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# Calculus exercises: part III

## Examples

Example. Consider $$f(x,y) = ( x^2 + ( y - 2 )^2 - y )^{\frac{1}{2}} + 1.$$

(a) Domain: $$x^2 + ( y - 2 )^2 - y \geq 0,$$ $$x^2 + ( y - 2 )^2 \geq 4,$$ i.e., the region with boundary: $$x^2 + ( y - 2 )^2 = 2^2.$$ This is the circle of radius $2$ with center at $(0, 2)$.

(b) Graph: $$f( 0, y ) = ( ( y - 2 )^2 - y )^{\frac{1}{2}} + 1 = z,$$ hence $$z = ( x^2 + ( y - 2 )^2 - y )^{\frac{1}{2}} + 1\ \Longrightarrow\ ( z - 1 )^2 = x^2 + ( y - 2 )^2.$$ Then with $x = 0$: $$( z - 1 )^2 = ( y - 2 )^2 \ \Longrightarrow\ z - 1 = \pm ( y - 2 ),$$ and with $y = 0$: $$( z - 1 )^2 = x^2 \ \Longrightarrow\ z - 1 = \pm x.$$ Then $z = y - 1$ and $z = -y + 3$ result in two lines. These cross-sections give us the graph.

(c) This is funnel, a truncated cone. $\square$

Example. To prove that $|| x ||$ is continuous, use $$|| x || = ( x_1^2 + ... + x_n^2 )^{\frac{1}{2}},$$ and show that $$|| x - a || < \delta \ \Longrightarrow\ \bigg| || x || - || a || \bigg| < \delta,$$ by the Triangle Inequality. $\square$

Example. Suppose $A$ and $B$ are path-connected and $a$ is in the intersection. If $P$ and $Q$ belong to the union, find a path from $P$ to $a$, from $a$ to $Q$. This gives you a path from $P$ to $Q$. $\square$

Example. Show that $$T(x) = x_2^2 + 2x_2 + 1$$ is the best linear approximation of the function $$f(x) = x_1^4 + x_2^2.$$ Consider $$\begin{array}{} \displaystyle\lim_{x \to a} \frac{f(x) - T(x)}{|| x - a ||} &= \displaystyle\lim_{x \to (0,-1)} \frac{x_1^4 + x_2^2 + 2x_2 + 1}{|| ( x_1, x_2) - ( 0, -1)||} \\ &= \displaystyle\lim_{x \to (0,-1)} \frac{x_1^4 + x_2^2 + 2x_2 + 1}{( ( x_1 - 0 )^2 + ( x_2 + 1 )^2)^\frac{1}{2}} \\ &= \displaystyle\lim_{x_2 \to (-1)} \frac{x_2^2 + 2x_2 + 1}{| x_2 + 1 |} \\ &= \displaystyle\lim_{x_2 \to -1} \frac{(x_2 + 1 )^2}{| x_2 + 1 |} &\text{ by canceling } ( x_2 + 1 ) \\ &= \displaystyle\lim_{x_2 \to -1} | x_2 + 1 | \\ &= 0. \end{array}$$ $\square$

Example. Let $$h(t) = ( \sin e^t, \cos e^t).$$ Then $h$ is the composition of

• $f: {\bf R} \to {\bf R}$ given by $f(t) = e^t$, and
• $g: {\bf R} \to {\bf R}^2, g(x) = ( \sin x, \cos x)$.

Then $$f'(t) = e^t, {\nabla}g = (\cos x, -\sin x),$$ and by the Chain Rule, $$h' = ( g {\circ}f )' = {\nabla}g \cdot f' = ( \cos e^t, -\sin e^t ) e^t.$$ $\square$

Example. Describe the curve which results from the vector valued function $$r(t) = ( \cos 2t, \sin 2t, t ),$$ 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. $\square$

Example. The position of a particle at time $t$ is $( x, y )$, where $$x = \sin \ t,$$ $$y = \sin^2 t.$$ 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 $( \sin a, \sin^2 a )$ to $( \sin b, \sin^2 b)$. $\square$

Example. Find the limit $$\displaystyle\lim_{x \to 1} \frac{x^2 - x}{x - 1}.$$

Solution: It is $$\frac{x \cdot ( x - 1 )}{x - 1} = x {\rm \hspace{3pt} for \hspace{3pt}} x \neq 1.$$ Hence $$\displaystyle\lim_{x \to 1} \frac{x \cdot ( x - 1 )}{x - 1} = \displaystyle\lim_{x \to 1} x = 1.$$ $\square$

Example. Find the limit $$\displaystyle\lim_{x \to \infty} \frac{x}{1 + x}.$$

Solution: Rewrite $$\frac{x}{1 + x} = \frac{1}{1 + \frac{1}{x}}.$$ Now it is $$\displaystyle\lim_{x \to \infty} \frac{1}{1 + x} = 1 \neq 0.$$ Therefore, $$\displaystyle\lim_{x \to \infty} \frac{x}{1 + x} = \displaystyle\lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}} = \frac{1}{1} = 1.$$ $\square$

Example. Consider $f: {\bf R} \to {\bf R}^3$. Is $$f( t ) = ( t^2 + 1, \cos( 2t ),\sin( 3t) )$$ continuous?

