definite integral calc problem

Question

An object moves up and down with initial height and velocity of

h(0) = 15 in dh/dt= −12sin(2 t) in/s

What is Δh on the interval [0,2]? (Be accurate to at least two decimal places.)

______________

What is the first instant (t > 0) that this object reaches a height of 8 inches?

(Be accurate to at least three decimal places.)

________________

Answer 1

First part:

To find Δh (the change in height) over the interval [0,2], we simply integrate dh/dt over that interval. So:

∫dh = ∫₀² - 12sin(2t) dt

Making this a bit easier to solve:

Δh = -12 ∫₀² sin(2t) dt

We can see here that I need to use u-substitution so I will set u equal to 2t:

u = 2t

du = 2 dt

du/2 = dt

So plugging back in, I plug u in for 2t and du/2 for dt. Don't forget to change the bounds on the integral (in this case the lower bound stays 0, but the upper bound becomes 2*2=4).

Δh = -12 ∫₀⁴ sin(u) ⋅ du/2

I'll move the 1/2 outside the integral:

Δh = -6 ∫₀⁴ sin(u) du

Finally Δh = -6[-cos(u)]₀⁴ = 6(cos(4) - cos(0)) = -9.92

Second part:

In this situation we no longer want Δh, but we want the function h(t). To find this, we just take the antiderivative of dh/dt, just like before:

h(t) = ∫dh/dt = ∫-12sin(2t)dt

It's the same integral as before, except this time there are no bounds and I have to add + C to the end:

h(t) = 6cos(2t) + C (remember that u = 2t, so I can just plug that back in)

Now I just have to calculate C. The initial problem tells us that h(0) = 15, so I plug in 0 for t and 15 for h(t):

15 = 6 cos(0) + C

15 = 6 + C

C = 9

So my final function is:

h(t) = 6 cos(2t) + 9

To find the first instant where height, meaning h(t), equals 8, I set the equation equal to 8:

8 = 6 cos(2t) + 9

I'm going to assume since this problem allows you to use a calculator, that you can just use the numeric solver function on your graphing calculator to solve for t (there are several solutions to the above equation so set your guess to 0 to find the first possible instant):

t = 0.869