Implicit differentiation, related rates and further derivatives

1StandardNon-calculatorHL only4 marks

A curve is defined implicitly by $x^2+xy+y^2=7$. The point $A(1,2)$ lies on the curve.

Find the gradient of the curve at $A$.

Differentiates implicitly, with $\dfrac{d}{dx}(y^2)=2y\dfrac{dy}{dx}$M1
$2x+y+x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=0$ (product rule on $xy$)A1
Substitutes $x=1,\ y=2$: $4+5\dfrac{dy}{dx}=0$M1
$\dfrac{dy}{dx}=-\dfrac45$A1
Answer: $-\dfrac45$
Examiner tip: The term $xy$ needs the product rule, and forgetting its $x\,dy/dx$ part is the usual lost mark.

2StandardCalculatorHL only5 marks

A spherical balloon is inflated so that its volume increases at a constant rate of $20\ \text{cm}^3\,\text{s}^{-1}$.

(a) Find the rate of increase of the radius when the radius is 5 cm. [3]

(b) Hence find the rate of increase of the surface area of the balloon at this instant. [2]

$V=\dfrac43\pi r^3$, so $\dfrac{dV}{dr}=4\pi r^2$A1
Chain rule: $\dfrac{dr}{dt}=\dfrac{dV/dt}{dV/dr}=\dfrac{20}{4\pi(25)}$M1
$\dfrac{dr}{dt}=\dfrac{1}{5\pi}\approx0.0637\ \text{cm s}^{-1}$ (accept exact $\dfrac{1}{5\pi}$)A1
$A=4\pi r^2$, $\dfrac{dA}{dt}=8\pi r\dfrac{dr}{dt}$M1
$8\pi(5)\times\dfrac{1}{5\pi}=8\ \text{cm}^2\,\text{s}^{-1}$ (allow $8.00$ from using $0.0637$ or similar rounded value)A1
Answer: (a) $\dfrac{1}{5\pi}\approx0.0637\ \text{cm s}^{-1}$; (b) $8\ \text{cm}^2\,\text{s}^{-1}$
Examiner tip: Write the chain rule line $\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}$ explicitly so the method mark is secure.

3StandardNon-calculatorHL only4 marks

(a) Differentiate $f(x)=\arctan(2x)$. [2]

(b) Hence find the exact value of $\displaystyle\int_0^{1/2}\dfrac{1}{1+4x^2}\,dx$. [2]

Uses $\dfrac{d}{dx}\arctan u=\dfrac{1}{1+u^2}\times u'$M1
$f'(x)=\dfrac{2}{1+4x^2}$A1
Reverses (a): antiderivative $\dfrac12\arctan(2x)$M1
$\dfrac12\arctan1=\dfrac{\pi}{8}$A1
Answer: (a) $\dfrac{2}{1+4x^2}$; (b) $\dfrac{\pi}{8}$
Examiner tip: In (b) the answer to (a) is double the integrand, so divide by 2.

4HardNon-calculatorHL only7 marks

A ladder of length 5 m rests with its top against a vertical wall and its foot on horizontal ground. The foot is pulled away from the wall at a constant speed of $0.3\ \text{m s}^{-1}$. Let the foot be $x$ m from the wall and the top be $y$ m above the ground.

(a) Find the rate at which the top is sliding down the wall at the instant when $x=3$. [5]

(b) Find the value of $x$ at which the top and the foot move at the same speed. [2]

$x^2+y^2=25$M1
Differentiates with respect to $t$: $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$M1
$y=4$ when $x=3$A1
Substitutes $x=3,\ y=4,\ \dfrac{dx}{dt}=0.3$M1
$\dfrac{dy}{dt}=-0.225$, i.e. sliding down at $0.225\ \text{m s}^{-1}$A1
Equal speeds: $\left|\dfrac{dy}{dt}\right|=\dfrac{dx}{dt}$ gives $x=y$, so $2x^2=25$M1
$x=\dfrac{5}{\sqrt2}\ (\approx3.54)$ mA1
Answer: (a) $0.225\ \text{m s}^{-1}$ downwards; (b) $x=\dfrac{5}{\sqrt2}\approx3.54$ m
Examiner tip: Do not substitute $x=3$ before differentiating; $x$ and $y$ both change with $t$.

