Implicit differentiation, related rates and optimisation: practice worksheet

1. Question 1warm-up · no calculator

   The curve $C$ has equation $x^{2} + xy + y^{2} = 7$.

   Find the gradient of $C$ at the point $(1, 2)$.[3]

   Markscheme

   attempts implicit differentiation, including product rule on $xy$ and chain rule on $y^{2}$	M1
   $2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$ (or $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$)	A1
   gradient $= -\frac{4}{5}$	A1

   Award M1A0 if the chain rule factor $\frac{dy}{dx}$ is omitted from $y^{2}$ but present on $xy$. Substitution of $x = 1$, $y = 2$ before rearranging is acceptable.

   Answer: $-\frac{4}{5}$

2. Question 2warm-up · no calculator

   The edge length $x$ cm of a cube is increasing at a constant rate of $0.05$ cm s$^{-1}$.

   (a)Find the rate at which the volume of the cube is increasing at the instant when $x = 8$.[2]

   Markscheme

   $\frac{dV}{dt} = 3x^{2}\frac{dx}{dt}$	M1
   $= 3(64)(0.05) = 9.6$ (cm$^{3}$ s$^{-1}$)	A1

   Accept $\frac{48}{5}$.

   Answer: $9.6$ cm$^{3}$ s$^{-1}$

   (b)Find the rate at which the total surface area of the cube is increasing at the same instant.[2]

   Markscheme

   $S = 6x^{2}$ so $\frac{dS}{dt} = 12x\frac{dx}{dt}$	M1
   $= 12(8)(0.05) = 4.8$ (cm$^{2}$ s$^{-1}$)	A1

   Award M1A0 for $S = x^{2}$ leading to $0.8$.

   Answer: $4.8$ cm$^{2}$ s$^{-1}$

3. Question 3exam-style · no calculator

   The curve $C$ has equation $x^{2} - xy + y^{2} = 12$.

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

   Markscheme

   attempts implicit differentiation with product rule on $xy$	M1
   $2x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$	A1
   $\frac{dy}{dx}(2y - x) = y - 2x$ leading to $\frac{dy}{dx} = \frac{y - 2x}{2y - x}$	A1
   	AG

   The rearrangement must be seen for the final A1.

   Answer: $\frac{dy}{dx} = \frac{y - 2x}{2y - x}$

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

   Markscheme

   sets numerator equal to zero: $y = 2x$	M1
   substitutes into the equation of $C$: $x^{2} - 2x^{2} + 4x^{2} = 12 \Rightarrow 3x^{2} = 12$	M1
   $(2, 4)$ and $(-2, -4)$	A1

   Award A0 if only one point is given or if extra points are included.

   Answer: $(2, 4)$ and $(-2, -4)$

   (c)Hence, or otherwise, find the coordinates of the points on $C$ at which the tangent is vertical.[2]

   Markscheme

   sets denominator equal to zero, $x = 2y$, and substitutes: $4y^{2} - 2y^{2} + y^{2} = 12$	M1
   $(4, 2)$ and $(-4, -2)$	A1

   Accept an argument from the symmetry of $C$ in the line $y = x$ provided it is stated.

   Answer: $(4, 2)$ and $(-4, -2)$

4. Question 4exam-style · calculator

   Sand falls from a conveyor belt at a constant rate of $0.6$ m$^{3}$ per minute and forms a pile in the shape of a right circular cone. The height $h$ metres of the pile is always three-quarters of its base radius $r$ metres.

   (a)Show that the volume $V$ m$^{3}$ of the pile is given by $V = \frac{16\pi h^{3}}{27}$.[2]

   Markscheme

   $r = \frac{4h}{3}$ substituted into $V = \frac{1}{3}\pi r^{2} h$	M1
   $V = \frac{1}{3}\pi \cdot \frac{16h^{2}}{9} \cdot h = \frac{16\pi h^{3}}{27}$	A1
   	AG

   Answer: $V = \frac{16\pi h^{3}}{27}$

   (b)Find the rate at which the height of the pile is increasing at the instant when $h = 1.5$. Give your answer in metres per minute correct to three significant figures.[4]

   Markscheme

   $\frac{dV}{dh} = \frac{16\pi h^{2}}{9}$	A1
   uses $\frac{dV}{dt} = \frac{dV}{dh}\cdot\frac{dh}{dt}$	M1
   $\frac{dV}{dh} = 4\pi$ when $h = 1.5$ (seen anywhere)	A1
   $\frac{dh}{dt} = \frac{0.6}{4\pi} = 0.0477$ (m min$^{-1}$)	A1

   Accept $\frac{3}{20\pi}$. Award M1 for an equivalent chain rule statement such as $0.6 = \frac{16\pi h^{2}}{9}\frac{dh}{dt}$.

   Answer: $\frac{3}{20\pi} \approx 0.0477$ m min$^{-1}$

   (c)Find the rate at which the base radius is increasing at the same instant.[1]

   Markscheme

   $\frac{dr}{dt} = \frac{4}{3}\cdot\frac{dh}{dt} = 0.0637$ (m min$^{-1}$)

   A1

   FT from part (b). Accept $\frac{1}{5\pi}$.

   Answer: $\frac{1}{5\pi} \approx 0.0637$ m min$^{-1}$

5. Question 5exam-style · no calculator

   A farmer has $60$ m of fencing and wishes to enclose a rectangular pen on the bank of a straight river. No fencing is needed along the river. The side of the pen parallel to the river has length $x$ metres. Local regulations require that $x \geq 40$.

