var{b} \\]

Now to find the gradient at the point where $D=\\var{d_1}$ we need to substitute the value in the equation we found in the previous step.

\\[ \\frac{dx}{dD}=\\simplify{2*{a}}\\times \\var{d_1} +\\var{b} \\\\ \\implies \\frac{dx}{dD}= \\simplify{2*{a}*{d_1}}+\\var{b} \\\\ \\implies \\frac{dx}{dD}= \\simplify{2*{a}*{d_1}+{b}} \\]

b) The discharge that results in the maximum average length of the fish can be calculated by solving the equation:

\\[ \\frac{dx}{dD}=0\\]

Therefore:

\\[ \\simplify{2*{a}}D+\\var{b}=0 \\\\ \\implies \\simplify{2*{a}}D= - \\var{b} \\\\ \\implies D = \\frac{-\\var{b}}{\\simplify{2*{a}}} \\\\ \\implies D= \\simplify{{-{b}}/{2*{a}}} \\]

Or, if rounded,

\\[ D=\\var{ans_b_rounded}\\]

c) Now that we know that $D=\\var{ans_b_rounded}$ is the discharge that results in the maximum average length we can substitute this value in the original equation of the curve and calculate the maximum length.

So,

\\[ x_{max}= \\var{a} \\times \\var{ans_b_rounded}^2 + \\var{b} \\times \\var{ans_b_rounded} + \\var{c} \\\\ \\implies x_{max}=\\var{a} \\times \\var{ans_b_rounded^2}+ \\var{b} \\times \\var{ans_b_rounded}+\\var{c} \\\\ \\implies x_{max}=\\simplify{{a*ans_b_rounded^2}}+ \\simplify{{b*ans_b_rounded}}+\\var{c} \\\\ \\implies x_{max}=\\simplify{{a*ans_b_rounded^2+b*ans_b_rounded+c}}\\]

So, $x_{max}=\\var{ans_c_rounded}$.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "Problem posing", "definition": "-random ( 50 .. 200)", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Problem posing", "definition": "random(50..200)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Problem posing", "definition": "Random(10..80)", "description": "", "templateType": "anything", "can_override": false}, "d_1": {"name": "d_1", "group": "Problem posing", "definition": "random (0.2 .. 0.75 #0.05 )", "description": "", "templateType": "anything", "can_override": false}, "ans_a": {"name": "ans_a", "group": "Answers", "definition": "2*{a}*{d_1}+{b}", "description": "", "templateType": "anything", "can_override": false}, "ans_b": {"name": "ans_b", "group": "Answers", "definition": "-b/(2a)", "description": "", "templateType": "anything", "can_override": false}, "ans_c": {"name": "ans_c", "group": "Answers", "definition": "a*ans_b_rounded^2+b*ans_b_rounded+c", "description": "", "templateType": "anything", "can_override": true}, "ans_b_rounded": {"name": "ans_b_rounded", "group": "Answers", "definition": "precround(ans_b,2)", "description": "", "templateType": "anything", "can_override": false}, "ans_c_rounded": {"name": "ans_c_rounded", "group": "Answers", "definition": "precround({ans_c},2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "(ans_c_rounded<100) and (ans_a>0)", "maxRuns": "100"}, "ungrouped_variables": [], "variable_groups": [{"name": "Problem posing", "variables": ["a", "b", "c", "d_1"]}, {"name": "Answers", "variables": ["ans_a", "ans_b", "ans_c", "ans_b_rounded", "ans_c_rounded"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the gradient of this curve at the point where $D=\\var{d_1}$.

", "answer": "{ans_a}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the discharge which results in the largest average length of the fish.

Give your answer rounded to 2 decimal places.

", "answer": "{ans_b_rounded}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the maximum average length of the fish.

Give your answer rounded to 2 decimal places.

", "answer": "{ans_c_rounded}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Quadratic functions: Working with the Formula - Dilution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Evi Papadaki", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18113/"}], "tags": [], "metadata": {"description": "

The proportion of the sodium carbonate, $p$, which has dissolved by time $t$ seconds is given by the formula $ p=\\frac{bt-at^2}{c}$. The aim is to calculate the proportion of sodium carbonate in a solution at a given time and vice versa.

", "licence": "None specified"}, "statement": "

A small volume of sodium carbonate is added to a beaker of water. The proportion of the sodium carbonate, $p$, which has dissolved by time $t$ seconds is given by:

\\[p=\\frac{\\var{b}t-\\var{a}t^2}{\\var{c}}\\]

", "advice": "

a) We need to substitute $t=\\var{t_1}$
\\[ p=\\frac{\\var{b}\\times\\var{t_1}-\\var{a}\\times \\var{t_1}^2}{\\var{c}}\\\\ \\implies p=\\frac{\\var{b}\\times\\var{t_1}-\\var{a}\\times \\var{t_1^2}}{\\var{c}} \\\\ \\implies p=\\frac{\\simplify{{b}*{t_1}}-\\simplify{{a}*{t_1^2}}}{\\var{c}} \\\\ \\implies p=\\frac{\\simplify{{b}*{t_1}-{a}*{t_1^2}}}{\\var{c}} \\\\ \\implies p=\\simplify{{b*t_1-a*t_1^2}/{c}} ~~~~ \\text{or} ~~~~ \\var{ans_a_rounded}~\\text{(2d.p.)} \\]

b) We first need to transform the percentage value which we were given into a proportion written as a decimal.

