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# Vector calculus: exercises

## Exercises

Introduction

Exercise 1. Describe the curve which results from the vector valued function

$$r(t) = ( {\rm cos \hspace{3pt}} 2t, {\rm sin \hspace{3pt}} 2t, t )$$

where $t \in {\bf R}$.

Solution: The first two components indicate that for

$$r(t) = ( x(t), y(t), z(t) ),$$

the pair $( x(t), y(t) )$ traces out a circle. While it is doing so, $z(t)$ is moving at a steady rate in the positive direction. Therefore, the curve which results is a cork screw shape, i.e. a helix.

Exercise 2. The position of a particle at time $t$ is $( x, y )$, where

$$x = {\rm \hspace{3pt} sin \hspace{3pt}} t,$$

$$y = {\rm sin}^2 t.$$

Describe the motion of the particle as t varies over the time interval $[ a, b ]$.

Solution: We can eliminate $t$ to see that the motion of the object takes place on the parabola $y = x^2$. The orientation of the curve is from $( {\rm sin \hspace{3pt}} a, {\rm sin}^2 a )$ to $( {\rm sin \hspace{3pt}} b, {\rm sin}^2 b)$.

Limits

Exercise 1. Find the limit

$$\displaystyle\lim_{x \rightarrow 1} \frac{x^2 - x}{x - 1}.$$

Solution: It is

$$\frac{x \cdot ( x - 1 )}{x - 1} = x {\rm \hspace{3pt} for \hspace{3pt}} x \neq 1.$$

Hence

$$\displaystyle\lim_{x \rightarrow 1} \frac{x \cdot ( x - 1 )}{x - 1} = \displaystyle\lim_{x \rightarrow 1} x = 1.$$

Exercise 2. Find the limit

$$\displaystyle\lim_{x \rightarrow \infty} \frac{x}{1 + x}.$$

Solution: Rewrite

$$\frac{x}{1 + x} = \frac{1}{1 + \frac{1}{x}}.$$

Now it is

$$\displaystyle\lim_{x \rightarrow \infty} \frac{1}{1 + x} = 1 \neq 0.$$

Therefore,

$$\displaystyle\lim_{x \rightarrow \infty} \frac{x}{1 + x} = \displaystyle\lim_{x \rightarrow \infty} \frac{1}{1 + \frac{1}{x}} = \frac{1}{1} = 1.$$

Continuity

Exercise 1. Consider $f: {\bf R} {\rightarrow} {\bf R}^3$. Is

$$f( t ) = ( t^2 + 1, {\rm cos}( 2t ), {\rm sin}( 3t) )$$

continuous?

Solution: All components are continuous for all $t \in {\bf R}$, so $f(t)$ is continuous on ${\bf R}$.

Exercise 2. Let $f: {\bf R} {\rightarrow} {\bf R}^2$. Is

$$f( t ) = ( ( t + 1 )^{\frac{1}{2}}, {\rm tan}( t ) )$$

continuous?

Solution: The first component is defined and continuous for

$$t \geq -1,$$

while the second component is defined and continuous for

$$t \neq \frac{\pi}{2} + k {\pi}, k \in {\bf Z}.$$

Hence it is continuous on ${\bf R}$ for

$t \geq -1$ and $t \neq \frac{\pi}{2} + k {\pi}, k \in {\bf Z}.$

Exercise 3. Let $f: {\bf R} {\rightarrow} {\bf R}^3$,

$$f( t ) = \left( \frac{1}{t^2 - 1}, ( 1 - t^2)^{\frac{1}{2}}, \frac{1}{t} \right).$$

If $f$ continuous?

Solution: The first component is defined and continuous for

$$t \neq \pm 1.$$

The second component is defined and continuous for

$$t \in [ -1, 1 ],$$

while the third component is defined and continuous for

$$t \neq 0.$$

Hence, $f$ is continuous on ${\bf R}$ for

$$t \in ( -1, 0 ) \cup ( 0, 1 ).$$

Exercise 4. Let $g: {\bf R} {\rightarrow} {\bf R}^4$,

$$g( t ) = ( {\rm cos}(4t), 1 - ( 3t + 1 )^{\frac{1}{2}}, {\rm sin}(5t), {\rm sec}(t) ).$$

If $g$ continuous?

