Ap Calculus BC 2002 Form B Question 3?

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Ap Calculus BC 2002 Form B Question 3?

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Answer 1 · 2018-04-29T05:07:21

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

$\frac{3}{4} x = 4 x - x^{3} + 1$ $3/4x = 4x - x^3 + 1$

Solve using a graphing calculator to get

$x = 1.940$ $x = 1.940$

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

$I = \int_{0}^{a} 4 x - x^{3} + 1 - \frac{3}{4} x d x \approx 4.515$ $I = int_0^a 4x - x^3 + 1 - 3/4x dx ~~ 4.515$

Thus the area will be $4.515$ $4.515$ square units.

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

$V = π \int_{b}^{c} {(f (x))}^{2} - {(g (x))}^{2} d x$ $V = pi int_b^c (f(x))^2 - (g(x))^2 dx$

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

$V = π \int_{0}^{a} {(4 x - x^{3} + 1)}^{2} - {(\frac{3}{4} x)}^{2} d x$ $V = pi int_0^a (4x -x^3 + 1)^2 - (3/4x)^2 dx$

Once again using a calculator to evaluate we get

$V = 57.463$ $V = 57.463$ cubic units

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

$A = \int_{b}^{c} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ $A = int_b^c sqrt(1 + (dy/dx)^2)dx$

$P = \int_{0}^{a} \sqrt{1 + {(4 x - 3 x^{2})}^{2}} d x + \int_{0}^{a} \sqrt{1 + {(\frac{3}{2})}^{2}} d x + 1$ $P = int_0^a sqrt(1 + (4x - 3x^2)^2)dx + int_0^a sqrt(1 + (3/2)^2) dx + 1$

$P = 7.528$ $P = 7.528$ 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!

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Ap Calculus BC 2002 Form B Question 3?

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