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{\bf Solutions to MATH 251 (4-6,11 ), Dr. Z.'s ,  Exam 1}

\bigskip

{\bf 1.} (10 points) Use the chain rule to find $f_u$ and $f_v$ if
$$
f(x,y)=x^3+y^3 \quad , \quad x=e^{u+v} \quad , \quad y=2u+3v \quad .
$$
Express your answer in terms of $u$ and $v$.

\bigskip
{\bf Sol.}
\bigskip
The {\bf types} of the answer are: functions of $u$ and $v$.
\bigskip
The chain rule (for two variables) says
$$
f_u=(f_x)(x_u)+(f_y)(y_u) \quad , \quad  f_v=(f_x)(x_v)+(f_y)(y_v) \quad  .
$$
So we have
$$
f_u=(f_x)(x_u)+(f_y)(y_u)=(3x^2)(e^{u+v})+ (3y^2)(2)=3x^2e^{u+v}+ 6y^2 \quad ,
$$
$$
f_v=(f_x)(x_v)+(f_y)(y_v)=3x^2e^{u+v}+ 9y^2 \quad .
$$

Finally, replacing $x$ and $y$ by their expressions in $(u,v)$, we get

$$
f_u=3(e^{u+v})^2e^{u+v}+ 6(2u+3v)^2 = 3e^{3u+3v}+ 6(2u+3v)^2 \quad ,
$$
$$
f_v=3(e^{u+v})^2e^{u+v}+ 9(2u+3v)^2 = 3e^{3u+3v}+ 9(2u+3v)^2 \quad .
$$


{\bf Ans.} $f_u=3e^{3u+3v}+ 6(2u+3v)^2$, $f_v=3e^{3u+3v}+ 9(2u+3v)^2$ .

\vfill\eject
{\bf 2.} (10 points) Find an equation of the tangent plane to the surface
$$
xz+2x^2y+y^2z^3=11 \quad
$$
at the point $(2,1,1)$.

\bigskip

{\bf Sol.}

\bigskip

The {\bf type} of the answer is: Equation of a plane.
\bigskip
Since this surface is given in {\bf implicit format} $F(x,y,z)=C$, where $F(x,y,z)$ is a function
of $(x,y,z)$ (not to be confused with the {\bf explicit format} $z=f(x,y)$), the
relevant formula is
$$
\g F \cdot \langle x-x_0, y-y_0, z-z_0 \rangle =0 \quad ,
$$
where $\g F$ is the gradient at the specified point.

{\bf First}, let's make sure that the problem makes sense, by plugging $x=2,y=1,z=1$ into the equation getting:
$$
2 \cdot 1+2 \cdot 2^2 \cdot 1+ 1^2 \cdot 1^3=11 \quad ,
$$
since this is correct, we must go on. Our $F(x,y,z)$ is $xz+2x^2y+y^2z^3-11$, so
We have
$$
\g F= \langle F_x , F_y , F_z \rangle = \langle z+4xy , 2x^2 + 2yz^3 , x+ 3y^2z^2 \rangle \quad .
$$

At the specified point $(x,y,z)=(2,1,1)$ this is
$$
\g F= \langle 1+4(2)(1) , 2(2)^2 + 2(1)(1)^3 , 2+ 3(1)^2(1)^2 \rangle = \langle 9, 10, 5 \rangle
$$

Plugging it into the general formula we get
$$
\langle 9,10,5 \rangle \cdot \langle x-2, y-1, z-1 \rangle =0 \quad ,
$$
that spells out to
$$
9(x-2)+10(y-1)+5(z-1)=0 \quad
$$
This is OK for the final answer, but it is even better to further simplify:
$$
9x+10y+5z=33 \quad .
$$

\bigskip

{\bf Ans.} : $9x+10y+5z=33$.

\bigskip

{\bf Comment}: it is crucial to plug-in the point into $\g F$. If you don't you get
a {\bf monster}, that is not a plane at all, and you would get zero points.



