Applications of differentiation
Differentiation
- A-level
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Find the coordinates of the stationary points on the curve $y = x^3 - 3x + 1$.
3 marks
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Classify the stationary points on the curve $y = x^3 - 3x$.
3 marks
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Find the values of $x$ for which $y = x^3 - 6x^2 + 9x$ is increasing.
4 marks
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The radius of a circle is increasing at $0.2$ cm s$^{-1}$. Find the rate of increase of the area when the radius is $5$ cm.
4 marks
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Find and classify the stationary point on the curve $y = xe^{-x}$.
5 marks
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Air is pumped into a spherical balloon at $12$ cm$^3$ s$^{-1}$. Find the rate at which the radius is increasing when the radius is $2$ cm.
5 marks
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A rectangle has a perimeter of $40$ cm. Find the maximum possible area, and the side lengths that give it.
6 marks
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An open box is made from a square sheet of side $12$ cm by cutting a square of side $x$ cm from each corner and folding up the sides. Find the value of $x$ that maximises the volume, and find that maximum volume.
7 marks
Mark scheme
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Answer. $(1, -1)$ and $(-1, 3)$
- $\dfrac{dy}{dx} = 3x^2 - 3 = 0$, so $x = \pm 1$.
- The points are $(1, -1)$ and $(-1, 3)$.
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Answer. There is a maximum at $(-1, 2)$ and a minimum at $(1, -2)$.
- $\dfrac{d^2y}{dx^2} = 6x$.
- At $x = -1$ the second derivative is negative, and at $x = 1$ it is positive.
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Answer. $x \le 1$ or $x \ge 3$
- $\dfrac{dy}{dx} = 3(x - 1)(x - 3)$.
- The derivative is at least $0$ when $x \le 1$ or $x \ge 3$.
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Answer. $2\pi$ cm$^2$ s$^{-1}$
- $A = \pi r^2$, so $\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$.
- $2\pi \times 5 \times 0.2 = 2\pi$.
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Answer. There is a maximum at $\left(1, \dfrac{1}{e}\right)$.
- $\dfrac{dy}{dx} = e^{-x}(1 - x)$.
- The stationary point is at $x = 1$, where $y = \dfrac{1}{e}$.
- The second derivative is $e^{-x}(x - 2)$, which is negative at $x = 1$.
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Answer. $\dfrac{3}{4\pi}$ cm s$^{-1}$
- $V = \dfrac{4}{3}\pi r^3$, so $\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}$.
- $12 = 4\pi \times 4 \times \dfrac{dr}{dt}$.
- $\dfrac{dr}{dt} = \dfrac{3}{4\pi}$.
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Answer. The maximum area is $100$ cm$^2$, when the rectangle is a square of side $10$ cm.
- If one side is $x$, the adjacent side is $20 - x$.
- $A = 20x - x^2$, so $\dfrac{dA}{dx} = 20 - 2x = 0$ when $x = 10$.
- The second derivative is $-2$, so this is a maximum, and the area is $100$.
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Answer. $x = 2$, and the maximum volume is $128$ cm$^3$.
- $V = x(12 - 2x)^2$.
- $\dfrac{dV}{dx} = (12 - 2x)(12 - 6x)$.
- The relevant solution is $x = 2$, since $x = 6$ gives a box of height with base zero.
- The volume is $2 \times 8^2 = 128$.
- The derivative changes from positive to negative at $x = 2$, so the volume is a maximum.