\\[ \\var{p_1*100}\\%=\\var{p_1}\\]

Then, substituting $p=\\var{p_1}$ we get the following equation to solve:

\\[ \\var{p_1}=\\frac{\\var{b}t-\\var{a}t^2}{\\var{c}} \\\\ \\implies \\var{c}\\times
\\var{p_1}=\\var{b}t-\\var{a}t^2 \\\\ \\implies \\simplify[unitFactor]{{a}t^2-{b}t+{c*p_1}}=0\\]

Using the quadratic formula

\\[ t=\\frac{-b\\pm \\sqrt{b^2-4a c}}{2a}\\]

we get the solutions:

\\[ t_1=\\var{anst_1_rounded} \\\\ t_2=\\var{anst_2_rounded}\\]

However, we are solving the equation to find a time which is $0\\leq p\\leq \\var{tmax}$. Therefore, the final answer is only $t=\\var{ans_t}$.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "Problem posing variables", "definition": "random (1 ..5 )", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Problem posing variables", "definition": "Random(50..90)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Problem posing variables", "definition": "random(100..400 #10)", "description": "", "templateType": "anything", "can_override": false}, "p_1": {"name": "p_1", "group": "Problem posing variables", "definition": "random(0 .. 1 # 0.1 except [ 0 , 1 ] )", "description": "", "templateType": "anything", "can_override": false}, "t_1": {"name": "t_1", "group": "Problem posing variables", "definition": "random(2 .. 10 )", "description": "", "templateType": "anything", "can_override": false}, "ans_a": {"name": "ans_a", "group": "answering variables", "definition": "({b}*{t_1}-{a}*{t_1}^2)/{c}", "description": "", "templateType": "anything", "can_override": false}, "disc": {"name": "disc", "group": "answering variables", "definition": "(-b)^2-(4*a*(p_1*c))", "description": "", "templateType": "anything", "can_override": false}, "anst_1": {"name": "anst_1", "group": "answering variables", "definition": "(b+sqrt(disc))/(2*a)", "description": "", "templateType": "anything", "can_override": false}, "anst_2": {"name": "anst_2", "group": "answering variables", "definition": "(b-sqrt(disc))/(2*a)", "description": "", "templateType": "anything", "can_override": false}, "anst_1_rounded": {"name": "anst_1_rounded", "group": "answering variables", "definition": "precround(anst_1,2)", "description": "", "templateType": "anything", "can_override": false}, "anst_2_rounded": {"name": "anst_2_rounded", "group": "answering variables", "definition": "precround(anst_2,2)", "description": "", "templateType": "anything", "can_override": false}, "ans_a_rounded": {"name": "ans_a_rounded", "group": "answering variables", "definition": "precround(ans_a,2)", "description": "", "templateType": "anything", "can_override": false}, "Disc_max_t": {"name": "Disc_max_t", "group": "answering variables", "definition": "b^2-(4*a*c)", "description": "", "templateType": "anything", "can_override": false}, "tmax1": {"name": "tmax1", "group": "answering variables", "definition": "(b+sqrt(disc_max_t))/(2*a)", "description": "", "templateType": "anything", "can_override": false}, "tmax2": {"name": "tmax2", "group": "answering variables", "definition": "(b-sqrt(disc_max_t))/(2*a)", "description": "", "templateType": "anything", "can_override": false}, "tmax": {"name": "tmax", "group": "answering variables", "definition": "round (min(tmax1,tmax2))", "description": "", "templateType": "anything", "can_override": false}, "ans_t": {"name": "ans_t", "group": "answering variables", "definition": "if(anst_10)and(disc_max_t>0)and(ans_a<=1)", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [{"name": "Problem posing variables", "variables": ["a", "b", "c", "p_1", "t_1"]}, {"name": "answering variables", "variables": ["disc", "ans_a", "anst_1", "anst_2", "anst_1_rounded", "anst_2_rounded", "ans_a_rounded", "Disc_max_t", "tmax1", "tmax", "tmax2", "ans_t"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the proportion of the sodium carbonate which has dissolved after $\\var{t_1}$ seconds.

$p=$ [[0]]

Give your answer rounded to 2 decimal places.

Calculate the time taken for $\\simplify{{p_1}*100}$ % of sodium carbonate to dissolve.

$t=$ [[0]] sec.

Give your answer rounded to 2 decimal places.

Finding the stationary point (maximum) of a quadratic equation in a contextualised problem.

", "licence": "None specified"}, "statement": "

The temperature, $T$°C of the air at height $z$ metres inside a blast furnace is modelled by the equation:

\\[ T=\\var{a}z^2+\\var{b}z+\\var{c} \\]

", "advice": "

a) We first need to differentiate the formula of the curve in respect of $z$.

\\[ \\frac{dT}{dz} = 2 \\times (\\var{a}) z+ \\var{b}+0 \\\\ \\implies \\frac{dT}{dz}= \\simplify{2*{a}}z+\\var{b} \\]

Now to find the gradient at the point where $z=\\var{z_1}$ we need to substitute the value in the equation we found in the previous step.

\\[ \\frac{dT}{dz}=\\simplify{2*{a}}\\times \\var{z_1} +\\var{b} \\\\ \\implies \\frac{dT}{dz}= \\simplify{2*{a}*{z_1}}+\\var{b} \\\\ \\implies \\frac{dT}{dz}= \\simplify{2*{a}*{z_1}+{b}} \\]

b) The height at which the temperature reaches the maximum can be calculated by solving the equation:

\\[ \\frac{dT}{dz}=0\\]

Therefore:

\\[ \\simplify{2*{a}}z+\\var{b}=0 \\\\ \\implies \\simplify{2*{a}}z= - \\var{b} \\\\ \\implies z = \\frac{-\\var{b}}{\\simplify{2*{a}}} \\\\ \\implies z= \\simplify{{-{b}}/{2*{a}}} \\\\ z=\\var{ans_b_rounded} \\text{m (2.d.p.)} \\]

c) Now that we know that $z=\\var{ans_b_rounded}$m is the height that results in the maximum temperature, we can substitute this value in the original equation of the curve and calculate the maximum temperature.

\\[ T_{max}= \\var{a} \\times \\var{ans_b_rounded}^2 + \\var{b} \\times \\var{ans_b_rounded} + \\var{c} \\\\ \\implies T_{max}=\\var{a} \\times \\var{ans_b_rounded^2}+ \\var{b} \\times \\var{ans_b_rounded}+\\var{c} \\\\ \\implies T_{max}=\\simplify{{a*ans_b_rounded^2}}+ \\simplify{{b*ans_b_rounded}}+\\var{c} \\\\ \\implies T_{max}=\\simplify{{a*ans_b_rounded^2+b*ans_b_rounded+c}}\\]

So, if rounded, $T_{max}=\\var{ans_c_rounded}$.