Solution: The first and the third component are continuous on ${\bf R}$. The second component is continuous and defined on $( -\frac{1}{3}, \infty )$, while the secant ${\rm \hspace{3pt} sec}(t) = \frac{1}{{\rm cos}(t)}$ in the fourth component is defined and continuous for

$$t \neq \frac{\pi}{2} + k{\pi}, k \in {\bf Z}.$$

Hence, $g(t)$ is continuous as long as

$t \geq -\frac{1}{3}$ and $t \neq \frac{\pi}{2} + l{\pi}, l \in {\bf Z}.$

Tangents

Exercise 1. Let

$f( t ) = ( {\rm \hspace{3pt} sin \hspace{3pt}} t, t^2, t + 1 )$ for $t \in [ 0, 5 ].$

Find a tangent line to the curve parameterized by $f$ at the point $t=2$.

Solution: A direction vector has the same direction as $f'(2)$. Therefore, it suffices to simply use $f'(2)$ as a direction vector for the line. Further

$$f'(2) = ( {\rm cos \hspace{3pt}} 2, 4, 1 ).$$

Hence, a parameterized equation for the tangent line is

$$( x, y, z ) = ( {\rm sin \hspace{3pt}} 2, 4, 3) + t ( {\rm cos \hspace{3pt}} 2, 4, 1).$$

Exercise 2. Let

$f( t ) = ( {\rm \hspace{3pt} sin \hspace{3pt}} t, t^2, t + 1 )$ for $t \in [ 0, 5 ].$

Find the velocity vector for $t = 1$.

Solution: The velocity vector is simply $f'(1) = ( {\rm cos \hspace{3pt}} 1, 2, 1 )$.

Exercise 3. Consider $g: {\bf R} {\rightarrow} {\bf R}^2$,

$$g( t ) = ( t, t^2 ).$$

What is its velocity, speed and acceleration as it passes through $( 2, 4 )$?

Solution: It is

$g'( t ) = ( 1, 2t )$ and
$g' '( t ) = ( 0, 2 )$.

Further, at $( 2, 4 )$, we know $t = 2$. For the velocity at $t = 2$, we have $g'(2) = ( 1, 4 )$. The speed equals

$$|| g'(2) || = ( 1^2 + 4^2 )^{\frac{1}{2}} = \sqrt{17},$$

whereas the acceleration is equal to $g' '(2) = ( 0, 2 )$.

Exercise 4. An object has position

$r(t) = ( t^3, \frac{t}{1 + t}, ( t^2 + 2 )^{\frac{1}{2}} )$ kilometers,

where $t$ is given in hours. Find the velocity of the object in kilometers per hour when $t = 1$.

Solution: Since velocity at time t equals $r'(t)$, we calculate

$$\begin{array}{} r'(t) &= \left( 3t^2, \frac{1 ( 1 + t ) - t}{( 1 + t )^2}, 2t \cdot \frac{1}{2} ( t^2 + 2 )^{-\frac{1}{2}} \right) \\ &= \left( 3t^2, \frac{1}{( 1 + t )^2}, \frac{1}{( t^2 + 2 )^{\frac{1}{2}}} t \right). \end{array}$$

For $t = 1$, the velocity is hence

$r'(1) = \left( 3, \frac{1}{4}, \frac{1}{\sqrt{3}} \right)$ kilometers per hour.

Exercise 5. Consider $g: {\bf R} {\rightarrow} {\bf R}^2$,

$$g(t) = ( \frac{1}{t}, 2t )$$

defined on $( 0, \infty )$. Its image is one branch of a hyperbola (this can be seen by writing $g(t) = ( x, y )$, i.e. $x = \frac{1}{t}, y = 2t$, which yields $y = \frac{2}{x})$. Find the veolcity, the speed and the acceleration at time $t$.

Solution. The velocity equals

$$g'(t) = ( -\frac{1}{t^2}, 2 ),$$

the acceleration is equal to

$$g' '(t) = ( \frac{2}{t^3}, 0 ),$$

and the speed is

$$|| g'(t) || = || ( -\frac{1}{t^2}, 2 ) || = ( \frac{1}{t^4} + 4 )^{\frac{1}{2}}.$$

Algebraic Properties of the derivative

Exercise 1. For the function

$$f( t ) = ( 3t {\rm \hspace{3pt} cos}( 2t ), 4t {\rm \hspace{3pt} sin}( 2t ) ),$$

calculate $f'(t)$.

