# Couple of Calculus Problems

• Apr 18th 2010, 05:01 PM
Naples
Couple of Calculus Problems
1. If $\displaystyle x = sin(t)$ and $\displaystyle y = cos^2(t)$ , then $\displaystyle d^2y/dx^2$ at $\displaystyle t = pi/2$ is?

I got
$\displaystyle dy/dx = cos(t)/-2cos(t)sin(t)$
$\displaystyle dy/dx = -1/2sin(t)$
$\displaystyle d^2y/dx^2 = -2cos(t)/(2sin(t))^2$

at $\displaystyle t = pi/2$
$\displaystyle d^2y/dx^2 = -2cos(pi/2)/(2sin(pi/2))^2$
$\displaystyle =0/4=0$

However, the correct answer is -2

2. If $\displaystyle f(x) = |(x^2-12)(x^2+4)|$, how many numbers in the interval [-2,3] satisfy the conclusion of the Mean Value Theorem?

f is always negative on the interval so I got rid of the absolute value signs and multiplied it by -1.
Then I differentiated and got:
$\displaystyle 16x-4x^3 = f(3)-f(-2)/3+2$
$\displaystyle 16x-4x^3 = 39-64/5$
$\displaystyle 4x^3-16x+5=0$
x=.321, -2.141, 1.820

interval is [-2,3] so c=.321, 1.820

The thing is the answer is three numbers...

Any help is greatly appreciated
• Apr 18th 2010, 06:26 PM
TKHunny
$\displaystyle \frac{dy}{dt}\;=\;\frac{dy}{dx}\cdot \frac{dx}{dt}\;\implies\;\frac{dy}{dx}\;=\;\frac{\ frac{dy}{dt}}{\frac{dx}{dt}}$

$\displaystyle \frac{dx}{dt}\;=\;\cos(t)$

$\displaystyle \frac{dy}{dt}\;=\;-2\sin(t)\cos(t)\;=\;-sin(2t)$

$\displaystyle \frac{dy}{dx}\;=\;\frac{-2\sin(t)\cos(t)}{\cos(t)}\;=\;-2\sin(t)$

You seem to be upside down. Please examine the formulation more carefully and give it another go.

$\displaystyle \frac{d^{2}y}{dx^{2}}\;=\;\frac{d}{dx}\cdot \frac{dy}{dx}\;=\;\frac{d}{dt}\frac{dy}{dx}\cdot \frac{dt}{dx}$

Let's see what you get.
• Apr 18th 2010, 06:35 PM
Naples
Quote:

Originally Posted by TKHunny
$\displaystyle \frac{dy}{dt}\;=\;\frac{dy}{dx}\cdot \frac{dx}{dt}\;\implies\;\frac{dy}{dx}\;=\;\frac{\ frac{dy}{dt}}{\frac{dx}{dt}}$

$\displaystyle \frac{dx}{dt}\;=\;\cos(t)$

$\displaystyle \frac{dy}{dt}\;=\;-2\sin(t)\cos(t)\;=\;-sin(2t)$

$\displaystyle \frac{dy}{dx}\;=\;\frac{-2\sin(t)\cos(t)}{\cos(t)}\;=\;-2\sin(t)$

You seem to be upside down. Please examine the formulation more carefully and give it another go.

$\displaystyle \frac{d^{2}y}{dx^{2}}\;=\;\frac{d}{dx}\cdot \frac{dy}{dx}\;=\;\frac{d}{dt}\frac{dy}{dx}\cdot \frac{dt}{dx}$

Let's see what you get.

Wow, I feel kind of dumb for making such a stupid error, thanks for catching it though.