If x=3 t and y=3 t, then the value of d^2ydx^2 at t=4, is:

Question

If $x=3	an t$ and $y=3\sec t$, then the value of $\dfrac{d^2y}{dx^2}$ at $t=\dfrac\pi4$, is:

Accepted Answer

$\dfrac{dx}{dt}=3\sec^2t$, $\dfrac{dy}{dt}=3\sec t\tan t$, so $\dfrac{dy}{dx}=\sin t$.

$$\frac{d^2y}{dx^2}=\cos t\cdot\frac{1}{3\sec^2t}=\frac{\cos^3t}{3}$$

At $t=\dfrac\pi4$: $\cos\dfrac\pi4=\dfrac1{\sqrt2}$, so $\dfrac{d^2y}{dx^2}=\dfrac{1}{3\cdot2\sqrt2}=\dfrac{1}{6\sqrt2}$.

Solutions · 1