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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**

See also **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**

See also 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

**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

**Tangents**

See also Tangents

**Exercise 1.**
Let

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

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

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

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

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

**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**

See also 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**

See also 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**

See also 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**

See also 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

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

**Gradients**

See also Gradients

**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

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