Solution: Apply the product rule for both components:

$$f'(t) = ( 3 {\rm \hspace{3pt} cos}( 2t ) - 6t {\rm \hspace{3pt} sin} ( 2t ), 4 {\rm \hspace{3pt} sin}( 2t ) + 8t {\rm \hspace{3pt} cos} ( 2t ) ).$$

Exercise 2. For the function

$$f( t ) = ( 3t {\rm \hspace{3pt} cos}( 2t ), 4t {\rm \hspace{3pt} sin}( 2t ) ),$$

show that

$$\frac{d}{dt} || f(t) || \neq || f'(t) || .$$

Solution: It is

$$\begin{array}{} || f(t) || &= ( 9t^2 {\rm \hspace{3pt} cos}^2(2t) + 16t^2 {\rm \hspace{3pt} sin}^2(2t) )^{\frac{1}{2}} &= ( 9t^2 + 7t^2 {\rm \hspace{3pt} sin}^2(2t) )^{\frac{1}{2}}, \end{array}$$

and its derivative equals

$$\frac{d}{dt} || f(t) || = \frac{9t + 14t^2 {\rm \hspace{3pt} sin}(2t) {\rm \hspace{3pt} cos}(2t) + 7t sin^2(2t)}{9t^2 + 7t^2 {\rm sin}^2(2t) )^{\frac{1}{2}}}.$$

Further

$$f'(t)= ( 3{\rm \hspace{3pt} cos}(2t) - 6t {\rm \hspace{3pt} sin}(2t), 4{\rm \hspace{3pt} sin}(2t) + 8t {\rm \hspace{3pt} cos}(2t) ),$$

and thus

$$\begin{array}{} || f'(t) || &= ( ( 3 {\rm \hspace{3pt} cos}(2t) - 6t {\rm \hspace{3pt} sin}(2t) )^2 + (4 {\rm \hspace{3pt} sin}(2t) + 8t {\rm \hspace{3pt} cos}(2t))^2 )^{\frac{1}{2}} \\ &= ( 9 + 7 {\rm \hspace{3pt} sin}^2(2t) + 36 t^2 + 28 t^2 {\rm \hspace{3pt} cos}^2(2t) + 14t {\rm \hspace{3pt} sin}(4t) )^{\frac{1}{2}}. \end{array}$$

Now evaluating both functions at $t = {\pi}$ (for example) yields

$$(\frac{d}{dt} || f(t) ||)_{t= \pi} = \frac{9 \pi}{ \sqrt{9 \pi ^2}} = 3,$$

whereas

$$|| f'( \pi ) || = \sqrt{9 + 64{\pi}^2} \approx 25.31,$$

hence $\frac{d}{dt} || f(t) || \neq || f'(t) ||$.

Exercise 3. Let

$$r(t) = ( t^2, {\rm \hspace{3pt} sin \hspace{3pt}} t, {\rm \hspace{3pt} cos \hspace{3pt}} t )$$

and let

$$p(t) = ( t, {\rm \hspace{3pt} ln}( t + 1 ), 2t ).$$

Find $( r(t) \times p(t) )'$.

Solution: From the cross-product rule, it is

$$( r(t) \times p(t) )' = ( 2t, {\rm \hspace{3pt} cos \hspace{3pt}} t, -{\rm \hspace{3pt} sin \hspace{3pt}} t) \times ( t, {\rm \hspace{3pt} ln}( t + 1 ), 2t ) + ( t^2, {\rm \hspace{3pt} sin \hspace{3pt}} t, {\rm \hspace{3pt} cos \hspace{3pt}} t ) \times ( 1, \frac{1}{t + 1}, 2 ).$$

Vector fields

Exercise 1. Let

$$F = yi - xj + 2k = ( y, -x, 2 )$$

be a vector field. Verify that the path

$$g(t) = ( {\rm \hspace{3pt} sin \hspace{3pt}} t, {\rm \hspace{3pt} cos \hspace{3pt}} t, 2t )$$

is a flow line (i.e. a path such that the velocity along the path is a vector in the vector field, $g'(t) = F(g(t)) )$ for the vector field $F$.

Solution: With

$$( x, y, z ) = ( {\rm \hspace{3pt} sin \hspace{3pt}} t, {\rm \hspace{3pt} cos \hspace{3pt}} t, 2t )$$

it follows that

$$g'(t) = ( {\rm \hspace{3pt} cos \hspace{3pt}} t, -{\rm \hspace{3pt} sin \hspace{3pt}} t, 2 ),$$

and

$$F(g(t)) = ( y, -x, 2t ) = ( {\rm \hspace{3pt} cos \hspace{3pt}} t, -{\rm \hspace{3pt} sin \hspace{3pt}} t, 2t ).$$

Lengths as integrals

Exercise 1. Consider the path $g: {\bf R} {\rightarrow} {\bf R}^2$,

$$g(t) = ( {\rm \hspace{3pt} cos}^3 t, {\rm \hspace{3pt} sin}^3 t ),$$

which describes an astroid. For $t \in [ 0, \frac{\pi}{2} ]$, one fourth of the astroid is described. Calculate its arclength.