5CoreNon-calculatorHL only6 marks

Find $\dfrac{dy}{dx}$ for each of the following.

(a) $y=5^{2x}$ [2]

(b) $y=\log_3\left(x^2+1\right)$ [2]

(c) $y=\arccos(3x)$ [2]

(a) Uses $\dfrac{d}{dx}a^{u}=a^{u}\ln a\cdot\dfrac{du}{dx}$: a term $5^{2x}\ln5$ seen (or rewrites $5^{2x}=e^{2x\ln5}$)M1
$\dfrac{dy}{dx}=2\ln5\cdot5^{2x}$ (accept $\ln25\cdot5^{2x}$, $25^x\ln25$)A1
(b) Uses $\dfrac{d}{dx}\log_a u=\dfrac{1}{u\ln a}\dfrac{du}{dx}$ (or change of base to $\dfrac{\ln(x^2+1)}{\ln3}$): $\dfrac{1}{(x^2+1)\ln3}$ seenM1
$\dfrac{dy}{dx}=\dfrac{2x}{(x^2+1)\ln3}$A1
(c) Uses $\dfrac{d}{dx}\arccos u=-\dfrac{1}{\sqrt{1-u^2}}\dfrac{du}{dx}$ with $u=3x$: $\dfrac{\ldots}{\sqrt{1-9x^2}}$ seenM1
$\dfrac{dy}{dx}=-\dfrac{3}{\sqrt{1-9x^2}}$ (minus sign and factor 3 both required; do not accept $\sqrt{1-3x^2}$)A1
Answer: (a) $2\ln5\cdot5^{2x}$; (b) $\dfrac{2x}{(x^2+1)\ln3}$; (c) $-\dfrac{3}{\sqrt{1-9x^2}}$
Examiner tip: Students often drop the $\ln a$ factor for $a^x$ and $\log_a x$, or forget that arccos (unlike arcsin) has a minus sign; check the formula booklet.

6StandardNon-calculatorHL only8 marks

The diagram shows the curve $C$ with equation $x^2-xy+y^2=12$.

(a) Show that $\dfrac{dy}{dx}=\dfrac{2x-y}{x-2y}$. [3]

(b) Find the coordinates of the points on $C$ at which the tangent is horizontal. [3]

(c) Find the coordinates of the points on $C$ at which the tangent is vertical. [2]

Attempts implicit differentiation, with $2y\dfrac{dy}{dx}$ seen for $y^2$M1
Product rule correct on $xy$: $-\left(y+x\dfrac{dy}{dx}\right)$, giving $2x-y-x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=0$A1
Collects $\dfrac{dy}{dx}$ terms to $\dfrac{dy}{dx}(2y-x)=y-2x$ and rearranges to the given answer (AG; must see this line)A1
(b) Sets numerator $2x-y=0$, i.e. $y=2x$, and substitutes into the equation of $C$M1
$x^2-2x^2+4x^2=12\Rightarrow3x^2=12\Rightarrow x=\pm2$A1
$(2,4)$ and $(-2,-4)$ (both required)A1
(c) Sets denominator $x-2y=0$, i.e. $x=2y$, and substitutes: $3y^2=12$M1
$(4,2)$ and $(-4,-2)$ (both required)A1
Answer: (b) $(2,4)$, $(-2,-4)$; (c) $(4,2)$, $(-4,-2)$
Examiner tip: A vertical tangent comes from the denominator of $\dfrac{dy}{dx}$ being zero, not the numerator; many candidates also give only the positive root.

