Load the attached. The title gibbers, adding "inininin" to the end of it
!
john
--
"Yeah, I woke up in the day accidentally once, the moon was on fire for some
reason and I couldn't see very well and all the bandwidth disappeared, it was
very scary :("
- Orion
#This file was created by <dlj0> Thu Nov 20 17:46:00 1997
#LyX 0.11 (C) 1995-1997 Matthias Ettrich and the LyX Team
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\layout Title
Math 23, Fall 1997, Exam # 2 Solutions
\layout Date
November 11, 1997
\layout Enumerate
Find the directional derivative of
\begin_inset Formula \( f(x,y)=x^{2}y+2y^{2} \)
\end_inset
at the point
\begin_inset Formula \( (1,3) \)
\end_inset
in the direction
\begin_inset Formula \( \mathbf{u}=\frac{1}{2}\mathbf{i}-\frac{\sqrt{3}}{2}\mathbf{j}
\)
\end_inset
.
\emph on
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(9 points)
\begin_deeper
\layout Description
Solution:
\begin_inset Formula \( \frac{\partial f}{\partial x}=2xy \)
\end_inset
,
\begin_inset Formula \( \frac{\partial f}{\partial y}=x^{2}+4y \)
\end_inset
, so
\begin_inset Formula \( \left. \nabla f\right| _{(1,3)}=6\mathbf{i}+13\mathbf{j} \)
\end_inset
.
Then,
\begin_inset Formula \( \left. D_{\mathbf{u}}f\right| _{(1,3)}=\left(
6\mathbf{i}+13\mathbf{j}\right) \cdot \left(
\frac{1}{2}\mathbf{i}-\frac{\sqrt{3}}{2}\mathbf{j}\right) =3-\frac{13\sqrt{3}}{2}. \)
\end_inset
\end_deeper
\layout Enumerate
Find the gradient of the function
\begin_inset Formula \( f(x,y)=x^{2}+3xy+e^{(x+y)} \)
\end_inset
at the point
\begin_inset Formula \( (2,1) \)
\end_inset
.
\emph on
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(9 points)
\begin_deeper
\layout Description
Solution:
\begin_inset Formula \( \frac{\partial f}{\partial x}=2x+3y+e^{(x+y)} \)
\end_inset
,
\begin_inset Formula \( \frac{\partial f}{\partial y}=3x+e^{(x+y)} \)
\end_inset
, so
\begin_inset Formula \( \left. \nabla f\right|
_{(2,1)}=(7+e^{3})\mathbf{i}+(6+e^{3})\mathbf{j} \)
\end_inset
.
\end_deeper
\layout Enumerate
Find the tangent plane to the graph of
\begin_inset Formula \( f(x,y)=3x^{2}+y^{2} \)
\end_inset
at the point
\begin_inset Formula \( (1,2,7) \)
\end_inset
.
\emph on
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(9 points)
\begin_deeper
\layout Description
Solution: The easiest way to find this is to remember that the tangent plane
satisfies the equation
\begin_inset Formula \( z=f(x_{0},y_{0})+\left. \frac{\partial f}{\partial x}\right|
_{(x_{0},y_{0})}(x-x_{0})+\left. \frac{\partial f}{\partial y}\right|
_{(x_{0},y_{0})}(y-y_{0}) \)
\end_inset
(p.
767).
Then, in this case
\begin_inset Formula \( z=7+6(x-1)+4(y-2) \)
\end_inset
is the equation of the tangent plane.
\end_deeper
\layout Enumerate
Egbert left for work in the morning and left his electric blanket on.
It produces a temperature distribution on the bed given by
\begin_inset Formula \( T(x,y)=60+6\sin (\pi x)+3y \)
\end_inset
, in degrees Fahrenheit, where
\begin_inset Formula \( x \)
\end_inset
is the distance in feet from the corner of the bed marked as
\begin_inset Formula \( (0,0) \)
\end_inset
, towards the foot of the bed, and
\begin_inset Formula \( y \)
\end_inset
is the distance towards the other side,
\begin_inset Formula \( 0\leq x\leq 6 \)
\end_inset
and
\begin_inset Formula \( 0\leq y\leq 3 \)
\end_inset
.
His cat is sitting at the point
\begin_inset Formula \( (2,1) \)
\end_inset
.
Cats always move towards warmth.
(warmth-seeking cat...) Which direction should he (the cat) head from that
point, in order to feel the greatest increase in warmth?
