MATH101 Calculus Assessed Exercise 3 Solutions

A3.1.

(i) We write

$$f(x)=x^{1/2}-2x^{-4}+2\log x+\log 3,$$

so

$$f^{\prime}(x)=2^{-1}x^{-1/2}+8x^{-5}+2x^{-1}.$$

[1] mark

(ii) By the product rule, and then the chain rule, we have

$$\eqalign{{{d}\over{dx}}\Bigl(e^{2x^{3}}\cos x\Bigr)&=\Bigl({{d}\over{dx}}e^{2x^{3}}\Bigr)\cos x+e^{2x^{3}}{{d}\over{dx}}\cos x\cr&=e^{2x^{3}}\bigl(6x^{2}\cos x-\sin x\bigr).\cr}$$

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(iii) By the chain rule

$${{d}\over{dx}}\Bigl(\bigl(\cos 5x\bigr)^{3}+e^{2x^{7}}\Bigr)=-15\sin 5x\cos^{2}5x+14x^{6}e^{2x^{7}}.$$

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(iv) By the quotient rule,

$$\eqalign{{{d}\over{dx}}{{\sin x^{3}}\over{7x}}&={{({{d}\over{dx}}\sin x^{3})x-\sin x^{3}{{dx}\over{dx}}}\over{7x^{2}}}\cr&={{3x^{3}\cos x^{3}-\sin x^{3}}\over{7x^{2}}}.\cr}$$

[1] mark

Total [4] marks for question

A3.2 Note that

$$\eqalignno{{{d}\over{dx}}{\hbox{sech}}\,x&={{d}\over{dx}}{{1}\over{\cosh x}}\cr&={{-\sinh x}\over{\cosh^{2}x}};\cr}$$

so

$$\eqalignno{{{d}\over{dx}}f(x)&={{d}\over{dx}}\Bigl({\hbox{sech}}\,x\Bigr)^{2}\cr&=-2{\hbox{sech}}\,x({\hbox{sech}}\,x\tanh x)\cr&={{-2\sinh x}\over{\cosh^{3}x}};\cr}$$

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(any equivalent form is acceptable) hence

$$(f^{\prime}(x))^{2}={{4\sinh^{2}x}\over{\cosh^{6}x}},$$

while

$$\eqalignno{4f(x)^{2}(1-f(x))&=4{\hbox{sech}}^{4}x(1-{\hbox{sech}}^{2}x)\cr&=4{\hbox{sech}}^{4}x\tanh^{2}x\cr&={{4\sinh^{2}x}\over{\cosh^{6}x}}\cr}$$

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(any equivalent form is acceptable) hence

$$f^{\prime}(x)^{2}=4f(x)^{2}(1-f(x)).$$

Total [2] marks for question

A3.3 Let $M$ be the maximum charge, $Q$ be the charge at time $x$ and $R=M-Q$. Then

$${{dR}\over{dx}}=-kR,$$

where $k$ is to be found. The solution is

$$R=R_{0}e^{-kx}$$

where $R(0)=R_{0}$; hence

$$M-Q=(M-Q_{0})e^{-kx}$$

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where $Q_{0}=M-R_{0}=10M/100$, and $Q(1)=50M/100,$ so $R_{0}=90M/100$ and

$$50M=90Me^{-k},$$

so $k=\log(9/5).$ We look for $x$ such that $Q=90M/100$, so

$$10M=90Me^{-kx};$$

hence

$$x={{\log 9}\over{k}}={{\log 9}\over{\log(9/5)}}=3.781\quad{\hbox{hours}}.$$

[1]

[2] marks for question

A3.4 (i) By the quotient rule, we have

$$\eqalignno{{{d}\over{dx}}\coth x&={{d}\over{dx}}{{\cosh x}\over{\sinh x}}\cr&={{({{d}\over{dx}}\cosh x)\sinh x-\cosh x{{d}\over{dx}}\sinh x}\over{\sinh^{2}x}}\cr&={{\sinh^{2}x-\cosh^{2}x}\over{\sinh^{2}x}}\cr&={{-1}\over{\sinh^{2}x}}\cr&=-{\hbox{cosech}}^{2}\,x\qquad(x\neq 0).\cr}$$

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(ii) We have

$$\cosh^{2}y-\sinh^{2}y=1,$$

so

$$\coth^{2}y-1={\hbox{cosech}}^{2}\,y.$$

Let $y=\coth^{-1}x$ so $x=\coth y$ and

$${{dx}\over{dy}}=-{\hbox{cosech}}^{2}\,y$$

$${{dx}\over{dy}}=1-\coth^{2}y=1-x^{2},$$

so

$${{dy}\over{dx}}=1/{{dx}\over{dy}}={{-1}\over{1-x^{2}}}$$

hence

$${{dy}\over{dx}}={{1}\over{x^{2}-1}}$$

[1]

[2] marks for question

Total [10] marks

this is a standard calculus exercise