7StandardCalculatorHL only7 marks

A camera $C$ is on level ground 400 m from the launch point $L$ of a rocket $R$. The rocket rises vertically at a constant speed of $30\ \text{m s}^{-1}$. When the rocket is $h$ m above $L$, the camera points at it at an angle of elevation $\theta$ radians.

(a) Find the rate of change of $\theta$ when $h=300$. [4]

(b) Show that, at any height, $\dfrac{d\theta}{dt}=\dfrac{12000}{160000+h^2}$. [1]

(c) The camera can turn at a maximum rate of $0.05$ radians per second. Find the range of heights for which the camera cannot keep the rocket in view. [2]

Relates variables: $\tan\theta=\dfrac{h}{400}$M1
Differentiates with respect to $t$: $\sec^2\theta\,\dfrac{d\theta}{dt}=\dfrac{1}{400}\dfrac{dh}{dt}$ (accept $\theta=\arctan\frac{h}{400}$ differentiated correctly)A1
At $h=300$: $\tan\theta=0.75$, so $\sec^2\theta=1.5625$ (or hypotenuse 500 m, $\cos\theta=0.8$) and substitutes $\dfrac{dh}{dt}=30$M1
$\dfrac{d\theta}{dt}=0.048\ \text{rad s}^{-1}$ (accept $\dfrac{6}{125}$)A1
(b) Uses $\sec^2\theta=1+\dfrac{h^2}{160000}$ to obtain $\dfrac{d\theta}{dt}=\dfrac{30}{400}\cdot\dfrac{160000}{160000+h^2}$, simplified to the given answer (AG)A1
(c) Solves $\dfrac{12000}{160000+h^2}>0.05$ (or $=$), giving $h^2<80000$M1
$0\le h<283$ m (accept $h<200\sqrt2$; accept $\le$)A1
Answer: (a) $0.048\ \text{rad s}^{-1}$; (c) $0\le h<283$ m
Examiner tip: Do not substitute $h=300$ before differentiating; differentiate $\tan\theta=h/400$ first, then evaluate.

8HardCalculatorHL only8 marks

A vertical fence 2 m high stands on horizontal ground, 1 m from a vertical wall. A straight ladder rests with its foot $F$ on the ground, touches the top of the fence and leans against the wall. The ladder makes an angle $\theta$ with the ground, where $0<\theta<\dfrac{\pi}{2}$.

(a) Show that the length of the ladder is $L=2\operatorname{cosec}\theta+\sec\theta$. [2]

(b) Find $\dfrac{dL}{d\theta}$. [2]

(c) Hence show that $L$ is stationary when $\tan^3\theta=2$. [2]

(d) Find the length of the shortest ladder that can be used. [2]

Length from $F$ to the top of the fence is $\dfrac{2}{\sin\theta}$B1
Length from the top of the fence to the wall is $\dfrac{1}{\cos\theta}$; adds to obtain $L=2\operatorname{cosec}\theta+\sec\theta$ (AG)B1
(b) Uses derivatives of $\operatorname{cosec}\theta$ and $\sec\theta$ (one correct term)M1
$\dfrac{dL}{d\theta}=-2\operatorname{cosec}\theta\cot\theta+\sec\theta\tan\theta$ (accept $-\dfrac{2\cos\theta}{\sin^2\theta}+\dfrac{\sin\theta}{\cos^2\theta}$)A1
(c) Sets $\dfrac{dL}{d\theta}=0$ and writes in sines and cosines: $\dfrac{\sin\theta}{\cos^2\theta}=\dfrac{2\cos\theta}{\sin^2\theta}$M1
$\sin^3\theta=2\cos^3\theta\Rightarrow\tan^3\theta=2$ (AG; no errors)A1
(d) $\theta=\arctan\sqrt[3]{2}=0.900$ (accept $51.6^\circ$) substituted into $L$M1
$L=4.16$ m (accept $(2^{2/3}+1)^{3/2}$)A1
Answer: (d) $4.16$ m (at $\theta=0.900$ rad)
Examiner tip: In (b) the derivative of $\operatorname{cosec}\theta$ is negative; a missed sign gives no stationary point in the interval.