\emph on
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(9 points)
\begin_deeper
\layout Standard
\added_space_top 0.3cm \added_space_bottom 0.3cm \align center
\begin_inset Figure size 223 144
file bed.eps
width 2 3.1
height 2 2
flags 9
\end_inset
\layout Description
Solution: Since the gradient points in the direction of the maximum increase
of temperature, all we need find is
\begin_inset Formula \( \left. \nabla T\right| _{(2,1)} \)
\end_inset
.
But,
\begin_inset Formula \( \frac{\partial T}{\partial x}=6\pi \cos (\pi x) \)
\end_inset
, and
\begin_inset Formula \( \frac{\partial T}{\partial y}=3 \)
\end_inset
, so
\begin_inset Formula \( \left. \nabla T\right| _{(2,1)}=6\pi \mathbf{i}+3\mathbf{j} \)
\end_inset
, which points in the direction the cat should head.
\end_deeper
\layout Enumerate
Find the absolute maximum and absolute minimum of the function
\begin_inset Formula \( f(x,y)=2x^{2}+4x+3y^{2} \)
\end_inset
on the region
\begin_inset Formula \( R \)
\end_inset
described by
\begin_inset Formula \( x^{2}+y^{2}\leq 4 \)
\end_inset
.
\emph on
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(9 points)
\begin_deeper
\layout Description
Solution: First, look for critical points in the interior.
\begin_inset Formula \( \nabla f=(4x+4)\mathbf{i}+6y\mathbf{j} \)
\end_inset
, so
\begin_inset Formula \( \partial f/\partial x=\partial f/\partial y=0 \)
\end_inset
only when
\begin_inset Formula \( x=-1 \)
\end_inset
and
\begin_inset Formula \( y=0 \)
\end_inset
.
\begin_inset Formula \( f(-1,0)=-2 \)
\end_inset
.
Now, you could try to decide whether that point is a local maximum or local
minimum, but it isn't necessary,
\newline
The next thing to do would be to find the critical points of the function
restricted to the boundary circle:
\begin_inset Formula \( x^{2}+y^{2}=4 \)
\end_inset
, which can be parameterized by
\begin_inset Formula \( x=2\cos (t) \)
\end_inset
,
\begin_inset Formula \( y=2\sin (t) \)
\end_inset
,
\begin_inset Formula \( t\in [0,2\pi ] \)
\end_inset
.
\begin_inset Formula
\[
f(2\cos (t),2\sin (t))=8\cos ^{2}(t)+8\cos (t)+12\sin ^{2}(t)=8+8\cos (t)+4\sin
^{2}(t).\]
\end_inset
Setting
\begin_inset Formula \( \frac{d}{dt}(f(2\cos (t),2\sin (t)))=0 \)
\end_inset
results in
\begin_inset Formula
\[
0=-8\sin (t)+8\cos (t)\sin (t)=8\sin (t)\left( -1+\cos (t)\right) ,\]
\end_inset
so
\begin_inset Formula \( \sin (t)=0 \)
\end_inset
or
\begin_inset Formula \( \cos (t)=1 \)
\end_inset
.
\begin_inset Formula \( \sin (t)=0 \)
\end_inset
implies that
\begin_inset Formula \( t=0 \)
\end_inset
or
\begin_inset Formula \( t=\pi \)
\end_inset
, which means
\begin_inset Formula \( (x,y)=(2,0),\, \textrm{or}\, (-2,0) \)
\end_inset
.
\begin_inset Formula \( \cos (t)=1 \)
\end_inset
implies
\begin_inset Formula \( t=0 \)
\end_inset
only, so
\begin_inset Formula \( (x,y)=(2,0) \)
\end_inset
.
\newline
Now, just compare the values of
\begin_inset Formula \( f \)
\end_inset
on these points:
\begin_inset Formula \( f(-1,0)=-2 \)
\end_inset
,
\begin_inset Formula \( f(2,0)=16 \)
\end_inset
, and
\begin_inset Formula \( f(-2,0)=0 \)
\end_inset
, so the absolute maximum is
\begin_inset Formula \( 16 \)
\end_inset
at
\begin_inset Formula \( (2,0) \)
\end_inset
and the absolute minimum is
\begin_inset Formula \( -2 \)
\end_inset
at
\begin_inset Formula \( (-1,0) \)
\end_inset
.
\end_deeper
\layout Enumerate
Find the maximum value of the of the function
\begin_inset Formula \( f(x,y,z)=x+2y-3z \)
\end_inset
subject to the constraint
\begin_inset Formula \( x^{2}+2y^{2}+3z^{2}=6 \)
\end_inset
.
\emph on
\protected_separator
\protected_separator
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(9 points)
\begin_deeper
\layout Description
Solution: This is a Lagrange multipliers problem.
