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\medskip
\bigskip
MATH 251 (1-3,10 ), Dr. Z. ,  Solutions to Exam 1 , Thurs., Oct. 8, 2009, SEC 118
\bigskip
{\bf 1}. (10 pts.)  Find the curvature of
the curve 
$$
{\bf r}(t)=\langle e^t, \sin t, \cos t \rangle
$$
at the point $(1,0,1)$.
\bigskip

{\bf Sol.}: 

The type of the answer is {\bf number}. Since it is at a specific point.

\bigskip
The formula is
$$
\kappa (t)=
{{\vert {\bf r}'(t) \times {\bf r}''(t) \vert} \over
{\vert {\bf r}'(t) \vert^3}}
$$

We have
$$
{\bf r}'(t)=\langle e^t, \cos t, -\sin t \rangle
$$
$$
{\bf r}''(t)=\langle e^t, -\sin t, -\cos t \rangle
$$

The next thing to worry about is: {\bf What time is it?}, in other words what is $t$ when
the particle is at $(1,0,1)$.
We need $e^t=1$, to $t=0$. To make sure we plug it into ${\bf r}(t)$ and we get
$$
{\bf r}(t)=\langle e^0, \sin 0, \cos 0 \rangle=\langle 1,  0, 1 \rangle \quad,
$$
so $t$ is indeed $0$ right now.

{\bf Now} that we know that $t=0$, we plug-in and get
$$
{\bf r}'(0)=\langle e^0, \cos 0, -\sin 0 \rangle=\langle 1, 1, 0 \rangle
$$
$$
{\bf r}''(0)=\langle e^0, -\sin 0, -\cos 0 \rangle=\langle 1, 0, -1 \rangle
$$
$$
{\bf r}'(0) \times {\bf r}''(0)=
$$
$$
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
1 & 1 & 0 \cr
1 & 0 & -1 
} \right \vert \quad =
$$
$$
{\bf i} \left \vert \matrix{ 
1 & 0\cr
0 & -1 
} \right \vert 
-{\bf j} \left \vert \matrix{ 
1& 0\cr
1 & -1 
} \right \vert 
+{\bf k} \left \vert \matrix{ 
1 & 1\cr
1 & 0 
} \right \vert 
$$
$$
=-{\bf i}+{\bf j} -{\bf k} = \langle -1 , 1, -1 \rangle \quad.
$$

So $|{\bf r}'(0) \times {\bf r}''(0)|=|\langle -1 , 1, -1 \rangle|=\sqrt{(-1)^2+1^2+(-1)^2}=\sqrt{3}$.
Also $|{\bf r}'(0)|=|\langle 1 , 1,0 \rangle|=\sqrt{(1)^2+1^2+0^2}=\sqrt{2}$.
So $\kappa(0)={{\sqrt{3}} \over {(\sqrt{2})^3}}= {{\sqrt{6}} \over {4}}$.

{\bf Ans.} ${{\sqrt{6}} \over {4}}$.


{\bf Comments}: 1. Some people found the general $\kappa(t)$ and only at the end plugged-in $t=0$. This is correct,
but takes much longer. You should plug-in as soon as you can.

2. Some people didn't get that $t=0$.


\vfill\eject
{\bf 2.} (10 points) Find ${{\partial h} \over {\partial q}}$ at $(q,r)=(3,2)$ where
$h(u,v)=ue^v$, $u=q^3$, $v=qr^2$
\bigskip
{\bf Sol.}:

The {\bf type} of the answer is: {\bf number}.

\bigskip
$$
h_q=(h_u)(u_q)+(h_v)(v_q)=(e^v)(3q^2)+(ue^v)(r^2)= e^{qr^2} (3q^2)+ q^3 e^{qr^2} r^2=e^{qr^2}(3q^2+q^3r^2)
$$
Now we plug-in $q=3,r=2$ to get
$$
e^{3 \cdot 2^2}(3 \cdot 3^2+3^3 \cdot 2^2)=135 e^{12} \quad.
$$
{\bf Ans.}: $135 e^{12}$.

\vfill\eject
{\bf 3.} (10 points)
Find the directional derivative of $f(x,y,z)=xy+z^3$ at $P=(3,-2,-1)$ in the direction pointing to the
point $Q=(2,-3,-3)$.
\bigskip
{\bf Sol.}

The {\bf type} of the answer is: {\bf number}.

\bigskip
$$
\g f= \langle f_x, f_y, f_z \rangle = \langle y, x, 3z^2 \rangle  \quad .
$$
At the designated point (the point $(3,-2,-1)$) it is
$$
\g f= \langle f_x, f_y, f_z \rangle = \langle -2, 3, 3 \rangle  \quad .
$$
The direction is
$$
{\bf PQ}= \langle 2, -3,-3 \rangle -\langle 3, -2,-1 \rangle=\langle -1, -1,-2 \rangle
$$
and the {\bf unit directio} is 
${\bf u}=\langle -1, -1,-2 \rangle/|\langle -1, -1,-2 \rangle|$=
${{\langle -1, -1,-2 \rangle} \over {\sqrt{6}}}$.

