Parametric Differentiation

The gradient is still $\frac{dy}{dx}$dy/dx

On a parametric curve, x and y are both given in terms of a parameter t. The gradient at a point still means what it always means: $\frac{dy}{dx}$dy/dx, how fast y changes compared with x.

What is different is that neither equation says y in terms of x. Instead each says how its coordinate changes with t. Differentiating gives two rates, $\frac{dx}{dt}$dx/dt and $\frac{dy}{dt}$dy/dt, and the gradient has to be built from them.

Divide the two rates

The chain rule gives $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$dy/dx = dy/dt × dt/dx. The second factor is the rate of t compared with x, which is the reciprocal of the rate of x compared with t: $\frac{dt}{dx}=1/\left(\frac{dx}{dt}\right)$dt/dx = 1 / (dx/dt). So $\frac{dy}{dx}=\left(\frac{dy}{dt}\right)/\left(\frac{dx}{dt}\right)$dy/dx = (dy/dt) / (dx/dt).

Picture a small step in t. The point moves a small distance across, about $\left(\frac{dx}{dt}\right)$(dx/dt) times the step, and a small distance up, about $\left(\frac{dy}{dt}\right)$(dy/dt) times the step. The gradient is the distance up divided by the distance across, and the step cancels.

The rule needs $\frac{dx}{dt}\ne 0$dx/dt ≠ 0. Where $\frac{dx}{dt}=0$dx/dt = 0 and $\frac{dy}{dt}$dy/dt is not 0, the point is moving straight up or down for that instant, and the tangent is vertical.

An example

Take $x=t^2$x = t² and y = 2t. Differentiate each with respect to t: $\frac{dx}{dt}=2t$dx/dt = 2t and $\frac{dy}{dt}=2$dy/dt = 2. Divide: $\frac{dy}{dx}=2/2t=\frac{1}{t}$dy/dx = 2 / 2t = 1/t.

At t = 2 the point is (4, 4) and the gradient is $\frac{1}{2}$1/2. The tangent through (4, 4) with gradient $\frac{1}{2}$1/2 is $y-4=\frac{x-4}{2}$y − 4 = (x − 4)/2, which is $y=\frac{x}{2}+2$y = x/2 + 2.

Check it numerically. At t = 2.001 the point is (4.004001, 4.002). From t = 2 to t = 2.001 the point rises 0.002 and moves across 0.004001, and 0.002 ÷ 0.004001 = 0.499875, very nearly $\frac{1}{2}$1/2.

It agrees with the Cartesian equation too. Eliminating t gave $x=\frac{y^2}{4}$x = y²/4, so the upper half is $y=2\sqrt{x}$y = 2√x, whose gradient $\frac{1}{\sqrt{x}}$1/√x is $\frac{1}{2}$1/2 at x = 4.

One x, two gradients

The curve passes x = 4 twice: at t = 2, the point (4, 4), and at t = −2, the point (4, −4). The rule gives a gradient for each value of t, so the two points get different gradients: $\frac{1}{2}$1/2 at t = 2 and $\frac{1}{-2}=-\frac{1}{2}$1/(−2) = −1/2 at t = −2. A single formula in x could not give both.

At t = 0, $\frac{dx}{dt}=2\times 0=0$dx/dt = 2 × 0 = 0 while $\frac{dy}{dt}=2$dy/dt = 2. The rule $\frac{1}{t}$1/t has no value there, and the curve passes through the origin moving straight up: its tangent there is the y-axis.

The curve $x=t^2$x = t², y = 2t. The gold tangent at t = 2 has gradient $\frac{1}{2}$1/2 and the plain one at t = −2 has gradient $-\frac{1}{2}$−1/2. At the origin, where t = 0, the curve stands upright along the y-axis.

Two more curves

Take $x=t^2$x = t² and $y=t^3$y = t³. Then $\frac{dx}{dt}=2t$dx/dt = 2t and $\frac{dy}{dt}=3t^2$dy/dt = 3t², so $\frac{dy}{dx}=3t^2/2t=\frac{3t}{2}$dy/dx = 3t² / 2t = 3t/2. At t = 2 the point is (4, 8) and the gradient is 3, so the tangent is y − 8 = 3(x − 4), which is y = 3x − 4. Check: at t = 2.001, $\left(2.001^3-8\right)/\left(2.001^2-4\right)=3.00075$(2.001³ − 8) / (2.001² − 4) = 3.00075. Eliminating t agrees as well: $y=x^{\frac{3}{2}}$y = x^(3/2) has gradient $\left(\frac{3}{2}\right)\sqrt{x}$(3/2)√x, which is 3 at x = 4.

Take the circle x = 3 cos t, y = 3 sin t. Then $\frac{dx}{dt}=-3\sin t$dx/dt = −3 sin t and $\frac{dy}{dt}=3\cos t$dy/dt = 3 cos t, so $\frac{dy}{dx}=-\cos t/\sin t$dy/dx = −cos t / sin t. At $t=\frac{\pi }{4}$t = π/4 the point is $\left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)$(3/√2, 3/√2) and the gradient is −1: the tangent runs at right angles to the radius, as a tangent to a circle must.

At t = 0 and at $t=\pi$t = π, sin t = 0, so $\frac{dx}{dt}=0$dx/dt = 0 while $\frac{dy}{dt}=\pm 3$dy/dt = ±3. The tangents at (3, 0) and (−3, 0) are vertical. At $t=\frac{\pi }{2}$t = π/2, cos t = 0, so $\frac{dy}{dt}=0$dy/dt = 0 and the tangent at (0, 3) is level.

The circle x = 3 cos t, y = 3 sin t. The gold tangent at $t=\frac{\pi }{4}$t = π/4 has gradient −1. At t = 0, $\frac{dx}{dt}=0$dx/dt = 0, and the tangent through (3, 0) is the vertical line x = 3.

The usual mistakes

Dividing the wrong way up. For x = 3t, $y=t^2$y = t², $\frac{dy}{dt}=2t$dy/dt = 2t and $\frac{dx}{dt}=3$dx/dt = 3, so $\frac{dy}{dx}=\frac{2t}{3}$dy/dx = 2t/3. The upside-down $\frac{3}{2t}$3/(2t) is $\frac{dx}{dy}$dx/dy, the run for each unit of rise.

Multiplying the rates. 2t × 3 = 6t is not a gradient; the two time-derivatives divide, so that dt cancels.

Taking $\frac{dy}{dt}$dy/dt as the gradient. $\frac{dy}{dt}$dy/dt says how fast y changes with time. For $x=t^2$x = t², y = 4t at t = 1, $\frac{dy}{dt}=4$dy/dt = 4, but the gradient is 4 ÷ 2 = 2.

A fairground carriage

In the application below, a carriage on a curved rail is at x = 3m and $y=2m^2$y = 2m², both in meters, after m seconds. Its two rates give its speed by Pythagoras' theorem and the gradient of the rail by dividing one by the other.