The condition
\begin_inset Formula \( \nabla f=\lambda \nabla g \)
\end_inset
becomes
\begin_inset Formula \( \mathbf{i}+2\mathbf{j}-3\mathbf{k}=\lambda \left(
2x\mathbf{i}+4y\mathbf{j}+6z\mathbf{k}\right) \)
\end_inset
, which can be solved by
\begin_inset Formula \( 1=2x\lambda \)
\end_inset
,
\begin_inset Formula \( 2=4y\lambda \)
\end_inset
,
\begin_inset Formula \( -3=6z\lambda \)
\end_inset
, so that
\begin_inset Formula \( x=\frac{1}{2\lambda } \)
\end_inset
,
\begin_inset Formula \( y=\frac{1}{2\lambda } \)
\end_inset
, and
\begin_inset Formula \( z=\frac{-1}{2\lambda } \)
\end_inset
,The other equation, that
\begin_inset Formula \( (x,y,z) \)
\end_inset
satisfy the constraint
\begin_inset Formula \( x^{2}+2y^{2}+3z^{2}=6 \)
\end_inset
, implies that
\begin_inset Formula
\[
\left( \frac{1}{2\lambda }\right) ^{2}+2\left( \frac{1}{2\lambda }\right) ^{2}+3\left(
\frac{-1}{2\lambda }\right) ^{2}=6,\]
\end_inset
or
\begin_inset Formula \( \frac{6}{4\lambda ^{2}}=6 \)
\end_inset
, which implies
\begin_inset Formula \( \lambda =\pm \frac{1}{2} \)
\end_inset
.
This then implies that
\begin_inset Formula \( (x,y,z)=(1,1,-1) \)
\end_inset
or
\begin_inset Formula \( (x,y,z)=(-1,-1,1) \)
\end_inset
.
Since
\begin_inset Formula \( f(1,1,-1)=6 \)
\end_inset
while
\begin_inset Formula \( f(-1,-1,1)=-6 \)
\end_inset
, the maximum value must be
\begin_inset Formula \( 6 \)
\end_inset
.
\end_deeper
\layout Enumerate
Evaluate
\begin_inset Formula \( \displaystyle \int _{0}^{1}\int _{x}^{1}e^{y^{2}}dy\, dx \)
\end_inset
.
\emph on
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(9 points)
\begin_deeper
\layout Description
Solution: You have to switch the order of integration on this one.
The region is bounded by the lines
\begin_inset Formula \( y=x \)
\end_inset
,
\begin_inset Formula \( y=1 \)
\end_inset
, and
\begin_inset Formula \( x=0 \)
\end_inset
.
The only place
\begin_inset Formula \( x=1 \)
\end_inset
occurs is at the vertex of the triangle, which is the intersection of
\begin_inset Formula \( y=x \)
\end_inset
and
\begin_inset Formula \( y=1 \)
\end_inset
.
So, written the other way around,
\begin_inset Formula
\begin{eqnarray*}
\int _{0}^{1}\int _{x}^{1}e^{y^{2}}dy\, dx & = & \int _{0}^{1}\int
_{0}^{y}e^{y^{2}}dx\, dy\\
& = & \int _{0}^{1}\left( \left. x\, e^{y^{2}}\right| _{x=0}^{x=y}\right) dy\\
& = & \int _{0}^{1}y\, e^{y^{2}}dy\\
& = & \left. \frac{1}{2}e^{y^{2}}\right| _{0}^{1}\\
& = & \frac{1}{2}\left( e-1\right) .
\end{eqnarray*}
\end_inset
\end_deeper
\layout Enumerate
Find the mass of the interior of the cardioid
\begin_inset Formula \( r=1+\cos \theta \)
\end_inset
, if the density
\begin_inset Formula \( \rho =r \)
\end_inset
.
\emph on
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(9 points)
\begin_deeper
\layout Description
Solution: Since the mass is the integral of the density over the region,
\begin_inset Formula
\begin{eqnarray*}
Mass & = & \int _{0}^{2\pi }\int _{0}^{1+\cos \theta }r\, rdrd\theta \\
& = & \int _{0}^{2\pi }\left( \left. \frac{1}{3}r^{3}\right| _{0}^{1+\cos \theta
}\right) d\theta \\
& = & \int _{0}^{2\pi }\frac{1}{3}\left( 1+\cos \theta \right) ^{3}d\theta \\
& = & \int _{0}^{2\pi }\left( \frac{1}{3}+\cos \theta +\cos ^{2}\theta
+\frac{1}{3}\cos ^{3}\theta \right) d\theta \\
& = & \frac{2\pi }{3}+0+\pi +0.