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Find the rate of decrease of the temperature with height at the point where $z=\\var{z_1}$.

When $z=\\var{z_1}$ then $\\frac{dT}{dz}=$ [[0]].

Find the height at which the temperature is highest.

The temperature is highest when $z=$ [[0]] m.

Give your answer rounded to 2 decimal places, if needed.

Find the maximum temperature in the blast furnace.

The maximum temperarure is $T_{max}=$[[0]]$^\\circ C$.

Give your answer rounded to 2 decimal places, if needed.

Calculating area under curves of the form $ax^2+bx$ and $ax^4+bx^3+cx^2+dx+e$ in a contextualised problem.

", "licence": "None specified"}, "statement": "", "advice": "

A cross-section is a cutting made across something, here, the fins of a fish. The curve y models the outline of the cross-section. The area of the cross-section can be found by calculating the definite integral between the two boundaries. Therefore,

a) The area of the dorsal fin is:

\\[ \\begin{split} \\simplify[!noLeadingMinus]{defint({-3*a}*x^2+{2*b}*x, x ,0 ,{x_upper})} &\\,= \\left[ \\simplify[!all]{{-3*a}/3 x^3+{2*b}/2 x^2}\\right]_0^\\var{x_upper} \\\\ &\\,= \\left[ \\simplify[all, !noLeadingMinus]{{-a}x^3+{b}x^2}\\right]_0^\\var{x_upper} \\\\ &\\,= \\left[ \\simplify[ !noLeadingMinus]{{-a}*{x_upper}^3+{b}*{x_upper}^2}\\right]-\\left[ \\simplify[ !noLeadingMinus]{{-a}*0^3+{b}*0^2}\\right] \\\\ &\\,= \\left[ \\simplify{{-a*{x_upper}^3}+{b*x_upper^2}}\\right]-\\left[ \\simplify{{-a}*0+{b}*0}\\right] \\\\ &\\,= \\var{ans_a} cm^2\\end{split}\\]

b)Similarly, the area of the pelvic fin is:

\\[ \\begin{split} \\simplify[!noLeadingMinus]{defint(-x^4+{4*c}*x^3-{3*d}*x^2+{2*k}*x-{f}, x ,1 ,3.5)} &\\,= \\left[ \\simplify[!all]{-{1/5}* x^5+{4*c}/4*x^4-{3*d}/3*x^3+{2*k}/2*x^2-{f}*x }\\right]_1^{3.5} \\\\ &\\,= \\left[ \\simplify[unitFactor]{-{1/5}* x^5+{4*c/4}*x^4-{3*d/3}*x^3+{2*k/2}*x^2-{f}*x }\\right]_1^{3.5} \\\\ &\\,= \\left[ \\simplify[unitFactor]{-{1/5}* 3.5^5+{4*c/4}*3.5^4-{3*d/3}*3.5^3+{2*k/2}*3.5^2-{f}*3.5 }\\right] - \\left[ \\simplify[unitFactor]{-{1/5}* 1^5+{4*c/4}*1^4-{3*d/3}*1^3+{2*k/2}*1^2-{f}*1 }\\right] \\\\ &\\,= \\left[ \\simplify[unitFactor, fractionNumbers]{-{1*3.5^5/5}+{4*c*3.5^4/4}-{3*d*3.5^3/3}+{2*k*3.5^2/2}-{f*3.5} }\\right] - \\left[ \\simplify[unitFactor, fractionNumbers]{-{1* 1^5/5}+{4*c*1^4/4}-{3*d*1^3/3}+{2*k*1^2/2}-{f*1} }\\right] \\\\ &\\,= \\left[ \\simplify{-{1*3.5^5/5}+{4*c*3.5^4/4}-{3*d*3.5^3/3}+{2*k*3.5^2/2}-{f*3.5} }\\right] - \\left[ \\simplify{-{1* 1^5/5}+{4*c*1^4/4}-{3*d*1^3/3}+{2*k*1^2/2}-{f*1} }\\right] \\\\ &\\,= \\simplify{ {-{1*3.5^5/5}+{4*c*3.5^4/4}-{3*d*3.5^3/3}+{2*k*3.5^2/2}-{f*3.5} } - {-{1* 1^5/5}+{4*c*1^4/4}-{3*d*1^3/3}+{2*k*1^2/2}-{f*1} }} \\\\ &\\,= \\var{ansb} cm^2 \\text{(2 d.p.)} \\end{split}\\]

The cross-section of the dorsal fin of a fish can be modelled by the curve $\\simplify[all,!noLeadingMinus]{y=-3*{a}x^2+2*{b}x}$ for $0\\le\\ x\\le \\var{x_upper}$, where all units are measured in centimetres.

Calculate the area of the dorsal fin of this fish.

Give your answer rounded to 2 decimal places, if needed.

[[0]] $\\mathrm{cm}^2$

The cross-section of the pelvic fin of this fish is modelled by the curve \\[ \\simplify[all,!noLeadingMinus]{ y=-x^4+4*{c}x^3-3*{d}x^2+2*{k}x-{f}}\\] for $1\\le\\ x\\le3.5$.

Calculate the area of the pelvic fin of the fish.

Give your answer rounded to 2 decimal places, if needed.

[[0]] $\\mathrm{cm}^2$

Integrating a polynomial functions which describe the rate of change of a population over time to find and use an equation that describes the total population according to time.

", "licence": "None specified"}, "statement": "

The growth rate, $r$ individuals per year, of a population in year $t$ after the start of observation is given by
$\\simplify {r=2*{a}t+{b}}$

", "advice": "

a) The function for the growth rate, $r$, is the derivative of the function that describes the total population, $n$, in respect of time, $t$. So, to find an expression of the function $n$ we need to integrate $r$ in respect of $t$.

Remember, indefinite integration (integration without limits) is the reverse of differentiation.

Therefore,

\\[ \\begin{split} \\simplify{n=int({2*a}*t+{b},t)} &\\,= \\simplify{ {2*a}/2*t^2+{b}*t+c} \\\\ &\\,= \\simplify{{a}*t^2+{b}*t+c} \\end{split} \\]

To find the exact expression that describes the total population we need to find the value of $c$. To do so, we use the fact that when $t=0$ the total population is $\\var{c}$.