\vfill\eject
{\bf 3.} (10 points)
Find the maximal rate of change of $f(x,y,z)=x^2y^3z^4$ at $(1,1,1)$,
and the {\bf unit} direction where it occurs.
\bigskip
{\bf Sol.}
\bigskip
The {\bf types} of the answers are: number and vector of numbers.
\bigskip
$$
\g f= \langle 2xy^3x^4, 3x^2y^2z^4,4x^2y^3z^3 \rangle
$$
At the specified point $(1,1,1)$ we have
$$
\g f= \langle 2,3,4 \rangle
$$
The {\it maximum rate of change} is the {\bf magnitude}
$$
|\g f|= |\langle 2,3,4 \rangle|=\sqrt{2^2+3^2+4^2}=\sqrt{4+9+16}=\sqrt{29} \quad .
$$
The {\bf unit} direction is the gradient divided by its length
$$
{ {\langle 2,3,4 \rangle} \over  {|\langle 2,3,4 \rangle|}}=
{ {\langle 2,3,4 \rangle} \over  {\sqrt{29}}}=
\langle {{2} \over {\sqrt{29}}} ,{{3} \over {\sqrt{29}}} ,{{4} \over {\sqrt{29}}}  \rangle \quad .
$$

\vfill\eject
{\bf 4.} (10 points) Compute $f_{xx}$, $f_{xy}$, and $f_{yy}$ if
$$
f(x,y)=e^{x^2+y^2} \quad .
$$
\bigskip
{\bf Sol.}
\bigskip
The {\bf type} of the answers is: functions of $x,y$.
\bigskip
We first need the {\it first derivatives}:
$$
f_x= 2x e^{x^2+y^2}
$$
$$
f_y= 2y e^{x^2+y^2}
$$

Now we can go on to the {\it second derivatives}. We need the {\bf product rule}:
$$
f_{xx}=(2x e^{x^2+y^2})'= (2x)'e^{x^2+y^2}+2x (e^{x^2+y^2})'
$$
(where at the present context $'$ means differentiation with respect to $x$) \quad .
This becomes
$$
(2x)'e^{x^2+y^2}+2x (e^{x^2+y^2})'=2e^{x^2+y^2}+2x (e^{x^2+y^2}) (2x)=
2e^{x^2+y^2}+4x^2 e^{x^2+y^2}= (2+4x^2) e^{x^2+y^2} \quad .
$$
\bigskip
To get $f_{xy}$ is a bit easier
$$
f_{xy}= {{\partial} \over {\partial y}} (2x e^{x^2+y^2})= 2x e^{x^2+y^2}(2y) = 4xy e^{x^2+y^2} \quad.
$$
Finally $f_{yy}$ is analogous to $f_{xx}$ getting 
$$
f_{yy}=(2+4y^2) e^{x^2+y^2} \quad .
$$
\bigskip
{\bf Ans.}: $f_{xx}=(2+4x^2) e^{x^2+y^2}$, $f_{xy}=4xy e^{x^2+y^2}$,
$f_{yy}=(2+4y^2) e^{x^2+y^2}$.

\bigskip

{\bf Comment}: Some people need to review the product rule and the chain rule for single-variable functions.

\vfill\eject
{\bf 5.} (10 points) Use implicit differentiation to find ${{\partial z} \over {\partial x}}$ and
${{\partial z} \over {\partial y}}$ if
$$
z^3+x^3+y^3+3xyz= 6 \quad .
$$

\bigskip
{\bf Sol.}
\bigskip
The {\bf types} of the answer are: functions of $x,y$ (but also involving the implicilty-defined function $z$)
\bigskip
The easiest way is to use the ready-made formulas
$$
{{\partial z} \over {\partial x}} = - {{F_x} \over {F_z}} \quad ,
$$
$$
{{\partial z} \over {\partial y}} = - {{F_y} \over {F_z}} \quad .
$$
In this problem $F(x,y,z)=z^3+x^3+y^3-3xyz-6$. So we have
$$
{{\partial z} \over {\partial x}} = - {{F_x} \over {F_z}}=- {{3x^2+3yz} \over {3z^2+3xy}}=
- {{3(x^2+yz)} \over {3(z^2+xy)}}=- {{x^2+yz} \over {z^2+xy}} \quad ,
$$
$$
{{\partial z} \over {\partial y}} = - {{F_y} \over {F_z}}=- {{3y^2+3xz} \over {3z^2+3xy}}=
- {{3(y^2+xz)} \over {3(z^2+xy)}}=- {{y^2+xz} \over {z^2+xy}} \quad .
$$
\bigskip
{\bf Ans.}: ${{\partial z} \over {\partial x}} =- {{x^2+yz} \over {z^2+xy}}$, 
$ {{\partial z} \over {\partial y}}  =- {{y^2+xz} \over {z^2+xy}}$ .




\vfill\eject
{\bf 6.} (10 points) Find a parametric equation of the line of intersection of the planes
$x+2y+3z=6$ and $3x+2y+z=6$.
\bigskip
{\bf Sol.}
\bigskip
The {\bf type} of the answer is: parametric equation of a line.
\bigskip
We need a {\bf direction} and (any!) {\bf point}.

