1. ## Antiderivative Word Problems

1) A ball is shot straight up in the air at the velocity of 42ft/sec. Assuming the air resistance can be ignored, how high does it go?

2) f"(x) = -4sin(2x), f ' (0)= -4, f(0) = -6
Find f(pi/5)?

For this problem I keep getting sin2x- 4x -6, then solving: I'm not sure what I'm doing wrong, can you show me step by step for this one? Thanks.

2. Originally Posted by helppalusmath
1) A ball is shot straight up in the air at the velocity of 42ft/sec. Assuming the air resistance can be ignored, how high does it go?
I'm going to make this longer than it needs to be, so feel free to skim.

Remember that velocity generally speaking is the derivative of displacement, and that acceleration is the derivative of velocity. There's probably some more information they gave you in your textbook or wherever you found this problem concerning the downward acceleration due to gravity. Someone correct me if my constants are wrong, but I believe gravity exerts an acceleration of $\displaystyle 16 \frac{ft}{sec^2}$. So let's take a look at the different functions laid out (Assuming positive indicates upward direction relative to ground):

Acceleration:
$\displaystyle A(x) = -32 \frac{ft}{sec^2}$
We can integrate this to find velocity:
$\displaystyle V(x) = -32x + C \frac{ft}{sec}$
You already have been given the initial value of velocity, 42ft/sec:
$\displaystyle V(x) = -32x + 42 \frac{ft}{sec}$

The ball will achieve maximum height when the initial velocity is finally canceled out by the downward acceleration due to gravity. Simply find when it's net velocity is zero to find the time of maximum height, and then plug that value into your formula for position (which can be found by integrating the velocity function and plugging in your initial height in for C, which should be zero).

3. Originally Posted by helppalusmath
2) f"(x) = -4sin(2x), f ' (0)= -4, f(0) = -6
Find f(pi/5)?

For this problem I keep getting sin2x- 4x -6, then solving: I'm not sure what I'm doing wrong, can you show me step by step for this one? Thanks.
$\displaystyle f''(x) = -4sin(2x)$
(Integrate & solve for C)
$\displaystyle f'(x) = 2cos(2x) + C$
$\displaystyle (-4) = 2(1) + C$
$\displaystyle f'(x) = 2cos(2x) - 6$
(Integrate & solve for C)
$\displaystyle f(x) = sin(2x) - 6x + C$
$\displaystyle (-6) = (0) - 6(0) + C$
$\displaystyle f(x) = sin(2x) - 6x - 6$

You probably made an error evaluating the constant for $\displaystyle f'(x)$, which threw you off the trail for the rest of the problem, I believe. Evaluating the function should be a piece of cake after this.