Question

Find an equation for the line tangent to the curve at the point defined by the given value of t.​ Also, find the value of d^2y / dx^2 at this point.
x = 3t^2 + 7, y = t^6, t = -1

Answer 1

The tangent line has slope $\frac{\mathrm dy}{\mathrm dx}$, which we can find using the chain rule:

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}$

We have

$x(t)=3t^2+7\implies\dfrac{\mathrm dx}{\mathrm dt}=6t$

$y(t)=t^6\implies\dfrac{\mathrm dy}{\mathrm dt}=6t^5$

so that

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{6t^5}{6t}=t^4$

When $t=-1$, the tangent line has slope 1.

To compute the second derivative $\frac{\mathrm d^2y}{\mathrm dx^2}$, first notice that the first derivative $\frac{\mathrm dy}{\mathrm dx}$ turns out to be a function of $t$. Denote this first derivative by $f(t)$. Then by the chain rule, we find

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\dfrac{4t^3}{6t}=\dfrac23t^2$

and at $t=1$, we get a value of 2/3.