Answer to Question #334610 in Calculus for peac_eboy

Question

1. The demand for a product, in dollars, is P = 2000 - 0.2x - 0.01x²

Find the consumer surplus when the sales level is 250.

2.1. Find the area of the region that lies inside the circle r = 3sin(x)and

outside the cardioid r = 1+ sin(x).

2.2 Find the length of the cardioid r = 1+ sin(x)

Answer 1

1. Since the number of products sold is $x=250,$ the coresponding price is

$$P=2000-0.2(250)-0.01(250)^2=1325$$

The consumer surplus is as follows

$$\displaystyle\int_{0}^{250}(p(x)-P)dx$$

$$=\displaystyle\int_{0}^{250}(2000-0.2x-0.01x^2-1325)dx$$

$$=[675x-0.1x^2-\dfrac{0.01x^3}{3}]\begin{matrix} 250 \\ 0 \end{matrix}$$

$$=675(250)-0.1(250)^2-\dfrac{0.01(250)^3}{3}$$

$$=\$110416.67$$

2.1

The cardioid intersects with the circle

$$3\sin\theta=1+\sin \theta$$

$$\sin \theta=\dfrac{1}{2}$$

The cardioid intersects with the circle at $(\dfrac{3}{2},\dfrac{\pi}{6}),(\dfrac{3}{2},\dfrac{5\pi}{6})$ and the pole.

The area of interest has been shaded above.

To find the area of a polar curve, we use

$$A=\dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}((3\sin \theta)^2-(1+\sin \theta)^2)d\theta$$

$$=\dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}(8\sin^2 \theta-2\sin\theta-1)d\theta$$

$$=\dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}(3-4\cos2 \theta-2\sin\theta)d\theta$$

$$=\dfrac{1}{2}[3\theta-2\sin2\theta+2\cos \theta]\begin{matrix} 5\pi/6\\ \pi/6 \end{matrix}$$

$$=\dfrac{1}{2}(\dfrac{5\pi}{2}+\sqrt{3}-\sqrt{3}-\dfrac{\pi}{2}+\sqrt{3}-\sqrt{3})$$

$$=\pi$$

$\pi$ square units.

2. 2

$$r(x)=1+\sin x, r'(x)=\cos x$$

$$(r)^2+(r')^2=(1+\sin x)^2+(\cos x)^2$$

$$=2+2\sin x$$

$$L=\displaystyle\int_{-\pi/2}^{3\pi/2}\sqrt{(r(x))^2+(r'(x))^2}dx$$

$$=\displaystyle\int_{-\pi/2}^{3\pi/2}\sqrt{2+2\sin x}dx$$

$$=2\displaystyle\int_{-\pi/2}^{3\pi/2}|\cos(\dfrac{x}{2}-\dfrac{\pi}{4})|dx$$

$$=4\bigg[\sin(\dfrac{x}{2}-\dfrac{\pi}{4})\bigg]\begin{matrix} 3\pi/2 \\ -\pi/2 \end{matrix}=4(1+1)=8(units)$$

8 units