7.9 Implicit and parametric differentiation — Practise

Work through these on your own. For an implicit relation, differentiate both sides with respect to $x$ and collect the $\dfrac{dy}{dx}$ terms. For a parametric curve, remember that the gradient is $\dfrac{dy/dt}{dx/dt}$ .

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 25$ .
Find the gradient of $x^2 + y^2 = 100$ at the point $(6, 8)$ .
Find $\dfrac{dy}{dx}$ for $xy = 12$ .
Find $\dfrac{dy}{dx}$ for $x^2 + xy = 6$ .
Find $\dfrac{dy}{dx}$ for $x^2 + 3xy - y^2 = 5$ .
Find $\dfrac{dy}{dx}$ for $x^3 + y^3 = 9$ .
A curve is given by $x = t^2$ , $y = t^3$ . Find $\dfrac{dy}{dx}$ .
A curve is given by $x = t^2$ , $y = t^4$ . Find $\dfrac{dy}{dx}$ .
A curve is given by $x = \cos t$ , $y = \sin t$ . Find $\dfrac{dy}{dx}$ .
A curve is given by $x = 2t$ , $y = t^2$ . Find $\dfrac{dy}{dx}$ .
Find the gradient of the tangent to $x^2 + y^2 = 25$ at the point $(4, 3)$ .
A curve is given by $x = t + 1$ , $y = t^3$ . Find $\dfrac{dy}{dx}$ .

Answers and concise solutions

1. The derivative

Answer: Differentiate both sides: $2x + 2y\dfrac{dy}{dx} = 0$ . So $\dfrac{dy}{dx} = -\dfrac{x}{y}$ . The tangent is vertical where $y = 0$ .

2. The gradient

Answer: For this circle $\dfrac{dy}{dx} = -\dfrac{x}{y}$ . At $(6, 8)$ the gradient is $-\dfrac{6}{8} = -\dfrac{3}{4}$ . The tangent through $(6, 8)$ is $y = -\tfrac{3}{4}x + \tfrac{25}{2}$ .

3. The derivative

Answer: Differentiate $xy$ with the product rule: $y + x\dfrac{dy}{dx} = 0$ . So $\dfrac{dy}{dx} = -\dfrac{y}{x}$ . This matches the derivative of $y = \dfrac{12}{x}$ .

4. The derivative

Answer: Differentiate: $2x + y + x\dfrac{dy}{dx} = 0$ . Collect the derivative: $\dfrac{dy}{dx} = -\dfrac{2x + y}{x}$ .

5. The derivative

Answer: Differentiate term by term: $2x + 3\left(y + x\dfrac{dy}{dx}\right) - 2y\dfrac{dy}{dx} = 0$ . Collect the terms: $\dfrac{dy}{dx}(3x - 2y) = -(2x + 3y)$ , so $\dfrac{dy}{dx} = \dfrac{2x + 3y}{2y - 3x}$ .

6. The derivative

Answer: Differentiate: $3x^2 + 3y^2\dfrac{dy}{dx} = 0$ . So $\dfrac{dy}{dx} = -\dfrac{x^2}{y^2}$ .

7. The derivative

Answer: $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2$ , so $\dfrac{dy}{dx} = \dfrac{3t^2}{2t} = \dfrac{3t}{2}$ .

8. The derivative

Answer: $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 4t^3$ , so $\dfrac{dy}{dx} = \dfrac{4t^3}{2t} = 2t^2$ .

9. The derivative

Answer: $\dfrac{dx}{dt} = -\sin t$ and $\dfrac{dy}{dt} = \cos t$ , so $\dfrac{dy}{dx} = \dfrac{\cos t}{-\sin t} = -\cot t$ .

10. The derivative

Answer: $\dfrac{dx}{dt} = 2$ and $\dfrac{dy}{dt} = 2t$ , so $\dfrac{dy}{dx} = \dfrac{2t}{2} = t$ .

11. The gradient

Answer: For the circle $\dfrac{dy}{dx} = -\dfrac{x}{y}$ . At $(4, 3)$ the gradient is $-\dfrac{4}{3}$ , which is the negative reciprocal of the radius gradient $\dfrac{3}{4}$ .

12. The derivative

Answer: $\dfrac{dx}{dt} = 1$ and $\dfrac{dy}{dt} = 3t^2$ , so $\dfrac{dy}{dx} = \dfrac{3t^2}{1} = 3t^2$ .