How do you find the equations of the tangent and normal of the curve at x=t^2, y=t+3, t=1?

Jun 24, 2017

The Tangent equation is:

$y = \frac{1}{2} x + \frac{7}{2}$

The Normal equation is:

$y = - 2 x + 6$

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is $- 1$).

We have:

$x = {t}^{2}$
$y = t + 3$

Firstly, let us find the coordinates where $t = 1$:

$\implies x = 1$, $y = 4$ ie $\left(1 , 4\right)$

Then differentiating wrt $t$ we have:

$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}} = \frac{1}{2 t}$

So at the parametric coordinate $t = 1$, we have;

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2}$

So the tangent passes through $\left(1 , 4\right)$ and has gradient ${m}_{T} = \frac{1}{2}$, so using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the tangent equation we seek is;

$y - 4 = \frac{1}{2} \left(x - 1\right)$
$\therefore y - 4 = \frac{1}{2} x - \frac{1}{2}$
$\therefore y = \frac{1}{2} x + \frac{7}{2}$

Similarly, the normal passes through $\left(1 , 4\right)$ and has gradient ${m}_{N} = - 2$, so using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the normal equation we seek is;

$y - 4 = - 2 \left(x - 1\right)$
$\therefore y - 4 = - 2 x + 2$
$\therefore y = - 2 x + 6$

We can verify this graphically: