Practice Problems Midterm 1

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Determine whether the following vectors are orthogonal (i.e. perpendicular), parallel, or neither.

$\textbf {v} = \left \langle 1,2\right \rangle$ and $\textbf {w} = \left \langle 4,-2 \right \rangle$ are
.
$\textbf {v} = \left \langle 1,2,3 \right \rangle$ and $\textbf {w} = \left \langle -1,-3,-2 \right \rangle$ are
.
$\textbf {v} = \left \langle 1,-1,0,1 \right \rangle$ and $\textbf {w} = \left \langle -1,1,0,-1 \right \rangle$ are
.
$\textbf {v} = \left \langle 2,3,-1 \right \rangle$ and $\textbf {w} = \left \langle 5,-3,1 \right \rangle$ are
.

Which of the following expressions make sense as a mathematical expression and which do not? Here, $\textbf {a},\textbf {b},\textbf {c},\textbf {d}$ are all three dimensional vectors, $\cdot$ denotes the dot product, and $\times$ denotes the cross product.

$\textbf {a} \cdot (\textbf {b}\times \textbf {c})$
$\textbf {a} \times (\textbf {b}\cdot \textbf {c})$
$\textbf {a} \times (\textbf {b} \times \textbf {c})$
$\textbf {a} \cdot (\textbf {b}\cdot \textbf {c})$
$(\textbf {a} \cdot \textbf {b}) \times (\textbf {c} \cdot \textbf {d})$
$(\textbf {a} \times \textbf {b}) \cdot (\textbf {c} \times \textbf {d})$
$(\textbf {a} \cdot \textbf {b}) \times (\textbf {c} \times \textbf {d})$

The acute angle between the lines $L_1:\,2x-y=3$ and $L_2:\,3x+y=7$ is $\answer {\pi /4}$ .

Find direction vectors for the lines. A direction vector for $L_1$ is $\textbf {v}_1 = \left \langle 1,\answer {2} \right \rangle$ , and a direction vector for $L_2$ is $\textbf {v}_2 = \left \langle 1,\answer {-3} \right \rangle$ . The cosine formula says $\textbf {v}_1 \cdot \textbf {v}_2 = \| \textbf {v}_1 \| \| \textbf {v}_2 \| \cos (\theta ).$ Notice $\textbf {v}_1\cdot \textbf {v}_2 = \answer {-5},$ $\| \textbf {v}_1\| = \answer {\sqrt {5}},\quad \| \textbf {v}_2 \| = \answer {\sqrt {10}}.$ Now use the formula to get $\theta = \answer {3\pi /4}$ . This is an obtuse angle, and the answer wanted an acute angle, so we do $\pi$ minus this angle to get an answer of $\pi /4$ .

If $\textbf {v}= \left \langle 3,0,-1 \right \rangle$ and $\textbf {w}$ is a vector such that the angle between $\textbf {v}$ and $\textbf {w}$ is $\pi /3$ , then $\| \text {proj}_{\textbf {w}}\textbf {v} \| = \answer {\sqrt {10}/2}$ .

The length of the projection is $\displaystyle \frac {|\textbf {v} \cdot \textbf {w}|}{\| \textbf {w} \|}$ .

Let $S$ be a cube of side length $l$ .

The (acute) angle between the main diagonal of the cube and one of its edges is equal to $\answer {\arccos (1/\sqrt {3})}$ .
If we say one of the vertices of the cube is at $(0,0,0)$ , then the main diagonal is given by $\left \langle l,l,l \right \rangle$ , and one of its edges is, say, $\left \langle l,0,0 \right \rangle$ .
The angle between the main diagonal and the diagonal of one of its faces is equal to (you can and should leave your answer as $\arccos$ of a number): $\answer {\arccos (\sqrt {2/3})}$ .

The area of the parallelogram with vertices $A(-2,1)$ , $B(0,4)$ , $C(4,2)$ , and $D(2,-1)$ is equal to $\answer {16}$ .

The area of the triangle with vertices $A(1,0,0)$ , $B(1,2,2)$ , and $C(3,0,1)$ is equal to $\answer {3}$ .

Consider the four points $P(-2,1,0)$ , $Q(2,3,2)$ , $R(1,4,-1)$ , and $S(3,6,1)$ . The volume of the parallelepiped spanned by $\overrightarrow {PQ}$ , $\overrightarrow {PR}$ , and $\overrightarrow {PS}$ is equal to $\answer {16}$ .

Are the vectors $\textbf {v} = \left \langle 1,5,-2 \right \rangle$ , $\textbf {w} = \left \langle 3,-1,0 \right \rangle$ , and $\textbf {u} = \left \langle 5,9,-4 \right \rangle$ coplanar (i.e. do they lie on the same plane)?
Are the points $A(1,3,2)$ , $B(3,-1,6)$ , $C(5,2,0)$ and $D(3,6,-4)$ coplanar?

If $\textbf {v}$ and $\textbf {w}$ are unit vectors with $\textbf {v} \cdot \textbf {w} = 1/2$ , then the angle between $\textbf {v}$ and $\textbf {w}$ is equal to $\answer {\pi /3}$ .

True or False (true means always true, otherwise it is false): If $\textbf {v}$ and $\textbf {w}$ are nonzero, three-dimensional vectors, then $\textbf {v} \times (\textbf {v} \times \textbf {w})$ must be parallel to $\textbf {w}$ .

The statement above is true if the angle between $\textbf {v}$ and $\textbf {w}$ is $\answer {\pi /2}$ .

Consider the line $L_1$ given by $\textbf {r}_1(t) = \left \langle 1+t,-2+3t,4-t \right \rangle$ and $\textbf {r}_2(t) = \left \langle 2t,3+t,-3+4t \right \rangle$ . Are the lines $L_1$ and $L_2$ parallel, intersecting, or skew?

Let $L$ be the line through $(1,1,1)$ and $(2,1,0)$ . At how many points does $L$ intersect the $xz$ -plane? $\answer {0}$ .

An equation of the plane through $(1,3,2)$ , $(3,-1,6)$ , and $(5,2,0)$ in the form $ax+by+cz=d$ is $\answer {6x+10y+7z}=50.$

Let $L$ be the line given by $\textbf {r}(t)=\left \langle 2+3t,-4t,5+t \right \rangle$ , and let $P$ be the plane through $(0,0,-9)$ perpendicular to $\left \langle 4,5,-2 \right \rangle$ . The point at which $L$ intersects $P$ is $(\answer {-4},\answer {8},\answer {3}).$

Consider the planes $P_1:\, x+y+z=1$ and $P_2:\, x-2y+3z=1$ .

The acute angle between the planes is $\answer {\arccos (2/\sqrt {42})}$ .
Find a vector equation for the line of intersection between the planes.
One such line is $\textbf {r}(t) = \left \langle 1,0,0 \right \rangle + t\left \langle 5,-2,-3 \right \rangle$ .

The distance from the point $(2,1,2)$ to the plane given by $6x+2y+3z=6$ is equal to $\answer {2}$ .

Are the planes $x+3y-2z=2$ and $5x+15y-10z=30$ parallel or intersecting?

For each of the following, give the answer in the form $ax+by+cz=d$ . Find an equation of the plane:

Passing through $(1,2,1)$ and normal to $\left \langle 2,-1,3 \right \rangle$ : $\answer {2x-y+3z} = 3.$
Passing through $(1,2,3)$ and parallel to $5x+2y+z=1$ : $\answer {5x+2y+z}=12.$
Containing the line $x = 1+t$ , $y = 2-t$ , $z=4-3t$ and parallel to $5x+2y+z=1$ : $\answer {5x+2y+z} = 13.$
Passing through $(1,2,3)$ and containing the line $x=3t$ , $y = 1+t$ , $z = 2-t$ : $\answer {x-2y}+z=0.$

Consider the origin $O(0,0,0)$ and the plane $P:\,x+y+z=1$ .

The distance from $O$ to the plane is $\answer {1/\sqrt {3}}$ .
The point on $P$ which is closest to $O$ is $(\answer {1/3},\answer {1/3},\answer {1/3})$ .

Parametrize the following curves. Remember to give bounds on the parameter if necessary.