\\[ \\begin{split} \\simplify[!all]{{a}*0^2+{b}*0+c} &= \\var{c} \\\\ c&= \\var{c} \\end{split} \\]

Thus, the expression that describes the total population in year $t$ is:

\\[ n= \\simplify{{a}*t^2+{b}*t+{c}} \\]

b) The function $n= \\simplify{{a}*t^2+{b}*t+{c}}$ gives us the total population at year t. Thus, to find the population at year $\\var{t_1}$, I need to substitute the value $t=\\var{t_1}$ in the formula.

\\[ \\begin{split} n&= \\simplify[!all]{{a}*{t_1}^2+{b}*{t_1}+{c}} \\\\ &= \\simplify[!all]{{a}*{t_1^2}+{b}*{t_1}+{c}} \\\\ &= \\simplify[!all]{{a*t_1^2}+{b*t_1}+{c}}\\\\ &=\\var{ansB} \\end{split} \\]

Therefore, when $t=\\var{t_1}$, the population is $n=\\var{ansB}$.

If there are $\\var{c}$ individuals in the population in year $t=0$, find an expression for the total population, $n$ individuals, in year $t$.

$n=$[[0]]

Give your answer in the form $n=at^2+bt+c$.

Use your expression to find the total population at year $t=\\var{t_1}$.

When $t=\\var{t_1}$, $n=$[[0]]

Knowing the doubling time of a population and the population on day $t$, calculate the initial population and the number of days required for the population to reach a threshold.

", "licence": "None specified"}, "statement": "

The doubling time of a population of flies is $\\var{d}$ days.

$\\var{days[j]}$ days after the start of an experiment, there were $\\var{d1}$ flies in the population.

", "advice": "

a) We know that $\\var{ndays[j]}$ days after the start of the experience there are $\\var{d1}$ flies. We also know that the doubling time is ${\\var{d}}$ days.

In those $\\var{ndays[j]}$ days, the populations would have doubled $\\frac{\\var{ndays[j]}}{\\var{d}}=\\simplify{{ndays[j]}/{d}}$ times. Which means that the initial population, let's call it $n_0$, would have been multiplied by $2^{\\var{int}}$ to result in $\\var{d1}$ flies. So,

\\[ \\begin{split} \\simplify[!all]{n_0*2^{ndays[j]/d}}&=\\var{d1} \\\\ \\simplify{n_0*2^{ndays[j]/d}}&=\\var{d1} \\end{split} \\]

If we solve the equation we created, we can find the initial population of flies. Therefore,

\\[ \\begin{split} \\simplify{n_0*2^{ndays[j]/d}}&= \\var{d1} \\\\ n_0 &=\\frac{\\var{d1}}{\\var{2^int}} \\end{split} \\]

Which means that there are $n_0=\\frac{\\var{d1}}{\\var{2^int}}=\\var{n0}~ $ flies at the start of the experiment.

b) To calculate the number of days for the population to reach $\\var{n1}$ flies, we need an expression that connects the number of days and the size of the population.

Let's say it takes $t$ days for the fly population to reach $\\var{n1}$. Since the populatioin has doubling time of $\\var{d}$ days, this means that in $t$ days the population would have doubled $ \\frac{t}{\\var{d}}$ times. In other words, the initial population will have been multiplied by $2^{\\frac{t}{\\var{d}}}$.

So, we can now substitute this expression into our growth equation:

\\[ \\begin{split} \\var{n0}\\times 2^{\\frac{t}{\\var{d}}} &= \\var{n1} \\end{split} \\]

Now, if we solve the equaction for t we will find the number of days.

\\[ \\begin{split} \\var{n0}\\times 2^{\\frac{t}{\\var{d}}} &= \\var{n1} \\\\ 2^{\\frac{t}{\\var{d}}} &= \\frac{ \\var{n1}}{\\var{n0} } \\\\ 2^{\\frac{t}{\\var{d}}} &= \\var{n1divn0} \\end{split}\\]

Taking $log_2()$ of both sides:

\\[ \\begin{split} \\\\ {\\frac{t}{\\var{d}}} &= \\log_2(\\var{n1divn0}) \\\\ t &= \\var{d}\\times\\log_2(\\var{n1divn0}) \\\\ t &= \\var{adviceB} \\end{split}\\].

Therefore, it takes approximately $\\var{ansb}$ days for the population to reach $\\var{n1}$ flies.

Note: Alternatively, you might already know the formula the formula

\\[n=n_0 \\times 2^\\frac{t}{T} \\]

Where, $n$ represents the size of the population (e.g., number of flies) at time $t$, $n_0$ the initial population and $T$ is the length of each time step.

If so, you can use it to find the same results.

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Calculate the number of flies in the population at the start of the experiment.

There were [[0]]flies.

How many days after the start of the experiment did the population reach $\\var{n1}$ flies?

The population reached $\\var{n1}$ flies in [[0]] days.

If needed round your answer to the nearest integer.

Knowing the half-life of Carbon-14 and the initial mass of Carbon-14 when a tree was cut (a) write an expression that describes the relationship between the remaining mass and time, (b) calculate the remaining mass after $t$ years, and (c) given the remaining mass calculate how many years ago the tree was cut down.

", "licence": "None specified"}, "statement": "

Carbon-14 has a half-life of $5730$ years.
When a tree was cut down, it contained $\\var{m}$ $\\mathrm{gr}$ of carbon-14.

", "advice": "

a) We need to find an expression that connects the number of years and the mass of Carbon-14.

We know that Carbon-14 has a half-life of $5730$ years.We also know that the initial mass of Carbon-14 when the tree was cut down was $\\var{m}$ grams.

Let's say it takes $t$ years for the mass Carbon-14 to reach $m$ grams. Since the mass of Carbon-14 halves every $5730$ years, this means that in $t$ years the mass would have halved $\\frac{t}{5730}$ times.

