Can we find the equation of the normal to the curve when $t = 2$?

Can we find the equation of the normal to the curve when $t = 2$?

Question

A curve is represented parametrically by the equations

$$\begin{equation*} x = (3-2t)^2, \quad y = t^2 - 2t. \end{equation*}$$

Find

1. $dy/dx$ in terms of $t$,

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1. the co-ordinates of the stationary point of the curve,

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1. the equation of the normal at the point where $t = 2$.

Solution

We can write

$$\begin{equation*} \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}. \end{equation*}$$

We have that (by the

Chain Rule

)

$$\begin{equation*} \frac{dx}{dt} = 2(3-2t)(-2) = 4(2t-3), \quad \frac{dy}{dt} = 2t - 2 \end{equation*}$$

and therefore

$$\begin{equation*} \frac{dy}{dx} = \frac{2t - 2}{4(2t-3)} = \frac{t-1}{2(2t-3)}. \end{equation*}$$

The stationary point of the curve occurs when $\dfrac{dy}{dx} = 0$, i.e. when $t = 1$.

As $x = (3-2t)^2$ and $y = t^2 - 2t$, the co-ordinates of the stationary point are $(1,-1)$.

When

$t = 2$

, the gradient of the curve is

$$\begin{equation*} \frac{dy}{dx} = \frac{2-1}{2(2 \times 2 - 3)} = \frac{1}{2}. \end{equation*}$$

The gradient of the normal to the curve at this point is the negative of the reciprocal of this, i.e. $-2$.

Thus, the equation of the normal is $y = -2x + c$, for some constant $c$, where we require that the original curve crosses the normal at $t=2$.

When $t=2$, the curve has $x$-coordinate $(3 - 2 \times 2)^2 = 1$ and $y$-coordinate $2^2 - 2 \times 2 = 0$.

So, substituting these values into the equation for the normal, we require $c$ to be such that $0 = - 2 \times 1 + c$, i.e. $c = 2$.

Consequently, the equation of the normal is $y = -2x + 2$.