$y=e^x+x^3$
Possible parametrization: $\textbf {r}(t) = \left \langle t,e^t+t^3 \right \rangle$ .
$x = y^2+\sin (y)$
Possible parametrization: $\textbf {r}(t) = \left \langle t^2+\sin (t),t \right \rangle$ .
$\displaystyle \frac {(x+9)^2}{25}+\frac {(y-1)^2}{16}=64$
Possible parametrization: $\textbf {r}(t) = \left \langle 40\cos (t)-9,32\sin (t) +1\right \rangle$ for $0 \leq t \leq 2\pi$ .
The line segment between the points $(0,1,1)$ and $(1,2,-1)$
Possible parametrization: $\textbf {r}(t) = \left \langle t,t+1,1-2t \right \rangle$ for $0 \leq t \leq 1$ .
The intersection of the surface $\displaystyle x^2+\frac {y^2}{4}-z=1$ and the plane $y=2$ .
Possible parametrization: $\textbf {r}(t) = \left \langle t,2,t^2 \right \rangle$ .
The intersection of the cylinder $x^2+y^2=1$ and the plane $x+y+z=1$ .
Possible parametrization: $\textbf {r}(t) = \left \langle \cos (t),\sin (t),1-\cos (t)-\sin (t) \right \rangle$ for $0 \leq t \leq 2\pi$ .
The intersection of the surfaces $x^2+y^2-z = 0$ and $y^2+z=1$ .
Possible parametrization: $\textbf {r}(t) = \left \langle \cos (t),\frac {\sin (t)}{\sqrt {2}},1-\frac {1}{2}\sin ^2(t) \right \rangle$ for $0 \leq t \leq 2\pi$ .
The line tangent to the curve $\textbf {c}(t) = \left \langle \sin (2t),e^{3t},\sqrt {1+t} \right \rangle$ at the points where $t=0$ .
Possible parametrization: $\textbf {r}(t) = \left \langle 2t,1+3t,1+\frac {t}{2} \right \rangle$ .
The intersection of the cylinder $x^2+y^2=4$ and the cone $z = \sqrt {x^2+y^2}$ .
Possible parametrization: $\textbf {r}(t) = \left \langle 2\cos (t),2\sin (t),2 \right \rangle$ for $0 \leq t \leq 2\pi$ .

The length of the curve $\textbf {r}(t) = \left \langle \cos ^3(t),\sin ^3(t) \right \rangle$ from $t =0$ to $t = \pi /2$ is $\answer {3/2}$ .

Suppose $\textbf {r}''(t) = \left \langle e^t,t,\sin (2t) \right \rangle$ , that $\textbf {r}'(0) = \left \langle 1,0,0 \right \rangle$ and $\textbf {r}(0) = \left \langle 0,0,0 \right \rangle$ . Find $\textbf {r}(\pi )$ . $\textbf {r}(\pi ) = \left \langle \answer {e^{\pi }-1},\answer {\frac {\pi ^3}{6}},\answer {\frac {\pi }{2}} \right \rangle .$

Consider the curve given by $\textbf {r}(t) = \left \langle t^2+2t,t^3-12t \right \rangle$ . At how many points is the tangent vector to the curve horizontal? $\answer {2}$ .

List the values in increasing order: $t = \answer {-2},\quad t = \answer {2}.$

Suppose a particle is moving with position $\textbf {r}(t) = \left \langle t^2,1-2t,t \right \rangle$ . At what value of $t$ is the speed of the particle smallest? $t = \answer {0}$ .

Consider the curves $\textbf {r}_1(t) = \left \langle t,1-t,3+t^2 \right \rangle$ and $\textbf {r}_2(t) = \left \langle 3-t,t-2,t^2 \right \rangle$ . At what point do the curves intersect? $(\answer {1},\answer {0},\answer {4}).$

Suppose $z = f(\textbf {r}(t))$ , where $f(x,y)$ is a differentiable function. If $\textbf {r}(2) = \left \langle 1,2 \right \rangle$ , $\textbf {r}'(2)=\left \langle 3,0 \right \rangle$ , and $\nabla f(1,2) = \left \langle 5,-3 \right \rangle$ , then $\frac {dz}{dt}\Big |_{t=2} = \answer {15}.$

Consider the function $f(x,y) = e^{y-x^2-1}$ .