So, our expression would be

\\[ \\begin{split} m&=\\var{m}\\times \\left( \\frac{1}{2} \\right) ^\\frac{t}{5730} \\qquad \\text{or} \\qquad m&=\\var{m}\\times0.5^\\frac{t}{5730}\\end{split}\\]

b) To find how much Carbon-14 the tree contained after $\\var{t1}$ years, we need to use the expression found in part (a) (either of the suggested forms would work) and set $t=\\var{t1}$:

\\[ \\begin{split} m&=\\var{m}\\times 0.5 ^\\frac{\\var{t1}}{5730} \\\\ &=\\var{m}\\times \\simplify{0.5 ^{t1/5730}}\\\\ &=\\var{m}\\times \\simplify{{0.5 ^adv1}} \\\\ &=\\simplify{{m}*{0.5 ^adv1}} \\\\&=\\var{ansb} ~~\\text{[2 d.p.]} \\end{split}\\]

Therefore, the tree contained $\\var{ansb}$ grams of Carbon-14 after $\\var{t1}$ years.

c) To find how many years ago the tree was cut if it now contains $\\var{m1}$ grams of Carbon-14, we need to use the expression found in part (a) (either of the suggested forms would work), set $m=\\var{m1}$ and then rearrange to calculate:

\\[ \\begin{split} \\var{m1}&=\\var{m}\\times 0.5 ^\\frac{t}{5730} \\\\ \\frac{\\var{m1}}{\\var{m}}&=0.5 ^\\frac{t}{5730} \\\\ \\simplify{{m1/m}}&=0.5 ^\\frac{t}{5730} \\end{split}\\]

Taking $\\log_{0.5}( )$ of both sides:

\\[ \\begin{split} \\log_{0.5}\\left(\\simplify{{m1/m}}\\right)&=\\frac{t}{5730} \\\\ 5730 \\times\\log_{0.5}\\left(\\simplify{{m1/m}}\\right)&=t \\\\ \\var{adv2}&=t \\end{split}\\]

Therefore, the tree was cut approximately $\\var{ansc}$ years ago.

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Write down an expression for the mass of carbon-14, $m$ $\\mathrm{gr}$, in the tree $t$ years after it was cut down.

$m=$[[0]] $\\mathrm{gr}$

How much carbon-14 did the tree contain after $\\var{t1}$ years?

[[0]]

Give your answer in grams to 2 decimal places if needed.

The tree now contains $\\var{m1}$ grams of carbon-14. How many years ago was it cut down?

[[0]]

Give your answer to the nearest year, if needed.

Using basic derivatives to calculate the gradient function of a hill $y=-e^{x}+b\\ln{\\left(x\\right)+c$, and then substituting values to find the gradient at specific distance from the sea.

", "licence": "None specified"}, "statement": "

The cross-section of a hill with a steep cliff face is modelled by the equation:
\\[y=-e^{x}+\\var{b}\\ln{\\left(x\\right)+\\var{c}}\\]
where $y$ is the height of the ground above sea level, measured in decametres ($\\mathrm{dam}$), and $x$ is the horizontal distance from the sea, also measured in decametres.

The model applies for values of $x$ in the range $0.01\\le\\ x\\le3$.

", "advice": "

a) To find the gradient of the hill at any given horizontal distance from the sea, $x$, we need to differentiate the equation

\\[y=-e^{x}+\\var{b}\\ln{\\left(x\\right)+\\var{c}}\\]

with respect to x:

\\[ \\begin{split} \\frac{dy}{dx}&=-e^x+\\var{b}\\times\\frac{1}{x}+0 \\\\&=-e^x+\\frac{\\var{b}}{x} \\end{split}\\]

\\[ \\begin{split} \\frac{dy}{dx}&=-e^x+\\frac{1}{x}+0 \\\\&=-e^x+\\frac{\\var{b}}{x} \\end{split}\\]

Therefore, the formula that describes the gradient of the hill at any horizontal distance from the sea is:

\\[ \\begin{split} \\frac{dy}{dx}&=-e^x+\\frac{\\var{b}}{x} \\end{split}\\]

b) We can use the formula from part (a) to calculate the gradient at specific points.

Remember that the distance in this problem is measured in decametres ($\\mathrm{dam}$) and $1 ~~\\mathrm{dam}$ = $10~~ \\mathrm{m}$.

So, when $x=1$:

\\[ \\begin{split} \\frac{dy}{dx}&=-e^1+\\frac{\\var{b}}{1} \\\\&=-e+\\var{b} \\\\ &= \\var{ansb1} ~ \\text{[2 d.p.]}\\end{split}\\]

Thus, the gradient of the hill at a horizontal distance of $1~~\\mathrm{dam}$ from the sea is $\\frac{dy}{dx}=\\var{ansb1} ~~ \\mathrm{dam}$.

When $x=0.1$:

\\[ \\begin{split} \\frac{dy}{dx}&=-e^{0.1}+\\frac{\\var{b}}{0.1} \\\\ &= \\var{ansb2} ~ \\text{[2 d.p.]}\\end{split}\\]

Thus, the gradient of the hill at a horizontal distance of $0.1~~\\mathrm{dam}$ from the sea is $\\frac{dy}{dx}=\\var{ansb2} ~~ \\mathrm{dam}$.

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Find the formula that describes the gradient of the hill at any given horizontal distance from the sea.

$\\frac{dy}{dx}=$ [[0]].

What is the gradient of the hill at a horizontal distance of:

$10$ metres ($1 ~~\\mathrm{dam}$) from the sea?

[[0]] $\\mathrm{dam}$

$1$ metre ($0.1~~ \\mathrm{dam}$) from the sea?

[[1]] $\\mathrm{dam}$

Give your answers to 2 decimal places when necessary.

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Calculating the rate of change of the temperature during a chemical reaction using the chain rule in a function of the form $T=ate^{-t}$, and finding the maximum temperature of the reaction.

", "licence": "None specified"}, "statement": "

Iron is mixed with an acid in an insulated container. The temperature, $T$°C, above room temperature, at time $t$ minutes after the metal is added to the acid is given by:
\\[T=\\var{a}te^{-t}\\]

", "advice": "

a) To find the rate of change at a specific time, we need to first differentiate the function

\\[ T=\\var{a}te^{-t} \\]

with respect to $t$.

