Question

Ap Calculus BC 2002 Form B Question 3?

Answer 1

a) We must start by finding the intersection points of the two curves.

`3/4x = 4x - x^3 + 1`

Solve using a graphing calculator to get

`x = 1.940`

Thus our bounds of integration will be from `x = 0` to `x = 1.940`. Therefore, letting `a = 1.940`, we get

`I = int_0^a 4x - x^3 + 1 - 3/4x dx ~~ 4.515`

Thus the area will be `4.515` square units.

b) Recall the formula for volume around the x-axis:

`V = pi int_b^c (f(x))^2 - (g(x))^2 dx`

Where `f(x)` is the upper function and `g(x)` the lower. Thus in our case

`V = pi int_0^a (4x -x^3 + 1)^2 - (3/4x)^2 dx`

Once again using a calculator to evaluate we get

`V = 57.463` cubic units

c) The perimeter is given by adding the arc length of the linear function on `[0, 1.940]`, the arc length of the cubic function on `[0, 1.940]` and `y_2(0) - y_1(0) = 4(0) - 0^3 + 1 - 3/2(0) = 1`. We must recall the arc length formula:

`A = int_b^c sqrt(1 + (dy/dx)^2)dx`

`P = int_0^a sqrt(1 + (4x - 3x^2)^2)dx + int_0^a sqrt(1 + (3/2)^2) dx + 1`

`P = 7.528` units

The last couple of steps would have not been required on the exam because it states NOT TO EVALUATE the arc length.

Hopefully this helps!