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

Example. Let $f: {\bf R} \to {\bf R}^2$. Is $$f( t ) = ( ( t + 1 )^{\frac{1}{2}}, \tan( t ) )$$ 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 $$t \geq -1 \text{ and }t \neq \frac{\pi}{2} + k {\pi}, k \in {\bf Z}.$$ $\square$

Example. Let $f: {\bf R} \to {\bf R}^3$, $$f( t ) = \left( \frac{1}{t^2 - 1}, ( 1 - t^2)^{\frac{1}{2}}, \frac{1}{t} \right).$$ 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 $$t \in ( -1, 0 ) \cup ( 0, 1 ).$$ $\square$

Example. Let $g: {\bf R} \to {\bf R}^4$, $$g( t ) = ( \cos(4t), 1 - ( 3t + 1 )^{\frac{1}{2}}, \sin(5t), \sec(t) ).$$ 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 $\sec(t) = \frac{1}{\cos(t)}$ 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 $$t \geq -\frac{1}{3} \text{ and }t \neq \frac{\pi}{2} + l{\pi},\ l \in {\bf Z}.$$ $\square$

Example. Let $$f( t ) = ( \sin \ t, t^2, t + 1 ) \text{ for } t \in [ 0, 5 ].$$ Find a tangent line to the curve parametrized 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, $$f'(2) = ( \cos \ 2, 4, 1 ).$$ Hence, a parametrized equation for the tangent line is $$( x, y, z ) = ( \sin \ 2, 4, 3) + t ( \cos \ 2, 4, 1).$$ $\square$

Example. Let $$f( t ) = ( \sin t, t^2, t + 1 ) \text{ for } t \in [ 0, 5 ].$$ Find the velocity vector for $t = 1$.

Solution: The velocity vector is simply $f'(1) = ( \cos 1, 2, 1 )$. $\square$

Example. Consider $g: {\bf R} \to {\bf R}^2$, $$g( t ) = ( t, t^2 ).$$ 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 $g' '(2) = ( 0, 2 )$. $\square$

Example. An object has position $$r(t) = ( t^3, \frac{t}{1 + t}, ( t^2 + 2 )^{\frac{1}{2}} )\ km .$$ 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 $$r'(1) = \left( 3, \frac{1}{4}, \frac{1}{\sqrt{3}} \right) km / hour.$$ $\square$

Example. Consider $g: {\bf R} \to {\bf R}^2$, $$g(t) = ( \frac{1}{t}, 2t )$$ 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 velocity, 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 $$|| g'(t) || = || ( -\frac{1}{t^2}, 2 ) || = ( \frac{1}{t^4} + 4 )^{\frac{1}{2}}.$$ $\square$

Example. For the function $$f( t ) = ( 3t \cos( 2t ), 4t \sin\}( 2t ) ),$$ calculate $f'(t)$.

Solution: Apply the product rule for both components: $$f'(t) = ( 3 \cos( 2t ) - 6t \sin ( 2t ), 4\sin( 2t ) + 8t \cos ( 2t ) ).$$ $\square$

Example. For the function $$f( t ) = ( 3t \cos( 2t ), 4t \sin( 2t ) ),$$ show that $$\frac{d}{dt} || f(t) || \neq || f'(t) || .$$

Solution: It is $$\begin{array}{} || f(t) || &= ( 9t^2 \cos^2(2t) + 16t^2 \sin^2(2t) )^{\frac{1}{2}} &= ( 9t^2 + 7t^2 \sin\}^2(2t) )^{\frac{1}{2}}, \end{array}$$ and its derivative equals $$\frac{d}{dt} || f(t) || = \frac{9t + 14t^2 \sin(2t) \cos(2t) + 7t \sin^2(2t)}{9t^2 + 7t^2 \sin^2(2t) )^{\frac{1}{2}}}.$$ Further $$f'(t)= ( 3\cos(2t) - 6t \sin(2t), 4\sin(2t) + 8t \cos(2t) ),$$ and thus $$\begin{array}{lll} || f'(t) || &= ( ( 3 \cos(2t) - 6t \sin(2t) )^2 + (4 \sin(2t) + 8t \cos(2t))^2 )^{\frac{1}{2}} \\ &= ( 9 + 7 \sin^2(2t) + 36 t^2 + 28 t^2 \cos^2(2t) + 14t \sin(4t) )^{\frac{1}{2}}. \end{array}$$ 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 $\frac{d}{dt} || f(t) || \neq || f'(t) ||$. $\square$

Example. Let $$r(t) = ( t^2, \sin t, \cos t )$$ and let $$p(t) = ( t, \ln( t + 1 ), 2t ).$$ Find $( r(t) \times p(t) )'$.

