A-level Maths revision · Pure, year 2

Implicit and parametric gradients

The idea

Implicit: differentiate every term with respect to xx, and multiply each yy term by dydx\frac{dy}{dx} (chain rule). Parametric: dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

  1. Implicit: differentiate term by term; y2→2ydydxy^2 \to 2y\frac{dy}{dx}, xy→y+xdydxxy \to y + x\frac{dy}{dx}.
  2. Collect the dydx\frac{dy}{dx} terms and factorise.
  3. Parametric: find dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}, then divide.

Watch out yy depends on xx: ddx(y2)\frac{d}{dx}(y^2) is 2ydydx2y\frac{dy}{dx}, not 2y2y.

See it worked

Worked example

Find the gradient of x2+y2=25x^2 + y^2 = 25 at (3,4)(3, 4).

  1. 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0.
  2. dydx=−xy\frac{dy}{dx} = -\frac{x}{y}.
  3. At (3,4)(3, 4): −34-\frac{3}{4}.

Answer −34-\frac{3}{4}

Your turn

Question 1Non-calculator3 marks

A curve has x=t2x = t^2, y=2ty = 2t. Find dydx\frac{dy}{dx} when t=2t = 2.

Question 2Non-calculator3 marks

Find dydx\frac{dy}{dx} at (1,2)(1, 2) on the curve x2+xy+y2=7x^2 + xy + y^2 = 7.

Question 3Non-calculator3 marks

A curve has x=3cos⁡tx = 3\cos t, y=3sin⁡ty = 3\sin t. Find its Cartesian equation.

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