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{\bf CORRECTING  TYPOS and ERRORS POINTED OUT BY Aastha and AAsyushi Kasera}

\medskip
{\bf NAME: (print!)}  Dr. Z.
\bigskip

{\bf Section:}  ALL
\qquad
 {\bf E-Mail address:} DrZcalc3@gmail.com

\bigskip
MATH 251 (22,23,24 ) [Fall 2020], Dr. Z. ,  Makeup for Exam 1, Sunday, Oct. 25, 2020, 9:00-11:00am
\bigskip

\hrule

\bigskip

WRITE YOUR FINAL ANSWERS BELOW
\medskip
\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
\bigskip
1. \qquad  $(t,0,0)$, $-\infty<t<\infty$ ; $x$-axis .
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2.  \qquad  $272$
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3.   \qquad $(-3,3)$ \quad .
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4.   \qquad  $z=2x+12y-17$
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5.   \qquad $\frac{1}{2}$
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6.    \qquad  $\frac{24999}{5000}$ or $4.9998$
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7.    \qquad DNE
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8.    \qquad $(-1,-1,0)$
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9.    \qquad  $6$ \quad ; \quad $\langle \frac{1}{3}, -\frac{2}{3}, -\frac{2}{3} \rangle $
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10.    \qquad $10$
\bigskip
\hrule
\bigskip
{\bf Types:} Number, Function of $variable$(s), 2D vector of numbers, 3D vector of numbers,
2D vector of functions, 3D vector of functions, equation of a plane, parametric equation of a line,
equation of a line, equation of a surface, equation of a line, DNE (does not exist).
\bigskip
\hrule
\bigskip

Sign the following declaration:
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I  \quad \quad \quad \quad \quad \quad \quad \quad Hereby declare that all the work was done by myself. 
I was allowed to use Maple, calculators, the book, and all the material in the
web-page of this class but {\bf not} other resources on the internet.
\bigskip
I only spent (at most) 90 minutes on  doing the exam. The last 30 minutes were spent in checking
and double-checking the answers.
\bigskip
I also understand that I may be subject to a random short chat to verify that I actually did it all by myself.
\bigskip
Signed:


\vfill\eject
\medskip
{\bf 1}. (10 pts.)  Find a parametric representation, in terms of the parameter, $t$,  of the line of intersection
of the planes $y=0$ and $z=0$. What is the usual name for that line?

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\medskip
The {\bf type} of the answers is:  parametric equation of a line in 3D space
\medskip
\hrule


\hrule
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{\bf ans.}  $(t,0,0)$, $-\infty < t < \infty$ ; The $x$-axis
\bigskip
\hrule
\bigskip

The {\bf short way} (perfectly OK!) is to say that obviously a typical point on the line of intersection of $y=0$ and $z=0$
looks like $(x,0,0)$ for some $x$, but the traditional letter for a parameter is $t$ so it is 
the set of all points $(t,0,0)$ where $t$ goes from $-\infty$ to $\infty$


The {\bf long way} is to note that the vector normal to $y=0$ is ${\bf j}$  and the vector normal to $z=0$ is
${\bf k}$ so the line of intersection is in the direction ${\bf j} x {\bf k}={\bf i}$, so it is of the form
$point+ t(1,1,0)$ where $point$ is any point common to both planes, the simplest being $(0,0,0)$.



\vfill\eject

{\bf 2.} (10 points) Use any method to find ${{\partial^2 h} \over {\partial r^2}}$ at $(q,r)=(1,1)$ where
$h(u,v)=u^5\,v^6$, $u=q^2\,r$, $v=q\,r^2$
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\hrule
\medskip
The {\bf type} of the answer is: number
\medskip
\hrule

\hrule
\bigskip
{\bf ans.}  $272$
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\hrule
\bigskip
I told you {\it use any method}. 
Superficially  it looks like that you need the chain rule, and some people did
it that way, and got the right answer and full credit (but it took them a long time!).
Other people messed up.

In this simple problem, it is much more efficient to express the function {\bf directly} in terms of $q$ and $r$ and
then it is a piece of cake.

