// Numbas version: exam_results_page_options {"name": "Further Differentiation", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": "0", "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", ""], "variable_overrides": [[], []], "questions": [{"name": "Differentiation: Exponential and Logarithmic functions - Gradient of Hill", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Evi Papadaki", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18113/"}], "tags": [], "metadata": {"description": "

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.

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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.

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