Problem 1
Let $w=\sin (2x-y), \; x=r+\sin s, \; y = rs$.
$$
\frac{\partial w}{\partial s} \quad \text{ at } \quad (r,s)=(\pi,0).
$$
Solution:
\begin{align*}
w'_{s} &= (\sin (2x-y) )'_{s} = \cos (2x-y) \cdot (2x'_{s}-y'_{s})\\
&=\cos(2\cdot (r+\sin s) -rs) \cdot (2\cdot \cos s-r)
\end{align*}
Thus
\begin{align*}
w'_{s}(\pi,0) =\cos (2 \pi) \cdot (2-\pi)=2-\pi
\end{align*}
Problem 2
(a)
Find (give a mathematical description) and sketch the domain of the function
$$
f(x,y)=\ln(2x+y)
$$
Solution:
The function $\ln (2x+y)$ is properly defined when $2x+y >0$.
So you need to indicate this domain which is the halfplane located on one side of the line $2x+y=0$
(b)
Find all possible constants $c$ such that the function above satisfies the equation: $f''_{yy}-c^{2}f''_{xx}=0$
Solution:
See the solution to the
Problem 2 from the Quiz 4
Problem 3
Find the arc length of the function over the given interval:
\begin{align*}
\vec{r}(t)=(3 \cos t)\vec{i}+(3 \sin t)\vec{j}+2t^{3/2}\vec{k} \qquad 0\leq t\leq 3
\end{align*}
Solution:
\begin{align*}
\text{length} &= \int_{0}^{3}|\vec{r}'(t)|dt = \int_{0}^{3}\sqrt{ (-3\sin t)^{2}+(3\cos t)^{2}+(3t^{1/2})2 }dt=\\
&=\int_{0}^{3}\sqrt{9\sin^{2}t+9\cos^{2} t+9t}dt = \int_{0}^{3}\sqrt{9+9t}dt=\\
&=3 \int_{0}^{3}\sqrt{1+t}=2(1+t)^{3/2}|_{0}^{3}=2\cdot 4^{3/2}-2=2\cdot 8-2=14.
\end{align*}
Problem 4
(a)
Let $ f(x,y)=\frac{5}{x^{2}+y^{2}}.$ Find the equation of the tangent plane to $f(x,y)$ at $(-1,2)$.
Solution:
Equation of the tangent plane is
\begin{align*}
\langle f'_{x}(-1,2), f'_{y}(-1,2),-1 \rangle\cdot \langle x+1,y-2,z-f(-1,2)\rangle=0 \quad \text{or}\\
f'_{x}(-1,2)(x+1) + f'_{y}(-1,2)(y-2)-(z-f(-1,2))=0 \quad
\end{align*}
but
\begin{align*}
&f'_{x} =\frac{-10x}{(x^{2}+y^{2})^{2}} \quad \text{so} \quad f'_{x}(-1,2)=\frac{10}{(1+4)^{2}}=\frac{2}{5}\\
&f'_{y} = \frac{-10y}{(x^{2}+y^{2})^{2}} \quad \text{so} \quad f'_{y}(-1,2) = \frac{-20}{(1+4)^{2}}=-\frac{4}{5}\\
&f(-1,2)=\frac{5}{1+4}=1
\end{align*}
Thus we have
\begin{align*}
f'_{x}(-1,2)(x+1) + f'_{y}(-1,2)(y-2)-(z-f(-1,2)) = \frac{2}{5}(x+1) -\frac{4}{5}(y-2)-(z-1)=0.
\end{align*}
(b)
Use linear approximation to approximate $f(-1.05,2.1)$
Solution:
At point $(-1,2)$ we constructed equation of the tangent plane which is the best linear approximation of the of our function at that point.
They are asking us to find the approximate value at neighborhood, namely, at point $(-1.05, 2.1).$ Let's use the linear approximation of our function, tangent plane
that we have constructed in previous question. We have
$$
z= \frac{2}{5}(x+1) -\frac{4}{5}(y-2)+1
$$
On the other hand we know that $f(x,y)\approx z(x,y)$ (because $z$ is the linear approximation of $f$ (because it is tangent plane at that point)).
Thus $f(-1.05, 2.1)\approx z(-1.05, 2.1)$.
\begin{align*}
z(-1.05,2.1)=\frac{2}{5}(-1.05+1) -\frac{4}{5}(2.1-2)+1=\frac{2}{5}(-0.05) -\frac{4}{5}(0.1)+1 = \frac{-0.1-0.4}{5}+1=\frac{-0.5}{5}+1=1-\frac{1}{10}=\frac{9}{10}
\end{align*}
Problem 5
(a) Find the position vector $\vec{r}$, and the velocity vector, $\vec{v}$, of a ball that moves with acceleration vector function:
$\vec{a}(t) = \langle 0,-32\rangle ft/sec^{2}$, initial velocity $\vec{v}(0)=\langle5,2\rangle ft/sec$ and
initial position $\vec{r}(0)=\langle 0,3 \rangle ft.$ Please give the correct units for your answers.