The directional derivative is
$$
\g f \cdot {\bf u}= \langle -2, 3, 3 \rangle  \cdot  {{\langle -1, -1,-2 \rangle} \over {\sqrt{6}}}=
{{(-2)(-1)+(3)(-1)+3(-2)} \over {\sqrt{6}}} = {{-7} \over {\sqrt{6}}}={{-7 \sqrt{6}} \over {6}} \quad .
$$
{\bf Ans.}: ${{-7 \sqrt{6}} \over {6}}$.


\vfill\eject
{\bf 4.} (10 points)
Find an equation of the tangent plane at the given point
$$
H(s,t)=te^{s^2 t} \quad, (0,0) \quad .
$$
\bigskip


{\bf Sol.}: 

The {\bf type} of the answer is: Equation of a plane.

\bigskip
The relevant formula is
$$
z-z_0=H_s(s_0,t_0)(s-s_0)+H_t(s_0,t_0)(t-t_0) \quad .
$$
At $(0,0)$, $z_0=H(0,0)=0 e^{0}=0$. 

We have:

$$
H_s(s,t)=te^{s^2 t}(2st)=2st^2e^{s^2 t} \quad ,
$$

$$
H_t(s,t)=e^{s^2 t} + st^2e^{s^2 t} = (1+st^2)e^{s^2 t} \quad .
$$

At $(0,0)$ we have:

$H_s(0,0)=0e^{0^2 \cdot 0}(0)=0$, 

$H_t(0,0)=(1+0 \cdot 0^2)e^{0^2 \cdot 0}=1$.

So the equation is:
$$
z-0=0(s-0)+1(t-0)
$$
that becomes
$$
z=t \quad  .
$$

\bigskip
{\bf Ans.}: $z=t$, or $H=t$.
\bigskip
{\bf Comments}: The traditional variables are $x,y$, so people who wrote $z=y$ got 9.5 out of 10.

\vfill\eject

{\bf 5.} (10 points) Compute $f_{yy}(2,3)$ if $f(x,y)=x \ln (y^2)$.
\bigskip


{\bf Sol.}

The {\bf type} of the answer is: Number.
\bigskip
Before you differentiate, {\bf simplify!}. Since $\ln y^2= 2 \ln y$, by the $\ln$ rules,
$$
f(x,y)=2x \ln y \quad.
$$
So 
$$
f_y(x,y)={{2x} \over {y}}=2x y^{-1} \quad ,
$$
and
$$
f_{yy}(x,y)=2x(-1)y^{-2}= {{-2x} \over {y^2}} \quad.
$$
Finally, plugging in $x=2,y=3$, yields
$$
f_{yy}(2,3)={{-2 \cdot 2} \over {3^2}}=-{{4} \over {9}} \quad.
$$
{\bf Ans.}: $-{{4} \over {9}}$.

{\bf Comments}: Most people didn't do the first simplification, but instead did
$$
f_y(x,y)= {{(xy^2)'} \over {y^2}}= {{2yx} \over {y^2}}
$$
(here $'$ means diff. w.r.t. $y$). That's OK, it is only slighlty more complicated, 
if they immediately simplify this to
$$
{{2x} \over {y}}
$$
and proceed as before. But quite a few people didn'd do the algebraic simplification
and used the quotient rule for ${{2yx} \over {y^2}}$, and some even did it correctly, but
it took them longer than necessary.

\vfill\eject

%I am here

{\bf 6.} (10 points) Use the linearization of $f(x,y)=\sqrt{x+2y}$ to approximate
$f(3.01,2.96)$.
\bigskip


{\bf Sol.} 

The {\bf type} of the answer is: Number.

\bigskip
The relevant formula is 
$$
L(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) \quad ,
$$
where $(x_0,y_0)$ is ``nice''. Here both $3.01$ and $2.96$ are close to $3$, so $(x_0,y_0)=(3,3)$.