The function is $u^5 \, v^6= (q^2 r)^5 (q r^2)^6= q^{10} r^5 \cdot q^{6} r^{12}= q^{16} r^{17}$

So $h_r=17 q^{16} r^{16}$ and $h_{rr}=17 \cdot 16 \cdot q^{15} r^{16} =272$

{\bf Finally}, don't forget!, plug-in $q=1$ $r=1$, and the type of the final answer is {\bf number}.

\vfill\eject
{\bf 3.} (10 points)
Find the point  on the line $y\,=\,x+6$ (in the $xy$-plane) where the rate of change
of the function $f(x,y)=x^2+y^2$ in the direction $<1,1>$ is zero.
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\hrule
\medskip
The {\bf type} of the answer is: point
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{\bf ans.} $(-3,3)$
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\hrule
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The {\bf gradient} of $f$, $<f_x,f_y>$ equals $<2x, 2y>$. The directional derivative in direction $<1,1>$ is
$<2x,2y>.<1,1>/\sqrt{2}= \sqrt{2}(x+y)$. Setting this equal to $0$ we get that $x+y=0$. But out point also lies on
the line $y=x+6$ . Solving the simple system of two equations and two unknowns $x,y$:
$$
\{ x+y=0, y=x+6 \} \quad ,
$$ 
we get
$-x=x+6$ so $-2x=6$ so $x=-3$ and $y=-(-3)=3$. So the desired point is $(-3,3)$.


\vfill\eject
{\bf 4.} (10 points)
Find an equation of the tangent plane to the following surface at the given point
$$
z=x^2+y^3 \quad, \quad (1,2,9) \quad .
$$
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\hrule
\medskip
The {\bf type} of the answer is: Equation of a plane
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\hrule
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{\bf ans.} z= 2x+12y-17
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{\bf First way}: Use the fact that for a surface given {\bf explicitly} in the format $z=f(x,y)$ the equation of
the tangent plane through $(x_0,y_0,z_0)$ (provided of course the point is a good one, i.e. $f(x_0,y_0)=z_0$) is
$$
z-z_0= f_x(x_0,y_0)\,(x-x_0) + f_y(x_0,y_0)\,(y-x_0)  \quad.
$$

{\bf Step 0}: Verify that $f(1,2)=1^2+2^3=9$ (yea!)
\bigskip
Now

$$
f_x=2x \quad, \quad  f_y=3y^2
$$

So 

$$
f_x(1,2) =2 \quad, \quad  f_y(1,2) =3 \cdot 2^2= 12
$$


So we get
$$
z-9= 2(x-1)+12(y-2)
$$

Simplifying
$$
z-9= 2x-2+12y-24
$$

Simplifyiny more
$$
z= 2x+12y-17
$$

{\bf Second way}: Convert it to {\bf implicit format} $z-x^2-y^3=0$ and use the  formula $grad(f)(x_0,y_0,z_0) . <x-x_0,y-y_0,z-z_0>=0$.


\vfill\eject

{\bf 5.} (10 points) Compute $f_{xyy}(1,1)$ (aka $\frac{\partial^3 f}{\partial x \partial^2 y}(1,1)$),
if $f(x,y)=\ln(x^2 \,+ \, y)$.
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\hrule
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The {\bf type} of the answer is: Number
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{\bf ans.} $\frac{1}{2}$ \quad .
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$f_x=\frac{2x}{x^2+y}$ (by the chain rule of calc1). This can be written $f_x=2x \cdot (x^2+y)^{-1}$

Moving right along $f_{xy}= 2x \cdot (-1) (x^2+y)^{-2}$ and  $f_{xyy}= 2x \cdot (-1)(-2) (x^2+y)^{-3}$,
so 
$$
f_{xyy} (x,y) = \frac{4x}{(x^2+y)^3} \quad .
$$

{\bf Finally} (don't forget!)
$f_{xyy} (1,1) = \frac{4 \cdot 1}{(1^2+1)^3}= \frac{4}{8} = \frac{1}{2}$.