To differentiate the function, we need to notice that $T$ is of the form $T=u(t) \\times v(t)$. In other words $T$ is the product of the functions $u=\\var{a}t$ and $v=e^{-t}$.

Therefore, we can use the product rule:

\\[ \\begin{split} \\frac{dT}{dt}&= \\frac{du}{dt}\\times v+u\\times \\frac{dv}{dt} \\end{split}\\]

First, we need to calculate the derivatives $\\frac{du}{dt}$ and $\\frac{dv}{dt}$:

\\[ \\begin{split} u(t)&=\\var{a}t\\qquad \\text{and} \\qquad v(t)&=e^{-t} \\\\ \\frac{du}{dt}&=\\var{a}~~~~~\\qquad \\qquad \\frac{dv}{dt}&=-e^{-t}\\end{split} \\]

Substituting these results into the product rule formula, we can obtain the derivative of $T$:

\\[ \\begin{split} \\frac{dT}{dt}&= \\frac{du}{dt}v+u\\frac{dv}{dt} \\\\ &=\\var{a}e^{-t}+\\var{a}t \\left(-e^{-t} \\right) \\\\ &=\\var{a}e^{-t}-\\var{a}te^{-t}\\end{split}\\]

Factorising,

\\[ \\begin{split} \\frac{dT}{dt}&= \\var{a}e^{-t}(1-t)\\end{split}\\]

Now, we can substitute $t=\\var{t1}$ and calculate the rate of change of the temperature $\\var{t1}$ minutes after the metal is added:

\\[ \\begin{split} \\frac{dT}{dt}&= \\var{a}e^{-\\var{t1}}(1-\\var{t1}) \\\\&= \\simplify{{a}e^{-{t1}}(1-{t1})} \\\\ &=\\var{ansa} ~~\\text{[2.dp]}\\end{split}\\]

Therefore, the rate of change of the temperature when $t=\\var{t1}$ is $\\frac{dT}{dt}=\\var{ansa} ^\\circ C/min$.

b) When the temperature reaches maximum, the rate of change of the temperature will be $0$. So, to find the time when the temperature reaches its maximum we must solve $\\frac{dT}{dt}=0$. Using our answer from a):

\\[ \\begin{split} \\var{a}e^{-t}(1-t)&=0 \\end{split}\\]

Since, $\\var{a}e^{-t} > 0$ for every value of $t$, this equation can only be equal to zero when

\\[ \\begin{split} 1-t &=0 \\\\ t&=1 \\end{split}\\]

Thus, the maximum temperature will be reached when $t=1$.

Finally, to calculate the temperature we need to substitute $t=1$ in the original function $T$:

\\[ \\begin{split} T&=\\var{a}te^{-t} \\\\\\\\ T_{max}&=\\var{a}\\times 1\\times e^{-1} \\\\&=\\frac{\\var{a}}{e} \\\\ &=\\var{ansb} ~~\\text{[2.d.p.]} \\end{split}\\]

Thus, the maximum temperature of the mixture is $\\var{ansb} ^\\circ C$ above room temperature.

Calculate the rate of change of the temperature $\\var{t1}$ minutes after the metal is added.

When $t=\\var{t1}$ minutes then $\\frac{dT}{dt}=$ [[0]] $^\\circ \\mathrm{C}/\\mathrm{min}$.

Give your answer to two decimal places, if needed.

Calculate the maximum temperature of the mixture.

The maximum temperature of the mixture is [[0]] $^\\circ \\mathrm{C}$ above room temperature.

Give your answer to two decimal places, if needed.

Solving a separable differential equation that describes the population growth over time with a known initial condition to calculate the population after $n$ years.

", "licence": "None specified"}, "statement": "

A population of $N$ individuals grows over time, $t$ years, according to the equation:
$\\frac{\\mathrm{d}N}{\\mathrm{d}t}=\\left(\\var{a}+ e^{\\simplify{-{b}t}}\\right)N $

At time $t=0$, the population contains $\\var{N}$ individuals.

", "advice": "

To calculate the population $\\var{N}$ years later, we need an expression for how the population, $N$, changes in terms of time, $t$. This is given by the differential equation,

\\[ \\frac{dN}{dt}=\\left(\\var{a}+ e^{\\simplify{-{b}t}}\\right)N \\]

which we can solve using separation of variables.

First, we need to separate the variables between the two sides:

\\[ \\frac{1}{N}dN=\\left( \\var{a}+e^{\\simplify{-{b}t}}\\right)dt \\]

Now, we integrate both sides:

\\[ \\begin{split} \\int\\frac{1}{N}dN &=\\int\\left( \\var{a}+e^{\\simplify{-{b}t}}\\right)dt \\\\ \\ln{N} &= \\simplify{{a}*t}+\\left(-\\simplify{1/{b}*e}^{\\simplify{-{b}t}}\\right)+c \\\\ \\ln{N} &= \\simplify{{a}*t}-\\simplify{1/{b}*e}^{\\simplify{-{b}t}}+c \\end{split} \\]

To find the value of $c$, we can use the fact that when $t=0$, $N=\\var{N}$. By substituting in the values in the equation we get:

\\[ \\begin{split} \\ln(\\var{N}) &= \\var{a} \\times 0-\\simplify{1/{b}*e}^{-\\simplify{{b}* 0}}+c \\\\ \\ln(\\var{N}) &= -\\simplify{1/{b}*e}^{0}+c \\\\ \\ln(\\var{N}) &= -\\simplify{1/{b}* 1}+c \\\\ \\ln(\\var{N}) &= -\\simplify{1/{b}}+c \\\\ c&=\\simplify{ ln{{N}}+1/{b}}\\end{split} \\]

Therefore $c=\\simplify{ln{{N}}+1/{b}}=\\var{c}$. So,

\\[ \\begin{split} \\ln({N}) &= \\simplify{{a}*t}-\\simplify{1/{b}*e}^{-\\simplify{{b}*t}}+\\simplify{ln{{N}}+1/{b}} \\end{split} \\]

Taking the exponential of both sides:

\\[ \\begin{split} N &= e^{\\simplify{{a}*t}-\\simplify{1/{b}e}^{-\\simplify{{b}*t}}+\\simplify{ln{{N}}+1/{b}}} \\end{split} \\]

We can now substitute $t=\\var{t0}$ to calculate the population:

\\[ \\begin{split} N &= e^{\\var{a} \\times \\var{t0}-\\simplify{1/{b}*e}^{-\\var{b} \\times \\var{t0}}+\\simplify{ln{{N}}+1/{b}}} \\\\ N&=\\var{step3} \\end{split} \\]

Thus, the population $\\var{t0}$ years later will be $\\var{ans}$ individuals.