The normal of the first plane is $\langle 1,2,3 \rangle$, 
the normal of the second plane is $\langle 3,2,1 \rangle$.

Since our desired line lies on {\bf both} planes, it is perpendicular to both these
vectors, so its direction is the same direction as the {\bf cross product}.
$$
\langle 1,2,3 \rangle \times \langle 3,2,1 \rangle =
$$
$$
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
1 & 2& 3\cr
3 & 2 & 1 
} \right \vert \quad=
$$
$$
{\bf i} \left \vert \matrix{ 
2 & 3\cr
2 & 1 
} \right \vert 
-{\bf j} \left \vert \matrix{ 
1 & 3\cr
3 & -1 
} \right \vert 
+{\bf k} \left \vert \matrix{ 
1 & 2 \cr
3 & 2 
} \right \vert 
$$
$$
={\bf i}(-4)+{\bf j}8+{\bf k}(-4)
=\langle -4,8,-4 \rangle \quad.
$$
\bigskip
Now we still need a {\bf point}. Quite arbitrarily let's pick $y=1$ and plug it
into the two planes
$$
x+2+3z=6 \quad , \quad 3x+2+z=6 \quad ,
$$
yielding:
$$
x+3z=4 \quad , \quad 3x+z=4 \quad .
$$
Solving these equations gives $x=1,z=1$, yielding the point $(1,1,1)$.

So the equation of the common line is:
$$
\langle 1,1,1 \rangle + t \langle -4, 8, -4 \rangle \quad ( -\infty < t < \infty ) \quad .
$$
and in {\it scalar form} it is:
$$
x=1-4t \quad , \quad y= 1+ 8t \quad , \quad z= 1- 4t \quad ( -\infty < t < \infty ) \quad .
$$

{\bf Ans.} $x=1-4t \quad , \quad y= 1+ 8t \quad , \quad z= 1- 4t \quad  ( -\infty < t < \infty )$

\bigskip

{\bf Comments}: There are many are right answers, since there many choices of points.
Also since $ -\infty < t < \infty $ we can simplify further and divide the direction by $4$, so an
even nicer answer is
$x=1-t \quad , \quad y= 1+ 2t \quad , \quad z= 1- t \quad  ( -\infty < t < \infty )$
\vfill\eject
{\bf 7.} (10 points) A certain particle has law of motion
$$
{\bf r}(t) = \langle \, \sin t \, , \, \cos 2t \, , \, e^t \, \rangle \quad .
$$
Find its velocity, acceleration, and speed at $t=\pi/6$.
\bigskip
{\bf Sol.}
\bigskip
The {\bf types} of the answers are: two vectors of numbers and a number.
\bigskip
$$
{\bf v}(t)={\bf r}'(t) = \langle \, \cos t \, , \, -2\sin 2t \, , \, e^t \, \rangle \quad ,
$$
$$
{\bf a}(t)={\bf v}'(t) = \langle \, -\sin t \, , \, -4\cos 2t \, , \, e^t \, \rangle \quad ,
$$
Now it is time to plug-in $t=\pi/6$, we have
$$
{\bf v}(\pi/6)= \langle \, \cos \pi/6 \, , \, -2\sin \pi/3 \, , \, e^{\pi/6} \, \rangle=
\langle \, {{\sqrt{3}} \over {2}} \, , -2{{\sqrt{3}} \over {2}}  , \, , \, e^{\pi/6} \, \rangle=
\langle \, {{\sqrt{3}} \over {2}} \, , -\sqrt{3}  , \, , \, e^{\pi/6} \, \rangle
$$
$$
{\bf a}(\pi/6)= \langle \, -\sin \pi/6 \, , \, -4\cos \pi/3 \, , \, e^{\pi/6} \, \rangle =
\langle \, -{{1} \over {2}} \, , \, -4 \cdot {{1} \over {2}} \, , \, e^{\pi/6} \, \rangle =
\langle \, -{{1} \over {2}} \, , \, -2 \, , \, e^{\pi/6} \, \rangle  \quad .
$$
Finally the speed at $t=\pi/6$ is:
$$
|{\bf v}(\pi/6)|= |\langle \, {{\sqrt{3}} \over {2}} \, , -\sqrt{3}  , \, , \, e^{\pi/6} \rangle|=
\sqrt{ ({{\sqrt{3}} \over {2}})^2 + (-\sqrt{3})^2+ (e^{\pi/6})^2 }
=
\sqrt{ {{3} \over {4}} + 3+ (e^{\pi/6})^2 }=
\sqrt{ {{15} \over {4}} + e^{\pi/3} }
$$
\bigskip
{\bf Ans.}: The velocity at $t=\pi/6$ is
$\langle \, {{\sqrt{3}} \over {2}} \, , -\sqrt{3}  , \, , \, e^{\pi/6} \, \rangle$
and the acceleration then is
$\langle \, -{{1} \over {2}} \, , \, -2 \, , \, e^{\pi/6} \, \rangle$, while the
speed then is
$\sqrt{ {{15} \over {4}} + e^{\pi/3} }$.
\bigskip
{\bf Comments}: 1. Some people interpreted 
the phrase ``velocity, acceleration, and speed at $t=\pi/6$''
as ``velocity (at general time $t$), acceleration (at general time $t$), and speed at $t=\pi/6$''.
Strictly speaking, this is also a valid interpreation, and they got full credit.
2. People who didn't evaluate $\cos \pi/6$ etc. got 9 out of 10 points.