(b) Find the time $T$ when the ball hits the ground and find the horizontal distance it has traveled at the time $T$. (Please give the correct units).
Solution (a):
We have a differential equation $\vec{r}''(t) = \vec{a}(t)=\langle 0,-32\rangle$. Solution of this differential equation is (after two times of integration)
\begin{align*}
\vec{r}(t) = \langle 0,-32\rangle \frac{t^{2}}{2}+\vec{v}(0)t + \vec{r}(0)
\end{align*}
Therefore
\begin{align*}
\vec{r}(t) = \langle 0,-32\rangle \frac{t^{2}}{2}+\langle 5,2\rangle t + \langle 0,3 \rangle.
\end{align*}
Solution (b):
Ball hits the ground if and only if the vertical component of $\vec{r}(t)$ becomes zero. But it is zero if and only if when
\begin{align*}
-\frac{32}{2}t^{2}+2t+3&=0\\
-16t^{2}+2t+3&=0
\end{align*}
This quadratic equation has two solutions $t=\frac{1}{2}$ and $t=-\frac{3}{8}$. Time must be positive, therefore $t=\frac{1}{2} \; sec$.
Horizontal distance will be the value of the horizontal component of $\vec{r}(t)$ at that time i.e.
\begin{align*}
\text{distance} = 5 \cdot \frac{1}{2}=2.5 \; ft.
\end{align*}
Problem 6
Given the function $f(x,y)=4x^{2}+y^{2}-6x+1$.
(a) Find $\nabla f$
(b) Find the critical points.
(c) Find all local extreme values and the corresponding points where these values are obtained.
(d) Find the absolute maximum and the absolute minimum of $f(x,y)$ over the region $D = \{ (x,y) \; : \; x^{2}+y^{2}\leq 1 \}$
Solution (a):
\begin{align*}
\nabla f = \langle f'_{x},f'_{y} \rangle =\langle 8x-6, 2y \rangle
\end{align*}
Solution (b):
\begin{align*}
\nabla f = 0 \quad \text{whence} \quad y=0, \; x=\frac{6}{8}=\frac{3}{4}.
\end{align*}
So there is only one critical point $\left(\frac{3}{4},0\right)$
Solution (c):
We will show that it is local minimum. Indeed $f''_{xx} = 8 >0$ and $f''_{xx}f''_{yy}-(f''_{xy})^{2} =8\cdot 2 - 0 >0$.
Thus miniumum value will be
$$
f\left(\frac{3}{4},0\right)=4\cdot \left(\frac{3}{4}\right)^{2}-6\cdot\frac{3}{4}+1=-\frac{5}{4}
$$
Solution (d):
We know that inside the domain $D$ our function attains its local minimum, namely value $-\frac{5}{4}$. But we don't know whether it is minimum value \textcolor{blue}{globally}.
In order to check it, we need to calculate minimum value on the boundary of the domain $D$ and compare its value to the value $-\frac{5}{4}$.
\textcolor{blue}{What about the global maximum?}
Since we have nothing inside the domain (local maximum points), therefore global maximum is going to attain only on the boundary of the domain $D$. Thus in order to find the global maximum
it is enough to find maximum value of our function on the boundary of $D$.
Firstly, lets find the minimum value on the boundary of $D$.
On the boundary we have $x^{2}+y^{2}=1$. Therefore $y^{2}=1-x^{2}$. Let's plug $y^{2}$ into our function $f(x,y)$. So we will get
$f(x,y)=4x^{2}+y^{2}-6x+1 = 4x^{2}+(1-x^{2})-6x+1 = 3x^{2}-6x+2$. The fact that we walk on the boundary means that $-1 \leq x \leq 1$. But on this interval
the function $g(x)=3x^{2}-6x+2$ decrease. Indeed $g'(x) = 6x-6 = 6(x-1) \leq 0$. Therefore, it has maximum value at the point $-1$ which is $g(-1) = 11$ and minimum value at point $1$ which is
$g(1)=-1$. Note that minimum value on the boundary is bigger than the minimum value inside the domain $D$ which is $-5/4$ therefore $-5/4$ is global minimum.
Finally, as we already said the global maximum value is attained on the boundary and its value is $11$.
Problem 7
Let $f(x,y) = x^{2}e^{-3y}$. Find the unit direction $\vec{u}$ such that the function decreases most rapidly at the point $(1,0)$. Please epxlain your answer.
Solution:
Take the function $-f$. Then its gradients direction $\nabla (-f)$ shows the direction where the function $-f$
increase the most rapidly, which in other words means that the fucntion $f$ decrease the most rapidly. Howevere this direction is
\begin{align*}
\frac{\nabla (-f)}{|\nabla (-f)|} = \frac{-\nabla f}{|-\nabla f|} = -\frac{\nabla f}{|\nabla f|}
\end{align*}
So at point $(1,0)$ this direction will be
\begin{align*}
-\frac{\nabla f (1,0)}{|\nabla f(1,0)|} = -\frac{\langle 2,-3 \rangle }{\sqrt{2^{2}+(-3)^{2}}}=\frac{\langle -2,3\rangle}{\sqrt{13}}
\end{align*}