How many individuals are in the population $\\var{t0}$ years later?

The population is [[0]] individuals.

If needed, round your answer to the nearest integer.

Solving a separable differential equation that describes the rate of decay of radioactive isotopes over time with a known initial condition to calculate the mass of the isotope after a given time and the time taken for the mass to reach $m$ grams.

Decay Constant - Radioactivity - Nuclear Power (nuclear-power.com)

", "licence": "None specified"}, "statement": "

The rate of decay of the radioactive isotope $\\var{isotopes_names[j]}$ is given by:

\\[\\frac{\\mathrm{d}M}{\\mathrm{d}t}=-\\var{lambda}M\\]

where $M$ is the mass of the isotope in grams and $t$ is time measured in $\\var{timescale[j]}$.
A sample initially contains $\\var{m0}$ grams of the isotope.

", "advice": "

a) To calculate the mass of the isotope at a specific time, we need an expression for how the mass, $M$, changes in terms of time, $t$. This is given by the differential equation,

\\[\\frac{\\mathrm{d}M}{\\mathrm{d}t}=-\\var{lambda}M\\]

which we can solve using separation of variables.

First we need to separate the variables between the two sides:

\\[\\frac{1}{M}dM=-\\var{lambda}dt\\]

Now we can integrate both sides:

\\[ \\begin{split} \\int\\frac{1}{M}dM&=\\int -\\var{lambda}dt \\\\ \\ln{M}&= -\\var{lambda}t + c \\end{split} \\]

Taking the exponential of both sides:

\\[ \\begin{split} M&= e^{ -\\var{lambda}t + c} \\\\ M&=e^c\\times e^{-\\var{lambda}t }\\end{split} \\]

Now, $e^c$ is a constant which we can calculate using the initial contition that $m=\\var{m0}$ when $t=0$. By substituting in the values in the equation we get:

\\[ \\begin{split} \\var{m0}&=e^c\\times e^{-\\var{lambda}\\times 0 } \\\\ \\var{m0}&=e^c\\times e^0 \\\\ \\var{m0}&=e^c\\times 1 \\\\ \\var{m0}&=e^c \\end{split} \\]

So, $c=\\var{m0}$. Therefore,

\\[ \\begin{split} M&=\\var{m0} e^{-\\var{lambda}t }\\end{split} \\]

We can now substitute $t=\\var{t1}$ to calculate the mass at that time:

\\[ \\begin{split} M&=\\var{m0} e^{-\\var{lambda} \\times \\var{t1} } \\\\ M&= \\var{m0}\\times e^{\\simplify{{-decayconstant[j]*t1}}}\\\\ M&= \\var{m0} \\times \\simplify{{e^(-decayconstant[j]*t1)}} = \\var{ansa} ~~\\text{[3 s.f.]} \\end{split} \\]

b) To calculate how long it takes for the mass of the isotope in the sample to fall to $\\var{m1}$ gram, we will use the expression

\\[ \\begin{split} M&=\\var{m0} e^{-\\var{lambda}t }\\end{split} \\]

found in part (a). We can substitute $M=\\var{m1}$ and solve the equation for t.

\\[ \\begin{split} \\var{m1}&=\\var{m0} e^{-\\var{lambda}t } \\\\ \\frac{\\var{m1}}{\\var{m0}}&=e^{-\\var{lambda}t }\\end{split} \\]

Taking $\\ln()$ of both sides:

\\[ \\begin{split} \\ln \\left(\\frac{\\var{m1}}{\\var{m0}}\\right)&=-\\var{lambda}t \\\\ -\\frac{\\ln \\left(\\frac{\\var{m1}}{\\var{m0}}\\right)}{\\var{lambda}}&=t \\end{split} \\]

Thus, $m=\\var{m1}$ when $t=\\var{ansb}$ {timescale[j]}.

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The mass is [[0]]grams.

Round your answer to 3 significant figures, if needed.

How long does it take for the mass of the isotope in the sample to fall to $\\var{m1}$ gram?

It takes $t=$[[0]]$\\var{timescale[j]}$.

Round your answer to the nearest integer if needed.

Using given information to complete the equation $c= A \\cos{ \\left( \\frac{2 \\pi}{P} \\left( t-H \\right) \\right) }+V $ that describes the concentration, $c$, of perscribed drug in a patient's drug over time, $t$. Calculating the maximum concentration and the concentration at a specific time.

", "licence": "None specified"}, "statement": "

A patient takes a drug every $\\var{P}$ hours each day. The concentration, $c$, of the drug in the patient’s blood $t$ hours after the start of the treatment is modelled by the equation:
$c=\\var{A}\\cos{\\left(k\\left(t-\\var{H}\\right)\\right)}+\\var{V}$

", "advice": "

a) The equation

\\[ \\begin{split} c=\\var{A}\\cos{\\left(k\\left(t-\\var{H}\\right)\\right)}+\\var{V} \\end{split} \\]

is a wave function.

The general form of a wave function can be written as:

\\[ \\begin{split} c=A\\cos{\\left(\\frac{2 \\pi}{P}\\left(t-H\\right)\\right)}+V \\end{split} \\]

Where $V$ is the average value, $A$ is the amplitute, $H$ the phase and $P$ the period.