\vfill\eject
{\bf 8.} (10 points) Write a definite integral that describes the length of the curve
$$
{\bf r}(t)= \langle \, \sin 5t \, , \, e^{2t} \, ,  \, e^{3t} \, \rangle \quad , \quad 0 \leq t \leq \pi \quad .
$$
{\bf Do not try to evaluate the integral!}
\bigskip
{\bf Sol.}
\bigskip
The {\bf type} of the answer is: number given in terms of a definite integral.
\bigskip
The relevant formula is
$$
Length = \int_{t_0}^{t_1} |{\bf r}'(t)| \, dt \quad .
$$
In this problem
$$
{\bf r}'(t)= \langle \, 5\cos 5t \, , \, 2e^{2t} \, ,  \, 3e^{3t} \, \rangle 
$$
Taking the {\bf magnitude} we have
$$
|{\bf r}'(t)|= |\langle \, 5\cos 5t \, , \, 2e^{2t} \, ,  \, 3e^{3t} \, \rangle|=
\sqrt{ (5 \cos 5t)^2+ (2e^{2t})^2+ (3e^{3t})^2} 
=
\sqrt{ 25 \cos^2 5t +4e^{4t}+ 9e^{6t}} 
$$
Putting it in the formula for the length we have:
$$
Length = \int_{0}^{\pi} \sqrt{ 25 \cos^2 5t +4e^{4t}+ 9e^{6t}} \, dt
$$
That's it! (don't try to ``simplify'', or you will lose lots of points, perhaps
even all of them, if your ``simplifications'' are wrong.)
\bigskip
{\bf Ans.}: the length is $\int_{0}^{\pi} \sqrt{ 25 \cos^2 5t +4e^{4t}+ 9e^{6t}} \, dt$
\vfill\eject
{\bf 9.} (10 points, altogether) Do the following limits exist? If they do, find them. Explain!
\bigskip
{\bf a.} (5 points)
If
$$
\lim_{(x,y,z) \rightarrow (1,2,3)} f(x,y,z)=1 \quad, \quad
\lim_{(x,y,z) \rightarrow (1,2,3)} g(x,y,z)=2 \quad
$$
compute
$$
\lim_{(x,y,z) \rightarrow (1,2,3)} (f(x,y,z)+g(x,y,z))^3 e^{g(x,y,z)} \quad
$$