For which values of $c$ does the level curve $f(x,y) =c$ have no points? It is for all $c<\answer {0}$ .
What do the level curves of $f(x,y)$ look like?

An equation for the tangent plane to the surface $z = y/x$ at the point $(1,2,2)$ is (in the form $ax+by+cz=d$ ): $\answer {-2x+y-z}=-2.$

An equation for the tangent plane to the surface given implicitly as $x^3y-xz^2 = 1-y^2z$ at the point $(1,1,1)$ is (in the form $ax+by+cz=d$ ): $\answer {2x+3y-z} = 4.$

Suppose $x^3y-xz^2=1-y^2z$ . Then $\displaystyle \frac {\partial x}{\partial y}$ at the point $(1,1,1)$ is equal to $\answer {-3/2}$ .

For each of the following, calculate the partials $f_x$ , $f_y$ (and $f_z$ where appropriate).

$f(x,y) = \sin (x^2y)$ . $f_x = \answer {2xy\cos (x^2y)} ,\quad f_y = \answer {x^2\cos (x^2y)}.$
$\displaystyle f(x,y) = \tan (xy)+e^{xy^2}+\frac {\ln (y)}{x}$ . $f_x = \answer {y\sec ^2(xy)+y^2e^{xy^2}-\frac {\ln (y)}{x^2}} ,\quad f_y = \answer {x\sec ^2(xy)+2xye^{xy^2}+\frac {1}{xy}}.$
$\displaystyle f(x,y,z) = \frac {y\sin (xz)}{z}$ . $f_x = \answer {y\cos (xz)} ,\quad f_y = \answer {\frac {\sin (xz)}{z}},\quad f_z = \answer {\frac {y(xz\cos (xz)-\sin (xz))}{z^2}}.$
$f(x,y,z) = x^{yz}$ (restricted to the domain where $x>0$ ). $f_x = \answer {yzx^{yz-1}} ,\quad f_y = \answer {zx^{yz}\ln (x)},\quad f_z = \answer {yx^{yz}\ln (x)}.$

If $f(x,y) = e^{xy^2}$ , find $f_{xxy}$ . That is, find $\displaystyle \frac {\partial ^3 f}{\partial y \partial x \partial x}$ . $f_{xxy} = \answer {4y^3e^{xy^2}+2xy^5e^{xy^2}}.$

Let $S$ be the surface consisting of the set of points equidistant from the point $(0,0,1)$ and the plane $z=2$ .

An equation for the surface is $z = \answer {-\frac {x^2}{2}-\frac {y^2}{2}+\frac {3}{2}}.$
Let $(x,y,z)$ be on $S$ . Calculate the distance from $(x,y,z)$ to $(0,0,1)$ and from $(x,y,z)$ to $z=2$ . Set them equal and solve for $z$ .
The surface is a:

Suppose a surface $S$ contains the parametric curves $\textbf {r}_1(t) = \left \langle 1+5t,3-t^2,2+t-t^3 \right \rangle$ and $\textbf {r}_2(t) = \left \langle 3s-2s^2,s+s^3+s^4,s-s^2+2s^3 \right \rangle$ . An equation for the tangent plane to $S$ at $(1,3,2)$ is (in the form $ax+by+cz=d$ ): $\answer {-4x-13y+20z} = -3.$

Find two tangent vectors and cross them to get the normal vector for the tangent plane.

Let $\displaystyle f(x,y) = \frac {\sin ^2(x)}{y^2}$ .

Which of the following describes the level curve $f(x,y)=1$ ?
The partials $\frac {\partial f}{\partial x} = \answer {\frac {2\sin (x)\cos (x)}{y^2}},\quad \frac {\partial f}{\partial y} = \answer {\frac {-2\sin ^2(x)}{y^3}},\quad \frac {\partial ^2 f}{\partial y\partial x} = \answer {\frac {-4\sin (x)\cos (x)}{y^3}}.$
The equation of the tangent plane to $z=f(x,y)$ at the point $(\pi /4,1)$ is (in the form $ax+by+cz=d$ ): $\answer {x-y}-z = \answer {-\frac {3}{2}+\frac {\pi }{4}}.$
What is the rate of change of $f(x,y)$ at $(\pi /4,1)$ in the direction of $\textbf {u} = \left \langle 2,-1 \right \rangle$ ? $\answer {3/\sqrt {5}}$
Use the directional derivative formula.
The unit vector direction in which $f(x,y)$ increases most rapidly at the point $(\pi /4,1)$ is: $\left \langle \answer {\frac {1}{\sqrt {2}}},\answer {-\frac {1}{\sqrt {2}}} \right \rangle .$

