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Math Help - Calculus Problems Multiple Choice

  1. #1
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    Exclamation 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)
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  2. #2
    Grand Panjandrum
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    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 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:

    <br />
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}<br />

    RonL
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  3. #3
    Grand Panjandrum
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    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
    <br />
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)<br />

    Hence:

    <br />
F'(x)=-\sin(x)<br />

    RonL
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  4. #4
    Grand Panjandrum
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    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 y \cos(x).

    Alternativly:

    <br />
\frac{dy}{dx}= y \cos(x)<br />

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

    <br />
\int\frac{1}{y}\ dy=\int \cos(x)\ dx+C<br />
,

    so:

    <br />
\ln(y)=\sin(x)+C<br />
,

    or:

    <br />
y=\rm e^{\sin(x)+C}=K\ \rm e^{\sin(x)}<br />

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

    <br />
y=\rm e^{\sin(x)+C}=3\ \rm e^{\sin(x)}<br />

    RonL
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