i really need help with these problems?

Question

1. The position of a ball thrown up from the top of a building 150 ft tall with an initial velocity of 20 ft/sec is given by the function s(t) = 16t² + 20t + 150. What is the maximum height of the ball?

2. The expression [1/ secx-1 ] - [1/ secx + 1] may be simplified to... either 2 cot² x or tan² x. i'm not sure which one.

3. The smallest interval of theta that will produce a complete graph of the polar equation r= -3cos2theta is..... my answer is [0, pi] is that right?

thanks!

Answer 1

1. Maximum height is at the turning point where the ball is neither moving up or down. Thus, find the distance when velocity is zero.
It is straight forward if you can easily find an equation relating displacement with velocity.
Otherwise, differentiating the given function gives the relation between velocity and time, and then find the time at zero velocity, and then find the position at this time.

2. Just plug in a few arbitrary values for x in the expressions, the two that agree for these values are very very likely to be equivalent.

3. This one a bit less straight forward. Your answer is true for y = 3 cos2x on a Cartesian plane. However, on polar coordinates, I get [0, 2pi] from visualizing it in the air.

Answer 2

2. 1/(sec(x) - 1) - 1/(sec(x) + 1) =
((sec(x) + 1) - (sec(x) - 1))/(sec^2(x) - 1) =
2/(tan^2(x) = 2*cot^2(x) .

3. You are correct, [0,pi] .