\vfill\eject
{\bf 6.} (10 points) Use Linearization to approximate
$\sqrt{(3.001)^2+(3.999)^2}$. Part of the challenge is to find the appropriate function $f(x,y)$ and
the appropriate `nice' point $(x_0,y_0)$.

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\hrule
\medskip
The {\bf type} of the answer is: Number
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\hrule
\hrule
\bigskip
{\bf ans.} $\frac{24999}{5000}$ OR 4.9998 (both are acceptable)
\bigskip
\hrule
\bigskip
The underlying {\bf function} is
$$
f(x,y)= \sqrt{x^2+y^2} = (x^2+y^2)^{1/2} \quad .
$$
The point of interest is $(3.001, 3.999)$. The {\bf nice} point nearby is $(x_0,y_0)=(3,4)$.

The {\bf Linearization} of $f(x,y)$ at $(x_0,y_0)$ is
$$
L(x,y)= f_x(x_0,y_0) \, (x-x_0) \, + \, f_y(x_0,y_0) \, (y-y_0) \quad.
$$

$f_x= \frac{1}{2} (x^2+y^2)^{-1/2} \cdot 2x = \frac{x}{x^2+y^2}$ \quad .

$f_y= \frac{1}{2} (x^2+y^2)^{-1/2} \cdot 2y = \frac{y}{x^2+y^2}$ \quad .

So $f_x(3,4)= \frac{3}{\sqrt{3^2+4^2}}= \frac{3}{5}$, $f_y(3,4)= \frac{4}{\sqrt{3^2+4^2}}= \frac{4}{5}$.

Getting

$$
L(x,y)= 5+ \frac{3}{5} \, (x-3) \, + \, \frac{4}{5} \, (y-4) \quad.
$$

{\bf Warning}: do {\bf not} simplify. Leave it like that! So

$$
L(3.001,3.999)= 5+ \frac{3}{5} \, (3.001-3) \, + \, \frac{4}{5} \, (3.999-4) \, = \,
5+\frac{3}{5} \, (0.001) \, + \, \frac{4}{5} \, (-0.001) \, 
$$
$$
= \, 5-\frac{1/1000}{5}\, = \, \frac{24999}{5000}=4.9998 \quad .
$$

{\bf Remark}: The exact value (up to 10 decimal points) is   $4.999800196$ (not bad!)

\vfill\eject
{\bf 7.} (10 points) Decide whether the following limit exists. If it does find it.
If it does not {\bf Explain} why not?
$$
\lim_{(x,y,z)\rightarrow (1,1,1)} \,\, \frac{x+2y+3z-6}{3x+2y+z-6} \quad
$$
\bigskip
{\bf Hint}: To prove that a limit {\bf exists} at a designated point you first try to plug it in. If there are
no issues you are done, and the value is the limit. Failing this, you try to use {\bf algebra} to simplify and then
plug it in, if possible. Failing this you try to prove that it does {\bf not} exist. One way of doing it
is to pick any two {\bf different} lines through the point in question and show that if you approach
the designated point via these lines you get {\bf different} limits (in the sense of single-variable calculus).
\bigskip
{\bf Remark}: This was the most challenging problem, and most people did not get it. Please study it carefully.

Plugging in gives $0/0$. Trying to simplify also fails. So we are suspecting that the limit does not exist.

We can do some preliminary investigations plugging into the function 
$$
f(x,y,z)=\frac{x+2y+3z-6}{3x+2y+z-6} \quad
$$
Random values {\bf near} the {\bf designated point} $(1,1,1)$. For example
$$
f(1,1,1.0001)=3 \quad , \quad f(1,1.0001,1)=1
$$

$1$ and $3$ are very far apart! So this is an {\bf indication} that the limit {\bf probably} does not exist.

In order to prove it {\bf conclusively} we need to find any two {\bf different} lines 
{\bf passing through} the designated point $(1,1,1)$ and plug-it in and take the limits in the
sense of calc1 (that DO EXIST!, some people claimed that the calc 1 limits do not exist!).