\end{eqnarray*}
\end_inset
The last integrals might be a little tricky, but certainly
\begin_inset Formula \( \int _{0}^{2\pi }\cos \theta d\theta =0 \)
\end_inset
and
\begin_inset Formula \( \int _{0}^{2\pi }\cos ^{2}\theta d\theta =\pi \)
\end_inset
.
The last one,
\begin_inset Formula \( \int _{0}^{2\pi }\cos ^{3}\theta d\theta =0 \)
\end_inset
, can be solved by substituting
\begin_inset Formula \( \cos ^{2}\theta =(1-\sin ^{2}\theta ) \)
\end_inset
and substituting
\begin_inset Formula \( u=\sin \theta \)
\end_inset
.
\end_deeper
\layout Enumerate
Set up, but
\series bold
do not attempt to evaluate
\series default
, the double integrals needed to find the
\begin_inset Formula \( x \)
\end_inset
-coordinate
\begin_inset Formula \( \overline{x} \)
\end_inset
of the center of mass of the of the region bounded above by the parabola
\begin_inset Formula \( y=4-x^{2} \)
\end_inset
and bounded below by the line
\begin_inset Formula \( y=-x \)
\end_inset
, if the density
\begin_inset Formula \( \rho \)
\end_inset
is given by
\begin_inset Formula \( \rho =x^{2}+3y^{2} \)
\end_inset
.
\emph on
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(9 points)
\begin_deeper
\layout Description
Solution:
\begin_inset Formula
\[
\overline{x}=\frac{\int
^{\frac{1}{2}+\frac{\sqrt{17}}{2}}_{\frac{1}{2}-\frac{\sqrt{17}}{2}}\int
_{-x}^{4-x^{2}}x(x^{2}+3y^{2})dydx}{\int
^{\frac{1}{2}+\frac{\sqrt{17}}{2}}_{\frac{1}{2}-\frac{\sqrt{17}}{2}}\int
_{-x}^{4-x^{2}}(x^{2}+3y^{2})dydx}.\]
\end_inset
\end_deeper
\layout Enumerate
Set up, but
\series bold
do not attempt to evaluate
\series default
, the triple integral to find the integral of the function
\begin_inset Formula \( f(x,y,z)=e^{x}\cos (y+z) \)
\end_inset
over the region
\begin_inset Formula \( R \)
\end_inset
in space bounded by:
\begin_inset Formula \( z=0 \)
\end_inset
,
\begin_inset Formula \( y=1 \)
\end_inset
,
\begin_inset Formula \( x^{2}=y \)
\end_inset
and
\begin_inset Formula \( z=x^{2}+y^{2} \)
\end_inset
.
\emph on
\protected_separator
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(9 points)
\begin_deeper
\layout Description
Solution:
\begin_inset Formula
\[
\int \int \int _{R}f(x,y,z)dV=\int _{-1}^{1}\int _{x^{2}}^{1}\int
_{0}^{x^{2}+y^{2}}e^{x}\cos (y+z)dzdydx.\]
\end_inset
\end_deeper
\layout Enumerate
How much water will a conical cup with depth 6 inches and top-radius 2 inches
hold, if it is filled up to the brim? Find the volume
\emph on
using a triple integral
\emph default
,
\series bold
\emph on
not
\series default
\emph default
some formula you may remember.
\emph on
\protected_separator
\protected_separator
\protected_separator
\protected_separator
(9 points)
\begin_deeper
\layout Standard
\added_space_top 0.3cm \added_space_bottom 0.3cm \align center
\begin_inset Figure size 144 144
file cup.eps
width 2 2
height 2 2
flags 9
\end_inset
\layout Description
Solution: It's best to use cylindrical coordinates.
The region is then defined as that region between
\begin_inset Formula \( z=6 \)
\end_inset
and
\begin_inset Formula \( z=3r \)
\end_inset
,
\begin_inset Formula
\begin{eqnarray*}
Volume & = & \int _{0}^{2\pi }\int _{0}^{2}\int _{3r}^{6}rdzdrd\theta \\
& = & \int _{0}^{2\pi }\int _{0}^{2}\left. rz\right| _{z=3r}^{z=6}drd\theta \\
& = & \int _{0}^{2\pi }\int _{0}^{2}\left( 6r-3r^{2}\right) drd\theta \\
& = & \int _{0}^{2\pi }\left. \left( 3r^{2}-r^{3}\right) \right| _{0}^{2}d\theta \\
& = & \int _{0}^{2\pi }4d\theta \\
& = & 8\pi .
\end{eqnarray*}
\end_inset
\the_end
