Further differentiation & integration (implicit, related rates, by parts, substitution) (HL)

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 only5 marks

Use integration by parts to find the exact value of $\displaystyle\int_0^1 xe^{2x}\,dx$.

Chooses $u=x,\ \dfrac{dv}{dx}=e^{2x}$ and applies $uv-\int v\,du$M1
$\dfrac{x}{2}e^{2x}-\int\dfrac12e^{2x}dx$A1
$\dfrac{x}{2}e^{2x}-\dfrac14e^{2x}$A1
Substitutes limits $1$ and $0$M1
$\dfrac{e^2}{4}+\dfrac14=\dfrac{e^2+1}{4}$A1
Answer: $\dfrac{e^2+1}{4}$
Examiner tip: Choose $u=x$ so it differentiates away; the other way round makes the integral harder.

4HardNon-calculatorHL only5 marks

Use the substitution $u=x+1$ to find the exact value of $\displaystyle\int_0^3\dfrac{x}{\sqrt{x+1}}\,dx$.

$u=x+1$, $du=dx$, $x=u-1$M1
Limits become $u=1$ to $u=4$A1
Integrand $\dfrac{u-1}{\sqrt u}=u^{1/2}-u^{-1/2}$M1
$\dfrac23u^{3/2}-2u^{1/2}$A1
$\Big(\dfrac{16}{3}-4\Big)-\Big(\dfrac23-2\Big)=\dfrac83$A1
Answer: $\dfrac83$
Examiner tip: Change the limits with the substitution, or convert back to $x$ before substituting; mixing the two loses the final marks.

5StandardNon-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.

6StretchNon-calculatorHL only7 marks

(a) By using integration by parts twice, show that $\displaystyle\int e^x\sin x\,dx=\dfrac{e^x(\sin x-\cos x)}{2}+c$. [6]

(b) Hence find the exact value of $\displaystyle\int_0^{\pi}e^x\sin x\,dx$. [1]

Attempts first application of parts with $u=\sin x,\ \dfrac{dv}{dx}=e^x$ (allow $u=e^x$ consistently in both applications)M1
$I=e^x\sin x-\displaystyle\int e^x\cos x\,dx$A1
Attempts second application on $\displaystyle\int e^x\cos x\,dx$ with $u=\cos x,\ \dfrac{dv}{dx}=e^x$M1
$\displaystyle\int e^x\cos x\,dx=e^x\cos x+\int e^x\sin x\,dx$A1
Substitutes back and recognises the original integral: $I=e^x\sin x-e^x\cos x-I$M1
$2I=e^x(\sin x-\cos x)$, leading to $I=\dfrac{e^x(\sin x-\cos x)}{2}+c$ AG (the A1 is for $2I=\ldots$; the AG line earns no extra mark)A1
(b) $\dfrac{e^{\pi}(0+1)-e^0(0-1)}{2}=\dfrac{e^{\pi}+1}{2}$A1
Answer: (a) shown; (b) $\dfrac{e^{\pi}+1}{2}$
Examiner tip: Keep the same choice of $u$ (trigonometric) in both applications, otherwise the integral just undoes itself.

7HardNon-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 $t$.

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