We have
$$
f(x,y)=(x+2y)^{1/2} \quad,
$$
So
$$
f_x(x,y)=(1/2)(x+2y)^{-1/2} \cdot 1= {{1} \over { 2\sqrt{x+2y} }}  \quad,
$$
$$
f_y(x,y)=(1/2)(x+2y)^{-1/2} \cdot 2= {{1} \over {\sqrt{x+2y} }}  \quad,
$$
So
$$
f_x(3,3)= {{1} \over { 2\sqrt{3+2 \cdot 3} }} = {{1} \over {6}}  \quad,
$$
$$
f_y(3,3)= {{1} \over { \sqrt{3+2 \cdot 3} }} = {{1} \over {3}}  \quad,
$$
Also
$$
f(3,3)=\sqrt{3 +2 \cdot 3}=\sqrt{9}=3 \quad,
$$
So
$$
L(x,y)=3 +{{1} \over {6}} (x-3)+{{1} \over {3}} (y-3) \quad .
$$
Finally, plugging in the actual arguments $x=3.01,y=2.96$, we get
$$
L(3.01,2.96)=3 +{{1} \over {6}} (3.01-3)+{{1} \over {3}} (2.96-3) 
=3 +{{1} \over {600}} -{{4} \over {300}} =3 -{{7} \over {600}}= {{1793} \over {600}} \quad .
$$
{\bf Ans.} ${{1793} \over {600}}$.


\vfill\eject
{\bf 7.} (10 points, altogether) Do the following limits exist? If they do, find them. Explain!
\bigskip



\bigskip
{\bf a.} (4 points)
$$
\lim_{(x,y,z) \rightarrow (0,0,0)} { {x^3+y^3+z^3}  \over {1+x+y+z}}
$$
{\bf Sol.}: 
The {\bf type} of the answer is {\bf number} .
\bigskip
The first thing to try is to {\bf plug-it in}:
$$
\lim_{(x,y,z) \rightarrow (0,0,0)} { {x^3+y^3+z^3}  \over {1+x+y+z}}=
{ {0^3+0^3+0^3}  \over {1+0+0+0}}={ {0}  \over {1}} =0 \quad .
$$
This makes perfect sense ($0$ on the top is no problem!), so the answer is that the limit exists and is $0$.

{\bf Ans.}: The limit exists and is $0$.

{\bf b.} (6 points)
$$
\lim_{(x,y) \rightarrow (0,0)} { {x^2+2y^2}  \over {x^2+3y^2}} \quad .
$$

{\bf Sol.}

The {\bf type} of the answer is {\bf DNE} .

\bigskip
{\bf First way}:  Let's see what is going on when we go to the origin along the $y$ axis:
$$
\lim_{(x,y) \rightarrow (0,0), x=0} { {2y^2}  \over {3y^2}}= 
\lim_{y \rightarrow 0} { {2}  \over {3}}=  {{2} \over {3}} \quad 
$$
And along the $x$ axis:
$$
\lim_{(x,y) \rightarrow (0,0), y=0} { {x^2}  \over {x^2}}= 
\lim_{x \rightarrow 0} { {1}  \over {1}}=  1 \quad 
$$
Since we got two {\bf different} values for the limit along different ways of approaching $(0,0)$ the
limit {\bf DNE} (Does Not Exist).

{\bf Second Way}. Let's see what is going along a typical line with slope $m$:
$$
\lim_{(x,y) \rightarrow (0,0), y=mx} { {x^2+2(mx)^2}  \over {x^2+3(mx)^2}}=
\lim_{x \rightarrow 0} { {(1+2m^2)x^2}  \over {(1+3m^2)x^2}}=
\lim_{x \rightarrow 0} { {(1+2m^2)}  \over {(1+3m^2)}}=
{ {(1+2m^2)}  \over {(1+3m^2)}} \quad .
$$
This this {\bf depends} on $m$, the limit {\bf DNE}.

\vfill\eject
{\bf 8.} (10 points) Find the length of the curve
$$
{\bf r}(t)= \langle \, t \, , \, 2 \sin t \, ,  \, 2 \cos t \, \rangle \quad 0 \leq t \leq 3 \quad .
$$
\bigskip


{\bf Sol.} 

The {\bf type} of the answer is: {\bf number}.
\bigskip
The relevant forumula is
$$
Arc Length= \int_{t_0}^{t_1} |{\bf r}'(t)| \, dt \quad  .
$$
We have
$$
{\bf r}'(t)= \langle \, 1 \, , \, 2 \cos t \, ,  \, -2 \sin t \, \rangle \quad ,
$$
So
$$
|{\bf r}'(t)|= \sqrt{1^2+ (2 \cos t)^2 + (-2 \sin t)^2}=
$$
$$
\sqrt{1^2+ 4\cos^2 t + 4 \sin^2 t}=\sqrt{1^2+ 4(\cos^2 t + \sin^2 t)}=
\sqrt{1^2+ 4}=\sqrt{5} \quad .
$$
So we have
$$
Arc Length= \int_{0}^{3} \sqrt{5} \, dt 
= \sqrt{5} \int_{0}^{3} \, dt = \sqrt{5} \cdot 3 = 3 \sqrt{5} \quad .
$$
{\bf Ans.} $3 \sqrt{5}$.



