# Trigonometry, Stationary Points and the Second Derivative

• Apr 7th 2008, 03:34 PM
Flay
Trigonometry, Stationary Points and the Second Derivative
I'm having trouble getting the answer to this particular question out;

Given that the $\displaystyle \frac{d^2y}{dx^2}$ is equal to $\displaystyle 9\sin3x$;
1. Find y if there is a stationary point at $\displaystyle (\frac{\pi}{2}, 1)$
2. Show that $\displaystyle \frac{d^2y}{dx^2} + 9y = 0$
• Apr 7th 2008, 04:00 PM
Mathstud28
What you have here
Quote:

Originally Posted by Flay
I'm having trouble getting the answer to this particular question out;

Given that the $\displaystyle \frac{d^2y}{dx^2}$ is equal to $\displaystyle 9\sin3x$;
1. Find y if there is a stationary point at $\displaystyle (\frac{\pi}{2}, 1)$
2. Show that $\displaystyle \frac{d^2y}{dx^2} + 9y = 0$

is a differntial equation...here is what you do...since $\displaystyle \frac{d^2y}{dx^2}=9sin(3x)$......ok now seperate to get $\displaystyle d^2y=9sin(3x)dx$ then integrate to get $\displaystyle y'=-3cos(3x)+C$...now stop before we get to y we must find that c...we know there is a stationary point at c which means the slope is 0 at $\displaystyle \frac{\pi}{2{$ which means that y'=0...so thefore $\displaystyle 0=-3cos\bigg(3\cdot\frac{\pi}{2}\bigg)+C$ solve and you see c=0...next integrate again to get y and you see that $\displaystyle y=-sin(3x)+C$ now using that condition again we know that $\displaystyle 1=-sin\bigg(3\cdot\frac{\pi}{2}\bigg)+C_1$ solving we get x=0..therfore $\displaystyle y=-sin(3x)$

for part two just add nine times what we just found to what $\displaystyle \frac{d^2y}{dx^2}$ is to verify its zero
• Apr 7th 2008, 08:11 PM
Flay
Got it. I was making yet another stupid mistake.