   (a)Show that the area $A$ m$^{2}$ of the pen is given by $A = 30x - \frac{1}{2}x^{2}$, and state the domain of $x$.[2]

   Markscheme

   each side perpendicular to the river has length $\frac{60 - x}{2}$, so $A = x\cdot\frac{60 - x}{2} = 30x - \frac{1}{2}x^{2}$	M1
   	AG
   domain $40 \leq x < 60$	A1

   Condone $40 \leq x \leq 60$.

   Answer: $A = 30x - \frac{1}{2}x^{2}$, $40 \leq x < 60$

   (b)Find $\frac{dA}{dx}$ and show that $A$ has no stationary point within the domain.[3]

   Markscheme

   $\frac{dA}{dx} = 30 - x$	A1
   sets $\frac{dA}{dx} = 0$ to obtain $x = 30$	M1
   $30 < 40$, so the stationary point is outside the domain	R1

   Answer: $\frac{dA}{dx} = 30 - x$; the only stationary point is at $x = 30$, which is not in the domain

   (c)Hence determine the maximum possible area of the pen, justifying your answer.[3]

   Markscheme

   considers the sign of $\frac{dA}{dx}$ on the domain: $30 - x < 0$ for all $x \geq 40$	M1
   $A$ is decreasing throughout the domain, so the maximum occurs at the end point $x = 40$	R1
   $A = 30(40) - \frac{1}{2}(1600) = 400$ (m$^{2}$)	A1

   Do not award R0A1. A sketch of the parabola with vertex at $x = 30$ and the domain marked is an acceptable justification.

   Answer: $400$ m$^{2}$ (at the end point $x = 40$)

6. Question 6challenge · calculator

   The number of bacteria in a colony, $N$ thousand, after $t$ hours satisfies the differential equation $$\frac{dN}{dt} = 0.02N(60 - N), \quad 0 < N < 60.$$ Initially there are $10$ thousand bacteria in the colony.

   (a)By differentiating implicitly with respect to $t$, show that $\frac{d^{2}N}{dt^{2}} = 0.02(60 - 2N)\frac{dN}{dt}$.[3]

   Markscheme

   differentiates the right-hand side with respect to $t$ using the product rule, with a factor $\frac{dN}{dt}$ on each $N$ term	M1
   $\frac{d^{2}N}{dt^{2}} = 0.02\left[(60 - N)\frac{dN}{dt} - N\frac{dN}{dt}\right]$	A1
   $= 0.02(60 - 2N)\frac{dN}{dt}$	A1
   	AG

   METHOD 2: expands to $1.2N - 0.02N^{2}$ first, then $\frac{d^{2}N}{dt^{2}} = (1.2 - 0.04N)\frac{dN}{dt}$ (M1A1), then factorises to the given form (A1).

   Answer: $\frac{d^{2}N}{dt^{2}} = 0.02(60 - 2N)\frac{dN}{dt}$

   (b)Hence find the value of $N$ for which the rate of growth of the colony is greatest, and justify that this value gives a maximum.[3]

   Markscheme

   sets $\frac{d^{2}N}{dt^{2}} = 0$	M1
   $N = 30$ (since $\frac{dN}{dt} \neq 0$ for $0 < N < 60$)	A1
   $\frac{dN}{dt} > 0$ on the domain, so $\frac{d^{2}N}{dt^{2}} > 0$ for $N < 30$ and $\frac{d^{2}N}{dt^{2}} < 0$ for $N > 30$; the rate increases then decreases, hence a maximum	R1

   Do not award R0A1. Accept an argument that $\frac{dN}{dt} = 0.02N(60 - N)$ is a quadratic in $N$ with negative leading coefficient and vertex at $N = 30$.

   Answer: $N = 30$ (thousand)

   (c)Write down the maximum rate of growth of the colony.[1]

   Markscheme

   $\frac{dN}{dt} = 0.02(30)(30) = 18$ (thousand per hour)

   A1

   Units not required for the mark.

   Answer: $18$ thousand bacteria per hour

   (d)Solve the differential equation to show that $N = \dfrac{60}{1 + 5e^{-1.2t}}$.[5]

   Markscheme

   separates variables: $\int \frac{1}{N(60 - N)}\,dN = \int 0.02\,dt$	M1
   $\frac{1}{N(60 - N)} = \frac{1}{60}\left(\frac{1}{N} + \frac{1}{60 - N}\right)$	A1
   $\frac{1}{60}\ln\left(\frac{N}{60 - N}\right) = 0.02t + c$ (or $\ln\left(\frac{N}{60 - N}\right) = 1.2t + k$)	A1
   uses $N = 10$ when $t = 0$: $k = \ln\frac{1}{5}$, so $\frac{N}{60 - N} = \frac{1}{5}e^{1.2t}$	M1
   $5N = (60 - N)e^{1.2t} \Rightarrow N = \frac{60e^{1.2t}}{5 + e^{1.2t}} = \frac{60}{1 + 5e^{-1.2t}}$	A1
   	AG

   Condone missing modulus signs since $0 < N < 60$. The final rearrangement must be shown for the last A1.

   Answer: $N = \dfrac{60}{1 + 5e^{-1.2t}}$

   (e)Find the time, in hours correct to three significant figures, at which the rate of growth of the colony is greatest.[2]

   Markscheme

   sets $N = 30$: $1 + 5e^{-1.2t} = 2$	M1
   $t = \frac{\ln 5}{1.2} = 1.34$ (hours)	A1

   FT from part (b) only if a value of $N$ in the domain was obtained.

   Answer: $t = \dfrac{\ln 5}{1.2} \approx 1.34$ hours

Original questions and notes written by Pietro Meloni, PhD, from the IB syllabus. Not endorsed by or affiliated with the International Baccalaureate Organization. Found a mistake? Please tell me privately at pietro@pietromeloni.com. More questions on this topic: Calculus practice.