Cracking The Ap Calculus ab Exam 2018

Integrate the expression: h = −16t^2 + 64t + C. Now solve for the constant by plugging in t = 0 and h = 32.

32  =   −16 (0^2 )  +   64(0)   +   C   and C   =   32

Thus, the equation for height is h = −16t^2 + 64t + 32.

Part C: In order to find the maximum height, you need to take the derivative of the height with respect to
time and set it equal to zero. Notice that the derivative of height with respect to time is the velocity; just
set the velocity equal to zero and solve for t.

−32t    +   64  =   0,  so  t   =   2

Thus, at time t = 2, the height of the particle is a maximum. Now, plug t = 2 into the equation for height.

h   =   −16(2)^2    +   64(2)   +   32  =   96

Therefore, the maximum height of the particle is 96 feet.

Here are some solved problems. Do each problem, covering the answer first, then check your answer.

PROBLEM 1. If and y(0) = 10, find an equation for y in terms of x.

Answer: First, separate the variables.

2 y dy  =   3x  dx

Then, we take the integral of both sides.

∫^2 y dy = ∫^3 x dx

Next, integrate both sides.

y^2     =       +   C

Finally, solve for the constant.

102     =       +   C,  so  C   =   100

The solution is y^2 = + 100.

PROBLEM 2. If = 4xy^2 and y(0) = 1, find an equation for y in terms of x.