Solution: With

$$g'(t) = ( -3 {\rm \hspace{3pt} sin \hspace{3pt}} t {\rm \hspace{3pt} cos}^2 t, 3 {\rm \hspace{3pt} sin}^2t {\rm \hspace{3pt} cos \hspace{3pt}} t ),$$

$$|| g'(t) || = 3 {\rm \hspace{3pt} sin \hspace{3pt}} t {\rm \hspace{3pt} cos \hspace{3pt}} t = \frac{3}{2} sin 2t,$$

it is

$$\begin{array}{} \displaystyle\int_0^{\frac{\pi}{2}} || g'(t) || dt &= \displaystyle\int_0^{\frac{\pi}{2}} \frac{3}{2} {\rm \hspace{3pt} sin \hspace{3pt}} 2t dt \\ &= -\frac{3}{4} {\rm \hspace{3pt} cos}(2t) |_0^{\frac{\pi}{2}} \\ &= -\frac{3}{4} {\rm \hspace{3pt} cos \hspace{3pt}}{\pi} + \frac{3}{4} {\rm \hspace{3pt} cos \hspace{3pt}} 0 = \frac{3}{2}. \end{array}$$

Exercise 2. Consider the path $f: {\bf R} {\rightarrow} {\bf R}^2$,

$$f(t) = ( t^2, \frac{2}{3} ( 2t + 1 )^{\frac{3}{2}} ), 0 \leq t \leq 4.$$

Calculate its arclength.

Solution: It is

$$\begin{array}{} \displaystyle\int_0^4 || f'(t) || dt &= \displaystyle\int_0^4 ( 4t^2 + 4( 2t + 1 ) )^{\frac{1}{2}} dt &= \displaystyle\int_0^4 2( t + 1 ) dt &= t^2 + 2t |_0^4 &= 24. \end{array}$$

Exercise 3. Find the arclength of the helix

$$g(t) = ( {\rm cos \hspace{3pt}} t, {\rm \hspace{3pt} sin \hspace{3pt}} t, t )$$

from $t = 0$ to $t = 2{\pi}$.

Solution: The arclength equals

$$\begin{array}{} \displaystyle\int_0^{2 \pi} || g'(t) || dt &= \displaystyle\int_0^{2 \pi} ( (-{\rm sin \hspace{3pt}} t)^2 + {\rm cos}^2 t + 1 )^{\frac{1}{2}} dt \\ &= \displaystyle\int_0^{2 \pi} ( {\rm sin}^2 t + {\rm \hspace{3pt} cos}^2 t + 1 )^{\frac{1}{2}} dt \\ &= \displaystyle\int_0^{2 \pi} \sqrt{2} dt \\ &= \sqrt{2} (2{\pi} - 0 ) \\ &= 2 \sqrt{2}{\pi}. \end{array}$$

Exercise 4. Parametrize the helix

$$g(t) = ( {\rm \hspace{3pt} cos \hspace{3pt}} t, {\rm \hspace{3pt} sin \hspace{3pt}} t, t )$$

for $t \in [0,2{\pi}]$ by arclength.

Solution: It is

$$\begin{array}{} \displaystyle\int_0^t || g'(t) || dt = \displaystyle\int_0^t \sqrt{2} dt = \sqrt{2} t \end{array}$$

for all $t \in [0,2{\pi}]$. We can solve for $t$ in terms of $s$:

$$t = \alpha(s) = \frac{s}{\sqrt{2}},$$

and hence

$$g(s) = \left( {\rm cos \hspace{3pt}} \frac{s}{\sqrt{2}}, {\rm sin \hspace{3pt}} \frac{s}{\sqrt{2}}, \frac{s}{\sqrt{2}} \right)$$

for all $s \in [0,2 \sqrt{2}{\pi}]$.

Exercise 5. Find the arclength of the curve whose cylindrical coordiates are given by

$$r = e^t,$$

$$\theta = t,$$

$$z = e^t$$

for $t \in [0,1]$.