9HardNon-calculatorHL only8 marks

(a) (i) Express $\dfrac{3x}{x^2-x-2}$ in partial fractions. [2]

(ii) Hence find the exact value of $\displaystyle\int_3^5\dfrac{3x}{x^2-x-2}\,dx$, giving your answer in the form $\ln k$, where $k\in\mathbb{Q}$. [3]

(b) By completing the square, find $\displaystyle\int\dfrac{1}{x^2+4x+13}\,dx$. [3]

(a)(i) Factorises to $(x-2)(x+1)$ and writes $\dfrac{A}{x-2}+\dfrac{B}{x+1}$, attempting $A$ and $B$M1
$\dfrac{2}{x-2}+\dfrac{1}{x+1}$A1
(ii) Integrates to $2\ln|x-2|+\ln|x+1|$ (follow through their $A$, $B$)A1
Substitutes limits: $(2\ln3+\ln6)-(2\ln1+\ln4)$M1
$\ln\dfrac{27}{2}$ (accept $\ln13.5$)A1
(b) Completes the square: $(x+2)^2+9$M1
Uses $\displaystyle\int\dfrac{1}{u^2+a^2}du=\dfrac1a\arctan\dfrac ua$ with $a=3$: $\dfrac13\arctan(\ldots)$M1
$\dfrac13\arctan\left(\dfrac{x+2}{3}\right)+c$ (award only if $+c$ is present)A1
Answer: (a)(i) $\dfrac{2}{x-2}+\dfrac{1}{x+1}$; (ii) $\ln\dfrac{27}{2}$; (b) $\dfrac13\arctan\dfrac{x+2}{3}+c$
Examiner tip: The factor $\dfrac1a$ in front of arctan is the most commonly lost mark, together with $+c$ in an indefinite integral.

10StretchCalculatorHL only9 marks

A curve has equation $e^{y}+xy=e$. The point $P(0,\,1)$ lies on the curve.

(a) Show that $\dfrac{dy}{dx}=-\dfrac{y}{e^{y}+x}$. [2]

(b) Find the equation of the tangent to the curve at $P$. [2]

(c) Find the value of $\dfrac{d^2y}{dx^2}$ at $P$. [4]

(d) The tangent at $P$ is used to estimate the value of $y$ on the curve when $x=0.1$. State, with a reason, whether this estimate is an overestimate or an underestimate. [1]

Differentiates implicitly: $e^{y}\dfrac{dy}{dx}+y+x\dfrac{dy}{dx}=0$ (chain rule on $e^y$ and product rule on $xy$)M1
Factorises $\dfrac{dy}{dx}(e^y+x)=-y$ to reach the given answer (AG)A1
(b) Gradient at $P$: $-\dfrac{1}{e}$A1
$y=1-\dfrac{x}{e}$ (accept $y=-0.368x+1$)A1
(c) Differentiates $e^{y}y'+y+xy'=0$ again (or differentiates the quotient): $e^{y}(y')^2+e^{y}y''+2y'+xy''=0$ with at least three terms correctM1
All four terms correctA1
Substitutes $x=0$, $y=1$, $y'=-\dfrac1e$: $e\cdot\dfrac{1}{e^2}+e\,y''-\dfrac{2}{e}=0$M1
$\dfrac{d^2y}{dx^2}=\dfrac{1}{e^2}$ (accept $0.135$)A1
(d) Underestimate, because $\dfrac{d^2y}{dx^2}>0$ at $P$, so the curve is concave up and lies above its tangent near $P$ (R1: reason must refer to the sign of their second derivative or concavity)R1
Answer: (b) $y=1-\dfrac{x}{e}$; (c) $e^{-2}\approx0.135$; (d) underestimate (curve concave up)
Examiner tip: When differentiating a second time, $x\dfrac{dy}{dx}$ needs the product rule again, which gives a second $\dfrac{dy}{dx}$ term; leaving it out loses the accuracy marks.

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