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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 $t$.

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

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Therefore, 

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\\[ \\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 $\\var{c}$. 

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

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

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

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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.

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\\[ \\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} \\] 

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

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

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$n=$[[0]]

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

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

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

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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, 

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a) The area of the dorsal fin is:

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\\[ \\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}\\] 

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

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\\[ \\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}\\]

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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.

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Calculate the area of the dorsal fin of this fish.

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

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

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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 $1\\le\\ x\\le3.5$.

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Calculate the area of the pelvic fin of the fish.

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

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

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