\vfill\eject
{\bf 9.} (10 points) A certain particle has acceleration
$$
{\bf a}(t) = \langle \, 12t^2 \, , \, 20t^3 \, , \, 30t^4 \, \rangle \quad ,
$$
and at $t=1$ its velocity is $\langle 4,5,6 \rangle$ and its position vector is $\langle 1,1,1 \rangle$, find
its velocity and position vector at time $t=2$.
\bigskip


{\bf Sol.}


The {\bf type} of the answer is: {\bf vectors of numbers}.
\bigskip
$$
{\bf v}(t) = \int \langle \, 12t^2 \, , \, 20t^3 \, , \, 30t^4 \, \rangle \, dt=
\langle \, 4t^3 \, , \, 5t^4 \, , \, 6t^5 \, \rangle + {\bf C} \quad .
$$
$$
{\bf v}(1) = \langle \, 4 \, , \, 5 \, , \, 6 \, \rangle + {\bf C} 
$$
On the other hand, from the problem ${\bf v}(1) = \langle \, 4 \, , \, 5 \, , \, 6 \, \rangle$, so
${\bf C}={\bf 0}$, and we have
$$
{\bf v}(t) = \langle \, 4t^3 \, , \, 5t^4 \, , \, 6t^5 \, \rangle \quad .
$$
{\bf Next}:
$$
{\bf r}(t) =
\int \langle \, 4t^3 \, , \, 5t^4 \, , \, 6t^5 \, \rangle \, dt
=
\langle \, t^4 \, , \, t^5 \, , \, t^6 \, \rangle \ + {\bf C} \quad .
$$
Once again ${\bf C}= {\bf 0}$, so{
$$
{\bf r}(t) =\langle \, t^4 \, , \, t^5 \, , \, t^6 \, \rangle  \quad .
$$
\bf Finally}, plugging-in $t=2$, we get
$$
{\bf v}(2) = \langle \, 4 \cdot 2^3 \, , \, 5 \cdot 2^4 \, , \, 6 \cdot 2^5 \, \rangle=
\langle \, 32 \, , \, 80 \, , \, 192 \, \rangle
$$
$$
{\bf r}(2) =\langle \, 2^4 \, , \, 2^5 \, , \, 2^6 \, \rangle
=\langle \, 16 \, , \, 32 \, , \, 64 \, \rangle
$$
{\bf Ans.}: The velocity when the time was $t=2$ was $\langle \, 32 \, , \, 80 \, , \, 192 \, \rangle$ and
the position vector was $\langle \, 16 \, , \, 32 \, , \, 64 \, \rangle$

\bigskip


\vfill\eject
{\bf 10.} (10 points) Find an equation to the plane that passes through the points $(1,1,1)$, $(2,2,0)$, $(0,3,0)$.
\bigskip


{\bf Sol.}


The {\bf type} of the answer is: Equation of a plane.
\bigskip
Let's call
$P=(1,1,1)$, $Q=(2,2,0)$, $R=(0,3,0)$,
$$
{\bf PQ}=Q-P=\langle 2-1,2-1,0-1 \rangle = \langle 1,1,-1 \rangle \quad ,
$$
$$
{\bf PR}=R-P=\langle 0-1,3-1,0-1 \rangle = \langle -1,2,-1 \rangle \quad .
$$

$$
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
1 & 1& -1\cr
-1 & 2 & -1 
} \right \vert \quad=
$$
$$
{\bf i} \left \vert \matrix{ 
1 & -1\cr
2 & -1 
} \right \vert 
-{\bf j} \left \vert \matrix{ 
1 & -1\cr
-1 & -1 
} \right \vert 
+{\bf k} \left \vert \matrix{ 
1 & 1 \cr
-1 & 2 
} \right \vert 
$$
$$
={\bf i}+2{\bf j}+{3\bf k}
=\langle 1,2,3 \rangle \quad.
$$

This is the {\bf normal vector}
${\bf n}=\langle a,b,c \rangle $. So 
${\bf n}=\langle a,b,c \rangle=\langle 1,2,3 \rangle$.
The equation is:
$$
a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \quad .
$$
where $(x_0,y_0,z_0)$ is {\bf any} of the three points (doesn't matter which).
Picking $(1,1,1)$ we get:
$$
1(x-1)+2(y-1)+3(z-1)=0 \quad ,
$$
and doing the algebra, we get:
$$
x+2y+3z=6 \quad .
$$

{\bf Ans.} $x+2y+3z=6$ .

{\bf Comment}: TO check your answer you should plug-in each of the three points into the equation
and see if you get a right statement.

$(1,1,1)$:  $1+2 \cdot 1+3 \cdot 1=6$ , OK!

$(2,2,0)$:  $2+2 \cdot 2+0 \cdot 1=6$ , OK!

$(0,3,0)$:  $0+2 \cdot 3+0 \cdot 1=6$ , OK!








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