Calculus Problems Multiple Choice

• Apr 19th 2006, 09:08 PM
Nimmy
Calculus Problems Multiple Choice
22. The base of a solid is the region enclosed by the ellipse 4x^2+y^2=1. If all plane cross sections prependicular to the x-axis are semicircles, the its volume is

(A) Pi/3
(B) Pi/6
(C) Pi/2
(D) 2Pi/3
(E) Pi/4

23. If F(x) = (Sinx)^2/(1-cosx) then F'(x) =

(A) cos x
(B) Sin x
(C) -Sin x
(D) -cos x
(E) 2cosx

24. If dy/dx= y cos x and y=3 when x=0 then y=

(A) e^(sinx)+2
(B) e^(sinx) +3
(C) sinx +3
(D) sin x +3e^x
(E) 3e^(sin x)
• Apr 20th 2006, 01:41 AM
CaptainBlack
Quote:

Originally Posted by Nimmy
22. The base of a solid is the region enclosed by the ellipse 4x^2+y^2=1. If all plane cross sections prependicular to the x-axis are semicircles, the its volume is

(A) Pi/3
(B) Pi/6
(C) Pi/2
(D) 2Pi/3
(E) Pi/4

The volume is the integral over the range of $\displaystyle x$ that correspond to points
on the ellipse $\displaystyle (x \in [-1/2,1/2])$ of the areas of semi-circles of radius y. That is:

$\displaystyle V=\int_{-1/2}^{1/2} \frac{1}{2} \pi y^2\ dx=\int_{-1/2}^{1/2} \frac{1}{2} \pi (1-4x^2)\ dx=\frac{1}{2} \pi \left[ x - \frac{4}{3}r^3\right]_{-1/2}^{1/2}$$\displaystyle =\frac{\pi}{3} RonL • Apr 20th 2006, 01:45 AM CaptainBlack Quote: Originally Posted by Nimmy 23. If F(x) = (Sinx)^2/(1-cosx) then F'(x) = (A) cos x (B) Sin x (C) -Sin x (D) -cos x (E) 2cosx \displaystyle F(x)=\frac{(\sin(x))^2}{1-\cos(x)}=\frac{1-(\cos(x))^2}{1-\cos(x)}=\frac{(1-\cos(x))(1+\cos(x))}{1-\cos(x)}$$\displaystyle =1+\cos(x)$

Hence:

$\displaystyle F'(x)=-\sin(x)$

RonL
• Apr 20th 2006, 01:54 AM
CaptainBlack
Quote:

Originally Posted by Nimmy

24. If dy/dx= y cos x and y=3 when x=0 then y=

(A) e^(sinx)+2
(B) e^(sinx) +3
(C) sinx +3
(D) sin x +3e^x
(E) 3e^(sin x)

You can confirm that (E) is correct by checking the initial condition and
differentiating and confirming that its derivative is $\displaystyle y \cos(x)$.

Alternativly:

$\displaystyle \frac{dy}{dx}= y \cos(x)$

is a ODE of seperable type, so seperating the and variables integrating gives:

$\displaystyle \int\frac{1}{y}\ dy=\int \cos(x)\ dx+C$,

so:

$\displaystyle \ln(y)=\sin(x)+C$,

or:

$\displaystyle y=\rm e^{\sin(x)+C}=K\ \rm e^{\sin(x)}$

Imposing the initial condition $\displaystyle y=3$ when $\displaystyle x=0$ gives:

$\displaystyle y=\rm e^{\sin(x)+C}=3\ \rm e^{\sin(x)}$

RonL