Solution: From the cross-product rule, it is $$( r(t) \times p(t) )' = ( 2t, \cos t, -\sin t) \times ( t, \ln( t + 1 ), 2t ) + ( t^2, \sin t, \cos \ t ) \times ( 1, \frac{1}{t + 1}, 2 ).$$ $\square$

Example. Let $$F = yi - xj + 2k = < y, -x, 2 >$$ be a vector field. Verify that the path $$g(t) = ( \sin t, \cos t, 2t )$$ 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 $$( x, y, z ) = ( \sin t, \cos t, 2t )$$ it follows that $$g'(t) = ( \cos t, -\sin t, 2 ),$$ and $$F(g(t)) = ( y, -x, 2t ) = ( \cos t, -\sin t, 2t ).$$ $\square$

Example. Consider the path $g: {\bf R} \to {\bf R}^2$, $$g(t) = ( \cos^3 t, \sin^3 t ),$$ which describes an astroid. For $t \in [ 0, \frac{\pi}{2} ]$, one fourth of the astroid is described. Calculate its arc-length.

Solution: With $$g'(t) = ( -3 \sin t \cos^2 t, 3 \sin^2t\cos t ),$$ $$|| g'(t) || = 3 \sin t \cos t = \frac{3}{2} sin 2t,$$ it is $$\begin{array}{} \displaystyle\int_0^{\frac{\pi}{2}} || g'(t) ||\, dt &= \displaystyle\int_0^{\frac{\pi}{2}} \frac{3}{2}\sin 2t\, dt \\ &= -\frac{3}{4} \cos\}(2t) |_0^{\frac{\pi}{2}} \\ &= -\frac{3}{4} \cos \pi + \frac{3}{4}\cos 0 = \frac{3}{2}. \end{array}$$ $\square$

Example. Consider the path $f: {\bf R} \to {\bf R}^2$, $$f(t) = ( t^2, \frac{2}{3} ( 2t + 1 )^{\frac{3}{2}} ), 0 \leq t \leq 4.$$ Calculate its arc-length.

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

Example. Find the arc-length of the helix $$g(t) = ( \cos t, \sin t, t )$$ from $t = 0$ to $t = 2{\pi}$.

Solution: The arc-length equals $$\begin{array}{} \displaystyle\int_0^{2 \pi} || g'(t) || dt &= \displaystyle\int_0^{2 \pi} ( -(\sin t)^2 + \cos^2 t + 1 )^{\frac{1}{2}}\, dt \\ &= \displaystyle\int_0^{2 \pi} ( \sin^2 t + \cos^2 t + 1 )^{\frac{1}{2}} dt \\ &= \displaystyle\int_0^{2 \pi} \sqrt{2} dt \\ &= \sqrt{2} (2{\pi} - 0 ) \\ &= 2 \sqrt{2}{\pi}. \end{array}$$ $\square$

Example. Parametrize the helix $$g(t) = ( \cos t, \sin t, t )$$ for $t \in [0,2{\pi}]$ by arc-length.

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 $$g(s) = \left( \cos \frac{s}{\sqrt{2}}, \sin \frac{s}{\sqrt{2}}, \frac{s}{\sqrt{2}} \right)$$ for all $s \in [0,2 \sqrt{2}{\pi}]$. $\square$

Example. Find the arc-length of the curve whose cylindrical coordinates 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 $$\begin{array}{} \displaystyle\int_0^1 ( r'(t)^2 + r(t)^2 \theta'(t)^2 + z'(t)^2)^{\frac{1}{2}}\, dt &=\displaystyle\int_0^1 ( e^{2t} + e^{2t} (1) + e^{2t} )^{\frac{1}{2}}\, dt \\ &= \displaystyle\int_0^1 e^t \sqrt{3}\, dt \\ &= \sqrt{3} (e - 1). \end{array}$$ $\square$

Example. 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. $\square$

Example. Interpretation of the gradient. Consider a hill whose height above sea level 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. $\square$

Example. 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 $$\frac{df}{ds} = < 1, 4, 6> \cdot <-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 > = \frac{3}{\sqrt{2}}.$$ $\square$

Example. Let $f: {\bf R}^3 \to {\bf R}$, $$f(x,y,z) = x + \sin(xy) + z.$$ 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 $${\nabla}f(x,y,z) = ( 2x + \cos(xy) y, \cos(xy) x, 1),$$ and hence $${\nabla}f(1,0,1) = ( 2, 1, 1 ).$$ The directional derivative is obtained as $$<2,1,1> \cdot < \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} > = \frac{4}{3} \sqrt{3}.$$ $\square$

Example. Let $f: {\bf R}^3 \to {\bf R}$. 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 is: $$< 2, 4, 6> + ( x - 1, y - 1, z - 1 ) = 0,$$ or $$2x + 4y + 6z - 12 = 0.$$ $\square$

Example. Compute the gradient of the function $$f(x,y) = x^2 \sin(xy)$$ at $( {\pi}, 0 )$.

Solution: It is $${\nabla}f({\pi},0) = \left[ \begin{array}{} f_x({\pi},0) \\ f_y({\pi},0) \end{array} \right] = \left[ \begin{array}{c} 2x \sin(xy) + x^2 y \cos(xy) \\ x^3 \cos(xy) \end{array} \right]_{(\pi, 0)} = <0, \pi^3>.$$ $\square$