By comparying the equation with the general form we can notice that

\\[ k= \\frac{2 \\pi}{P} \\]

We know that the patient takes the drug every $\\var{P}$ hours each day. Therefore, the period is $P=\\var{P}$. Therefore,

\\[ \\begin{split} k&= \\frac{2 \\pi}{\\var{P}} \\\\ &=\\simplify{ {2*pi} / {P}} \\end{split} \\]

So, $k=\\simplify{ {2*pi} / {P}}$ and we can now rewrite the equation as:

\\[ \\begin{split} c=\\var{A}\\cos{\\left(\\simplify{ {2*pi} / {P}}\\left(t-\\var{H}\\right)\\right)}+\\var{V} \\end{split} \\]

b) We know that the wave function is a transformation of the trigonometric function cos(x).

Clink on the link for a visual representation of the wave function.

https://www.desmos.com/calculator/ssqdx7ys7k

We also know that the cosine function has a maximum value of 1. So, the maximum value of the wave function will occure when $ \\cos{\\left(\\simplify{ {2*pi} / {P}}\\left(t-\\var{H}\\right)\\right)}=1$. Therefore,

\\[ \\begin{split} c_{max}&=\\var{A}\\times 1 +\\var{V} \\\\ &= \\var{A} +\\var{V} \\\\ &=\\var{A+V} \\end{split} \\]

So, $c_{max}=\\var{max}$ mg/L.

c) To calculate the concentration of the drug $\\var{t1}$ hours after taking it for the first time, we need to subtitute $t=\\var{t1}$ in the equation

\\[ \\begin{split} c=\\var{A}\\cos{\\left(\\simplify{ {2*pi} / {P}}\\left(t-\\var{H}\\right)\\right)}+\\var{V} \\end{split} \\]

Therefore, when $t=\\var{t1}$ the equation become

\\[ \\begin{split} c_{\\var{t1}}&=\\var{A}\\cos{\\left(\\simplify{ {2*pi} / {P}}\\left(\\var{t1}-\\var{H}\\right)\\right)}+\\var{V} \\\\ &=\\var{A}\\cos{\\left(\\simplify{ {2*pi} / {P}} \\times \\var{t1-H}\\right)}+\\var{V} \\\\ &=\\var{A}\\cos{\\left(\\simplify{ {2*pi*(t1-H)}/{P}}\\right)}+\\var{V} \\\\ &=\\var{ansC} \\end{split} \\]

So, $ c_{\\var{t1}}=\\var{ansc} $ mg/L.

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Calculate the value of the constant $k$.

$k=$[[0]]

Give your answer as a fraction in terms of $\\pi$ (you can type: pi).

What is the maximum concentration of the drug in the patient’s bloodstream at any time?

$c_{max}=$[[0]] mg/L

What is the concentration of the drug $\\var{t1}$ hours after taking it for the first time?

$c_{\\var{t1}}=$[[0]]

Give your answer rounded to 2 decimal places, if needed.

Using the given information to complete the equation $y= A \\cos{ \\left( \\frac{2 \\pi}{P} x \\right) }+V $ that describes an electromagnetic wave and calculating the smallest angle, $x$, for which $y=y_0$.

", "licence": "None specified"}, "statement": "

The shape of an electromagnetic wave is modelled by the equation:
$y=\\var{a}\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)}+\\var{v}$

where $x$ is measured in radians.

", "advice": "

The equation

\\[ \\begin{split} y=\\var{a}\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)}+\\var{v} \\end{split} \\]

is a wave function.

The general form of a wave function can be written as:

\\[ \\begin{split} y=A\\cos{\\left(\\frac{2 \\pi}{P}\\left(x-H\\right)\\right)}+V \\end{split} \\]

Where $V$ is the average value, $A$ is the amplitute, $H$ the phase and $P$ the wavelenght (also called period).

By comparying the equation with the general form we can notice that:

a) The amplitude of the wave is $A=\\var{a}$

b)The period can be found by solving the equation

\\[ \\begin{split} \\frac{\\pi}{\\simplify{{p}/2}} &= \\frac{2 \\pi}{P} \\\\ \\pi P&= \\simplify{{p}/2}\\times 2\\pi \\\\ \\pi P&= {p}\\pi \\\\ P&=\\frac{{p} \\pi}{\\pi} \\\\ P &=\\var{p} \\end{split} \\]

So, the wavelenght is $P=\\var{p}$.

To find the smallest value of $x$ for which $y=\\var{y1}$. We need to substitute $y=\\var{y1}$ in the equation and solve for $x$.

\\[ \\begin{split} \\var{y1}&=\\var{a}\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)}+\\var{v} \\\\ \\var{y1}-\\var{v} &=\\var{a}\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)} \\\\ \\var{y1-v} &=\\var{a}\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)} \\\\ \\frac{\\var{y1-v}}{\\var{a}} &=\\cos{\\left(\\frac{\\pi x}{\\simplify{{p}/2}}\\right)}\\end{split} \\]

We take $\\cos^{-1}$ from both sides:

\\[ \\begin{split} \\cos^{-1} \\left(\\frac{\\var{y1-v}}{\\var{a}} \\right)&=\\frac{\\pi x}{\\simplify{{p}/2}} \\\\ \\simplify{{p}/2} \\cos^{-1} \\left(\\frac{\\var{y1-v}}{\\var{a}} \\right)&=\\pi x \\\\ \\frac{\\simplify{{p}/2} \\cos^{-1} \\left(\\frac{\\var{y1-v}}{\\var{a}} \\right)}{\\pi}&=x \\end{split} \\]

We can use the calculator to find that $x= \\frac{\\simplify{{p}/2} \\cos^{-1} \\left(\\frac{\\var{y1-v}}{\\var{a}} \\right)}{\\pi}=\\var{ansC}$ radians (rounded to 2 decimal places).

We can use the calculator to find that $x= \\frac{\\simplify{{p}/2} \\cos^{-1} \\left(\\frac{\\var{y1-v}}{\\var{a}} \\right)}{\\pi}=\\var{ansC}$ radians.

Write down the amplitude of this wave.

$A=$[[0]]

Calculate the wavelength of this wave.

The wavelength is [[0]].

Find the smallest value of $x$ for which $y=\\var{y1}$.

$x=$[[0]] radians.

Give your answer rounded to 2 decimal places, if needed.

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