\bigskip

A convenient line is $(1+t,1,1)$ ($-\infty <t <\infty$), getting that on {\bf this line}
$$
f(1+t,1,1) \, = \, \frac{(1+t)+2+3-6}{3(1+t)+2+1-6}\, = \, \frac{t}{3(1+t)+2+1-6} \, = \, \frac{t}{3t}= \frac{1}{3}
$$

So along this line the {\bf value} of the function t is {\bf always} $\frac{1}{3}$, and obviously the {\bf limit}
in the sense of calc1 {\bf exists} and equals $\frac{1}{3}$.



Another convenient line is $(1,1+t,1)$  ($-\infty <t <\infty$), getting that on {\bf this line}
$$
f(1,1+t,1) \, = \, \frac{1+2(1+t)+3-6}{3+2(1+t)+1-6}\, = \, \frac{2t}{2t} \, = \, \frac{t}{t} \, = \,  1.
$$

So along this line the {\bf value} of the function t is {\bf always} $1$, and obviously the {\bf limit}
in the sense of calc1 {\bf exists} and equals $1$.

But since these two calc1 limits do {\bf not} agree, this is proof that the limit (in the sense of calc3) {\bf does not exist}.
\medskip
{\bf Comments}: 
\medskip
{\bf 1.} You can take {\bf any} two lines . Of course it is good to take convenient line. But the lines
{\bf must  pass through the designated point}, {\bf not} the origin (unless the point in question happened to be, as often is the
case in the homework, and this confused many students) is the origin $(0,0,0$.

\bigskip

{\bf 2.} Another, more general approach is to consider an `abstract' (general) line {\bf passing through} $(1,1,1)$

$$
(1,1+ a t, 1+ bt) \quad
$$

For $a$ and $b$ arbitrary numbers, find the limit along that line as an {\bf expression} in $a,b$, and since
it {\bf depends} on $a,b$ the limit does not exist.

\bigskip

{\bf 3.} Yet another (legal!) approach , for all those who {\bf love} the origin, is to {\bf convert} the
problem to a limit about the origin doing a change of  variable
$$
\bar{x}=x-1 \quad , \quad
\bar{y}=y-1 \quad , \quad
\bar{z}=z-1 \quad , \quad
$$
Getting that we have to study
$$
\lim_{(x,y,z)\rightarrow (0,0,0)} \,\, \frac{x+2y+3z}{3x+2y+z} \quad
$$
and then take lines through the origin! (Here we {\bf renamed}  $\bar{x}$ by $x$ , $\bar{y}$ by $y$ , and $\bar{z}$ by $z$).
\bigskip
{\bf Warning}: I was disappointed that quite a few people `faked' it (unintenionaly) writing {\bf gibberish}.
If you are not sure, please exercise intellectual honesty and admit that you don't know how to do it.



\vfill\eject
{\bf 8.} (10 points)  Find the  minimum {\bf location(s)} (i.e the {\bf point(s)} where the function is the smallest)
of the function
$$
f(x,y,z) \, = \, x +2y-z
$$
in the {\bf closed pyramid} whose vertices are
$$
(1,1,0) \quad, \quad (-1,1,0) , \quad (1,-1,0) ,  \quad (-1,-1,0) ,  \quad (0,0,1)
$$

\bigskip

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The {\bf type} of the answer(s) is: point
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{\bf ans.} $(-1,-1,0)$
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The function is {\bf linear} so there are no critical points (not even along the edges). All we have
to do is plug-in the function at the five vertices and see who gives the minimal value.
$$
f(1,1,0) \, = \, 3 \quad , \quad
f(-1,1,0) \, = \, 1 \quad , \quad
f(1,-1,0) \, = \, -1 \quad , \quad
f(-1,-1,0) \, = \, -3 \quad , \quad
f(0,0,1)=-1 \quad .
$$

The minimum value is $-3$ at it {\bf happens} at the location $(-1,-1,0)$.

\vfill\eject
{\bf 9.} (10 points) A certain particle of mass $2$ kilograms has position function, expressed in
meters, where $t$ is time, expressed in seconds,
$$
{\bf r}(t) = \langle \, e^t, 2 \cos t, 2 \cos t \, \rangle \quad ,
$$
\bigskip
(i) Find the {\bf magnitude} of the force acting on it at time $t=0$. {\bf Explain!}
\bigskip
(ii) Find the {\bf unit direction} of that force.
\bigskip
{\bf Reminder from physics}: $F=ma$, i.e. the force equals the mass  times the acceleration.