Let $S$ be the surface consisting of all points $P(x,y,z)$ for which the distance between $P$ and the $x$ -axis is twice the distance from $P$ to the $yz$ -plane. The surface $S$ is a:

The equation of the surface is $\answer {4x^2-y^2}-z^2=0$ .

The function $f(x,y) = e^x\sin (y)$ satisfies which of the following equations (select all that apply).

The function $u(x,t)=\sin (x-at)+\ln (x+at)$ , where $a$ is a constant, satisfies which of the following equations (select all that apply).

If $z = x^2y+2xy^4$ and $x = \sin (2t)$ , $y = \cos (t)$ , then $\displaystyle \frac {dz}{dt}$ at $t=0$ is equal to $\answer {4}$ .

The temperature at a point is given by $T(x,y)$ , measured in Celsius. A bug crawls so that its position at $t$ seconds is $x = \sqrt {1+t}$ and $y = 2+\frac {1}{3}t$ , where $x$ and $y$ are in centimeters. The temperature satisfies $T_x(2,3) = 4$ and $T_y(2,3) = 3$ . How fast is the temperature rising on the bug’s path after $3$ seconds? $\answer {2}$

If $w(x,y,z) = xy+yz+xz$ and $x = r\cos (\theta )$ , $y = r\sin (\theta )$ , and $z = r\theta$ , then at the point where $r = 2$ and $\theta = \pi /2$ : $\frac {\partial w}{\partial r}\Big |_{(r,\theta ) = (2,\pi /2)} = \answer {2\pi },\quad \frac {\partial w}{\partial \theta }\Big |_{(r,\theta ) = (2,\pi /2)} = \answer {-2\pi }.$

At which points is the direction of fastest increase of the function $f(x,y) = x^2+y^2-2x-4y$ equal to $\displaystyle \frac {1}{\sqrt {2}}\textbf {i}+\frac {1}{\sqrt {2}}\textbf {j}$ :

Do the three points $A(2,4,2)$ , $B(3,7,-2)$ , and $C(1,3,3)$ lie on one line?

How many lines are there which pass through $(1,1,0)$ and are perpendicular to the vector $\left \langle 4,6,-2 \right \rangle$ ?

The set of all such lines forms a:

Consider the function $f(x,y) = x^3-xy+y^2$ . How many critical points does the function have? $\answer {2}$ .

The points happen at $x=0$ and $x = \answer {1/6}$ .

The point $(0,0)$ is a and the point $(1/6,1/12)$ is a .

The absolute maximum of the function $f(x,y) = x-y^2$ on the region $x^2+y^2 \leq 1$ is equal to $\answer {1}$ .

Consider $f(x,y) = x^3y-3xy^2$ . How many critical points does the function have? $\answer {1}$ .

The point happens at $x=\answer {0}$ .

The point $(0,0)$ is a .

Let $f(x,y) = x^3+y^2-3x-2y$ . How many critical points does the function have? $\answer {2}$ .

The points happen at $x=\answer {-1}$ and $x=1$ .

The point $(-1,\answer {1})$ is a and the point $(1,\answer {1})$ is a wordChoicelocal maximumlocal minimumsaddle point.

If $f(x,y) = x^3+y^2-3x-2y$ , find the absolute maximum and minimum of $f(x,y)$ on the rectangle with vertices $(0,0)$ , $(1,0)$ , $(1,1)$ , and $(0,1)$ . $\text {Absolute max is}: \answer {0},\quad \text {Absolute min is}: \answer {-3}.$

Consider the function $f(x,y) = 4x+4y-x^2-y^2$ . On the disc $x^2+y^2 \leq 2$ , the absolute maximum value is $\answer {6}$ and the absolute minimum value is $\answer {-10}$ .

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