Solution: With

$$r'(t) = e^t,$$

$$\theta '(t) = 1,$$

$$z'(t) = e^t$$

we obtain

$$\begin{array}{} \displaystyle\int_0^1 ( r'(t)^2 + r(t)^2 \theta'(t)^2 + z'(t)^2)^{\frac{1}{2}} dt &=\displaystyle\int_0^1 ( e^{2t} + e^{2t} (1) + e^{2t} )^{\frac{1}{2}} dt \\ &= \displaystyle\int_0^1 e^t \sqrt{3} dt \\ &= \sqrt{3} (e - 1). \end{array}$$

Curvature

Exercise 1. Let

$$g(t) = ( t, t^2 ).$$

Find the curvature at $t = 0$.

Solution: It is

$$g'(t) = ( 1, 2t ),$$

$$g' '(t) = ( 0, 2 ).$$

Hence

$$\kappa (0) = 2.$$

Exercise 1. Interpretation of the gradient. Consider a room in which the temperature is given by a scalar field $T$, so that at each point $(x,y,z)$ the temperature equals

$$T(x,y,z),$$

assuming the temperature does not change in time. How can the gradient be interpreted?

Solution: In this case, at each point in the room, the gradient of $T$ at that point will show the direction in which the temperature rises most quickly. The magnitude of the gradient will determine how fast the temperature rises in that direction.

Exercise 2. Interpretation of the gradient. Consider a hill whose height above sealevel at a point $(x,y)$ is $H(x,y)$. How can the gradient of $H$ be interpreted?

Solution: The gradient of $H$ at a point is a vector pointing in the direction of the steepest slope at that point. The steepness of the slope at that point is given by the magnitude of the gradient vector.

Exercise 3. Consider a room in which the temperature is given by

$$f(x,y,z) = x y^2 z^3.$$

Consider further that a fly is crawling at unit speed in the direction of the vector

$$v = ( -1, 1, 0 )$$

starting at the point

$$s = ( 2, 1, 1 ).$$

Compute that rate of temperature change the fly is about to experience.

Solution: It is

$${\nabla}f(x,y,z) = ( y^2 z^3, 2xyz^3, 3xy^2z^2 ),$$

hence

$${\nabla}f(2,1,1) = ( 1, 4, 6 ).$$

Further,

$$\frac{v}{|| v ||} = ( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 ),$$

and thus

$$\frac{df}{ds} = ( 1, 4, 6 ) \cdot ( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 ) = \frac{3}{\sqrt{2}}.$$

Exercise 4. Let $f: {\bf R}^3 {\rightarrow} {\bf R}$,

$$f(x,y,z) = x + {\rm \hspace{3pt} sin}(xy) + z.$$

Find the directional derivative $D_v f(1,0,1)$, where

$$v = ( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} ).$$

Solution: Note that $v$ is already a unit vector. Therefore, it is only necessary to find ${\nabla}f(1,0,1)$ and take the dot product. It is

$${\nabla}f(x,y,z) = ( 2x + {\rm \hspace{3pt} cos}(xy) y, {\rm \hspace{3pt} cos}(xy) x, 1),$$

and hence

$${\nabla}f(1,0,1) = ( 2, 1, 1 ).$$

The directional derivate is obtained as

$$(2,1,1) \cdot ( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} ) = \frac{4}{3} \sqrt{3}.$$

Exercise 5. Let $f: {\bf R}^3 {\rightarrow} {\bf R}$. Find the equation of the tangent plane to the level surface $f(x,y,z) = 6$ of the function

$$f(x,y,z) = x^2 + 2y^2 + 3z^2$$

at the point $( 1, 1, 1 )$.

Solution: First note that $(1,1,1)$ is a point on the level surface (i.e. $f(1,1,1) = 6$ ). To find the desired plane it suffices to find the normal vector to the proposed plane. But we see

$${\nabla}f(x,y,z) = ( 2x, 4y, 6z ),$$

and hence

$${\nabla}f(1,1,1) = ( 2, 4, 6 ).$$

Therefore, the equation of the tangent plane equals

$$( 2, 4, 6) = ( x - 1, y - 1, z - 1 ) = 0,$$

or

$$2x + 4y + 6z - 12 = 0.$$

Exercise 6. Compute the gradient of the function

$$f(x,y) = x^2 {\rm \hspace{3pt} sin}(xy)$$

at $( {\pi}, 0 )$.

Solution: It is

$${\nabla}f({\pi},0) = \left| \begin{array}{} f_x({\pi},0) \\ f_y({\pi},0) \end{array} \right| = \left| \begin{array}{c} 2x {\rm \hspace{3pt} sin}(xy) + x^2 y {\rm \hspace{3pt} cos}(xy) \\ x^3 {\rm \hspace{3pt} cos}(xy) \end{array} \right|_{(\pi, 0)} = [0, \pi^3]^T$$