Answer to Question #334610 in Calculus for peac_eboy

Question #334610

1. The demand for a product, in dollars, is = 2000 - 0.2- 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)


1
Expert's answer
2022-05-04T14:51:26-0400

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


P=20000.2(250)0.01(250)2=1325P=2000-0.2(250)-0.01(250)^2=1325

The consumer surplus is as follows


0250(p(x)P)dx\displaystyle\int_{0}^{250}(p(x)-P)dx

=0250(20000.2x0.01x21325)dx=\displaystyle\int_{0}^{250}(2000-0.2x-0.01x^2-1325)dx

=[675x0.1x20.01x33]2500=[675x-0.1x^2-\dfrac{0.01x^3}{3}]\begin{matrix} 250 \\ 0 \end{matrix}

=675(250)0.1(250)20.01(250)33=675(250)-0.1(250)^2-\dfrac{0.01(250)^3}{3}

=$110416.67=\$110416.67

2.1


The cardioid intersects with the circle


3sinθ=1+sinθ3\sin\theta=1+\sin \theta

sinθ=12\sin \theta=\dfrac{1}{2}

The cardioid intersects with the circle at (32,π6),(32,5π6)(\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=12π/65π/6((3sinθ)2(1+sinθ)2)dθA=\dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}((3\sin \theta)^2-(1+\sin \theta)^2)d\theta

=12π/65π/6(8sin2θ2sinθ1)dθ=\dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}(8\sin^2 \theta-2\sin\theta-1)d\theta

=12π/65π/6(34cos2θ2sinθ)dθ=\dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}(3-4\cos2 \theta-2\sin\theta)d\theta

=12[3θ2sin2θ+2cosθ]5π/6π/6=\dfrac{1}{2}[3\theta-2\sin2\theta+2\cos \theta]\begin{matrix} 5\pi/6\\ \pi/6 \end{matrix}

=12(5π2+33π2+33)=\dfrac{1}{2}(\dfrac{5\pi}{2}+\sqrt{3}-\sqrt{3}-\dfrac{\pi}{2}+\sqrt{3}-\sqrt{3})

=π=\pi

π\pi square units.


2. 2

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

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

=2+2sinx=2+2\sin x

L=π/23π/2(r(x))2+(r(x))2dxL=\displaystyle\int_{-\pi/2}^{3\pi/2}\sqrt{(r(x))^2+(r'(x))^2}dx

=π/23π/22+2sinxdx=\displaystyle\int_{-\pi/2}^{3\pi/2}\sqrt{2+2\sin x}dx

=2π/23π/2cos(x2π4)dx=2\displaystyle\int_{-\pi/2}^{3\pi/2}|\cos(\dfrac{x}{2}-\dfrac{\pi}{4})|dx

=4[sin(x2π4)]3π/2π/2=4(1+1)=8(units)=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


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