\bigskip
{\bf Sol.}
\bigskip
The {\bf type} of the answer is: number.
\bigskip
$$
\lim_{(x,y,z) \rightarrow (1,2,3)} (f(x,y,z)+g(x,y,z))^3 e^{g(x,y,z)} =
(1+2)^3 e^{2} =27 e^{2} \quad .
$$
{\bf Ans.}: The limit exists and equals $27e^2$.
\bigskip
{\bf b.} (5 points)
$$
\lim_{(x,y) \rightarrow (0,0)} { {x^4+y^4}  \over {2x^4+y^4}} \quad .
$$
\bigskip
{\bf Sol.}:
The {\bf type} of the answer is: DNE.
\bigskip
{\bf First Way}: The limit, {\bf only} going along the $y$ axis is:
$$
\lim_{(x,y) \rightarrow (0,0), x=0} { {y^4}  \over {y^4}}=
\lim_{y \rightarrow 0} 1= 1 \quad,
$$
The limit, {\bf only} going along the $x$ axis is:
$$
\lim_{(x,y) \rightarrow (0,0), y=0} { {x^4}  \over {2x^4}}=
\lim_{x \rightarrow 0} {{1} \over {2}}= {{1} \over {2}} \quad .
$$
Since these are {\bf different} the limit does not exist.
\bigskip
{\bf Second Way}:
Let's invesitage the limit along a line of general slope $m$, passing through the origin, given by $y=mx$:
$$
\lim_{(x,y) \rightarrow (0,0), y=mx} { {x^4+y^4}  \over {2x^4+y^4}}=
\lim_{x \rightarrow 0} { {x^4+(mx)^4}  \over {2x^4+(mx)^4}}=
\lim_{x \rightarrow 0} { {x^4(1+m^4)}  \over {x^4(2+m^4)}}=
\lim_{x \rightarrow 0} { {1+m^4}  \over {2+m^4}}=
{ {1+m^4}  \over {2+m^4}} \quad .
$$
Since this {\bf depends} on $m$, the limit {\bf does not exist}.
\bigskip
{\bf Ans.}: DNE


\vfill\eject
{\bf 10.} (10 pts.)  A force with magnitude $500N$ is moving a body of mass $5kg$ in the direction
$\langle 3,4,0 \rangle$. If at $t=0$ the body is at location $(1,2,3)$ and it is moving with
velocity $\langle 2,1,4 \rangle$, 

(i)find its position vector ${\bf r}(t)$ at time $t$ ;

(ii) find its speed at time $t$.
\bigskip
{\bf Sol.}
\bigskip

The {\bf types} of the answers are: (i) vector of functions of $t$ (ii) function of $t$ .

\bigskip

By Newton's Second Law the {\bf magnitude} of the acceleration is $500/5=100$.
The direction is $\langle 3,4,0 \rangle$. But we need the {\bf unit direction}.
Since $|\langle 3,4,0 \rangle|=\sqrt{3^2+4^2+0^2}=\sqrt{25}=5$, the
{\bf unit direction is}
$$
{{\langle 3,4,0 \rangle} \over {5}}= \langle {{3} \over {5}} \, , \,  {{4} \over {5}} \, , \, 0 \rangle \quad .
$$
Multiplying the magnitude of the acceleration by the unit direction gives the {\bf acceleration vector}:
$$
{\bf a} = 100 \langle {{3} \over {5}} \, , \,  {{4} \over {5}} \, , \, 0 \rangle=
\langle 60,80,0 \rangle \quad .
$$
To get ${\bf v}(t)$ we integrate
$$
{\bf v}(t)= \int \langle 60,80,0 \rangle \, dt = \langle 60t, 80t, 0 \rangle + {\bf C}
$$
Plugging-in $t=0$ gives
$$
{\bf v}(0)={\bf C} \quad .
$$
From the data of the problem ${\bf v}(0)= \langle 2,1,4 \rangle$, so
${\bf C}= \langle 2,1,4 \rangle$, and we get:
$$
{\bf v}(t)= \int \langle 60,80,0 \rangle \, dt = \langle 60t, 80t, 0 \rangle +\langle 2,1,4 \rangle=
\langle 60t+2, 80t+1, 4 \rangle 
$$
\bigskip
To get the {\bf position} vector, we integrate once again
$$
{\bf r}(t)= \int \langle 60t+2, 80t+1, 4 \rangle \, dt=
\langle 30t^2+2t, 40t^2+t, 4t \rangle + {\bf C} \quad
$$
Plugging-in $t=0$ gives ${\bf C}= \langle 1,2,3 \rangle$. Putting it above, we get
$$
{\bf r}(t)= 
\langle 30t^2+2t, 40t^2+t, 4t \rangle + \langle 1,2,3 \rangle=
\langle 30t^2+2t+1, 40t^2+t+2, 4t+3 \rangle \quad .
$$
This is the answer to part (i). For part (ii) we take the magnitude of ${\bf v}(t)$.
$$
Speed= 
|{\bf v}(t)|=
|\langle 60t+2, 80t+1, 4 \rangle |= \sqrt{ (60t+2)^2+ (80t+1)^2+4^2}=
\sqrt{ (60t+2)^2+ (80t+1)^2+16} \quad .
$$
This is acceptable (no need to expand the inside of the square-root).
\bigskip
{\bf Ans.}: (i) $\langle 30t^2+2t+1, 40t^2+t+2, 4t+3 \rangle $
(ii)$\sqrt{ (60t+2)^2+ (80t+1)^2+16}$.



\end


