1. 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)

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

$
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}$
$=\frac{\pi}{3}
$

RonL

3. 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
$
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)}$
$=1+\cos(x)
$

Hence:

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

RonL

4. 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 $y \cos(x)$.

Alternativly:

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

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

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

so:

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

or:

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

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

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

RonL