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The {\bf types} of the answers are (i) non-negative number (it is OK to say number) \qquad\qquad\qquad\qquad\qquad 
\medskip
(ii) vector (or even better unit vector)
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{\bf ans.}  (i) magnitude :    $6$ Newtons \quad \quad ; \quad 
(ii) unit direction:$ \langle \frac{1}{3}, -\frac{2}{3}, -\frac{2}{3} \rangle$
\bigskip
\hrule
\bigskip
$$
{\bf r}'(t)= \langle \, e^t, -2 \sin t, -2 \sin t \, \rangle \quad ,
$$
$$
{\bf r}''(t)= \langle \, e^t, -2 \cos t, -2 \cos t \, \rangle \quad ,
$$

$$
{\bf r}''(0)= \langle \, 1, -2 , -2  \, \rangle \quad ,
$$
$$
{\bf F} \, = \, 2 \langle \, 1, -2 , -2  \, \rangle 
\, = \,  \langle \, 2, -4 , -4  \, \rangle \quad ,
$$
$$
|{\bf F}| \, = \,  |\langle \, 2, -4 , -4  \, \rangle| \, = \, \sqrt{2^2+(-4)^2+(-4)^2}=\sqrt{36}=6 \quad .
$$

{\bf direction}:  $\langle \, 2, -4 , -4  \, \rangle / 6 = \langle \, \frac{1}{3}, -\frac{2}{3} , -\frac{2}{3}   \rangle $



\bigskip


\vfill\eject
{\bf 10.} (10 points) A certain function $f(x,y,z)$ depends on three variables called $x$, $y$ and $z$. 
At a certain time
the rate of change of this function with respect to time happens to be $140$.
It is also known, that at that same time

\medskip
$\bullet$ The rate of change of $f(x,y,z)$ with respect to $x$ is $1$
\medskip
$\bullet$ The rate of change of $f(x,y,z)$ with respect to $y$ is $2$
\medskip
$\bullet$ The rate of change of $f(x,y,z)$ with respect to $z$ is $3$
\medskip
$\bullet$ The rate of change of  $y$ with respect to time is 
{\bf two times} The rate of change of  $x$ with respect to time.
\medskip
$\bullet$ The rate of change of  $z$ with respect to time is 
{\bf three times} The rate of change of  $x$ with respect to time.
\medskip

What is the rate of change of $x$ with respect to time at that time?


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The {\bf type} of the answer is: Number
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{\bf ans.}  $10$
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\bigskip
We use the {\bf Chain rule} in its {\it abstract} form
$$
\frac{df}{dt} \, = \,
\frac{df}{dx} \cdot \frac{dx}{dt}  \, + \,
\frac{df}{dy} \cdot \frac{dy}{dt}  \, + \,
\frac{df}{dz} \cdot \frac{dz}{dt}  \quad.
$$

Incorporating this {\bf data}
$$
140 \, = \,
1 \cdot \frac{dx}{dt}  \, + \,
2 \cdot \frac{dy}{dt}  \, + \,
3 \cdot \frac{dz}{dt}  \quad.
$$

But
$$
\frac{dy}{dt}=2 \frac{dx}{dt} \quad, \quad
\frac{dz}{dt}=3 \frac{dx}{dt} \quad, \quad
$$

So going back

$$
140 \, = \,1 \cdot \frac{dx}{dt}  \, + \,2 \cdot (2 \frac{dx}{dt})  \, + \,3 \cdot (3 \frac{dx}{dt})  \quad,
$$
giving
$$
140 \, = \frac{dx}{dt} (1+4+9)\, = \, \frac{dx}{dt} \cdot 14 \quad
$$
Giving
$$
\frac{dx}{dt}= 10 \quad .
$$

Going back to English: The rate of change of $x$ with respect to time, at that above instant, is $10$.

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