Consider the curve $y=\frac{1}{x}$, $1\leq x<\infty$.

Gabriel’s horn is the surface that is obtained by rotating the above curve about the x-axis.

What’s interesting about this Gabriel’s horn is that its surface area is infinite while the volume of its interior is finite. Let us first calculate the volume. Using the disk method, the volume is given by \begin{align*}V&=\pi\int_1^\infty\left(\frac{1}{x}\right)^2dx\\&=\pi\lim_{a\to\infty}\int_1^a\frac{1}{x^2}dx\\&=\pi\lim_{a\to\infty}\left[-\frac{1}{x}\right]_1^a\\&=\pi\lim_{a\to\infty}\left[1-\frac{1}{a}\right]\\&=\pi\end{align*}Its surface area is obtained by calculating the integral \begin{align*}A&=2\pi\int_1^\infty\frac{1}{x}\sqrt{1+\left(-\frac{1}{x^2}\right)^2}dx\\&=2\pi\lim_{a\to\infty}\int_1^a\frac{1}{x}\sqrt{1+\frac{1}{x^4}}dx\end{align*} We don’t actually have to evaluate this integral to see the area is infinite. Since $\sqrt{1+\frac{1}{x^4}}>1$, $$\int_1^a\frac{1}{x}\sqrt{1+\frac{1}{x^4}}dx\geq \int_1^a\frac{1}{x}dx=\ln a$$ Hence, $A=\infty$. The integral $\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}dx$ can be evaluated exactly. Using first the substitution $u=x^2$ and then the trigonometric substitution $u=\tan\theta$, \begin{align*}\int\frac{1}{x}\sqrt{1+\frac{1}{x^4}}dx&=\frac{1}{2}\int\frac{\sqrt{1+u^2}}{u^2}du\\&=\frac{1}{2}\int\frac{\sec^2\theta\sec\theta}{\tan^2\theta}d\theta\\&=\frac{1}{2}\int\frac{(1+\tan^2\theta)\sec\theta}{\tan^2\theta}d\theta\\&=\frac{1}{2}\left[\int\cot^2\theta\sec\theta d\theta+\int\sec\theta d\theta\right]\\&=\frac{1}{2}\left[\int\frac{\cos\theta}{\sin^2\theta}d\theta+\int\sec\theta d\theta\right]\\&=\frac{1}{2}\left[-\csc\theta+\ln|\sec\theta+\tan\theta|\right]\\&=\frac{1}{2}\left[-\frac{\sqrt{x^4+1}}{x^2}+\ln(\sqrt{x^4+1}+x^2)\right]\end{align*}

Another horn-shaped surface can be obtained by rotating the curve $y=e^{-x}$, $0\leq x<\infty$:

The volume of the interior is finite and \begin{align*}V&=\pi\int_0^\infty e^{-2x}dx\\&=\pi\lim_{a\to\infty}\int_0^\infty e^{-2x}dx\\&=-\frac{\pi}{2}\lim_{a\to\infty}[e^{-2x}]_0^a\\&=\frac{\pi}{2}\end{align*} Unlike Gabriel’s horn, the surface area is also finite. To see that, it is given by the improper integral \begin{align*}A&=\int_0^\infty 2\pi e^{-x}\sqrt{1+(-e^{-x})^2}dx\\&=2\pi\lim_{a\to\infty}\int_0^a e^{-x}\sqrt{1+e^{-2x}}dx\end{align*} Using the substitution $u=e^{-x}$ and then the trigonometric substitution $u=\tan\theta$ the integral $\int e^{-x}\sqrt{1+e^{-2x}}dx$ is evaluated to be \begin{align*}\int e^{-x}\sqrt{1+e^{-2x}}dx&=-\int\sqrt{1+u^2}du\\&=-\int\sec^3\theta d\theta\\&=-\frac{1}{2}[\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|]\\&=-\frac{1}{2}[u\sqrt{1+u^2}+\ln|\sqrt{u^2+1}+u|]\\&=-\frac{1}{2}[e^{-x}\sqrt{e^{-2x}+1}+\ln(\sqrt{e^{-2x}+1}+e^{-x})]\end{align*} (For the details on how to evaluate the integral $\int\sec^3\theta d\theta$, see here.) Hence, $A$ is computed to be $\pi[\sqrt{2}+\ln(\sqrt{2}+1)]$.

Update: Instead of doing substitutions twice as seen above, one can use a single substitution $e^{-x}=\tan\theta$. The area then turns into the integral $$A=2\pi\int_0^{\frac{\pi}{4}}\sec^3\theta d\theta$$

There is a particularly interesting horn-shaped surface (actually a shape of two identical horns put together as shown in the figure below) although it has both a finite volume and a finite surface area. It is called a pseudosphere. A pseudosphere is a surface with constant negative Gaussian curvature. The pseudosphere of radius 1 is obtained by revolving the tractrix $$t\mapsto (t-\tanh t,\mathrm{sech}\ t),\ -\infty<t<\infty$$ about its asymptote (the $x$-axis).

The resulting surface of revolution, the pseudosphere of radius 1, is seen in the following figure.

The volume of the interior of the pseudosphere is \begin{align*}V&=\pi\int_{-\infty}^\infty y^2dx\\&=2\pi\int_0^\infty y^2dx\\&=2\pi\int_0^\infty\mathrm{sech}^2\ t (1-\mathrm{sech}^2\ t)dt\\&=2\pi\int_0^\infty \mathrm{sech}^2\ t\tanh^2 tdt\\&=2\pi\int_0^1 u^2du\ (u=\tanh t)\\&=2\pi\left[\frac{u^3}{3}\right]_0^1\\&=\frac{2\pi}{3}\end{align*} The area of the pseudosphere is \begin{align*}A&=\int_{-\infty}^\infty 2\pi y ds\\&=2\int_0^\infty 2\pi y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\\&=4\pi\int_0^\infty\mathrm{sech}\ t\sqrt{(1-\mathrm{sech}^2\ t)^2+(-\mathrm{sech}\ t\tanh t)^2}dt\\&=4\pi\int_0^\infty\mathrm{sech}\ t\sqrt{\tanh^4 t+\mathrm{sech}^2 t\tanh^2 t}dt\\&=4\pi\int_0^\infty\mathrm{sech}\ t\tanh t dt\\&=4\pi\int_0^\infty\frac{\sinh t}{\cosh^2 t}dt\\&=4\pi\int_1^\infty\frac{1}{u^2}du\ (u=\cosh t)\\&=4\pi\end{align*} Notice that its volume is half of the volume of the unit sphere and its area is the same as the area of the unit sphere. Such volume and area relationships are still true for the pseudosphere of radius $r$, i.e. the volume and the area of the pseudosphere of radius $r$ are, respectively, $\frac{2}{3}\pi r^3$ and $4\pi r^2$ (we do not discuss it here, but the radius of a pseudosphere is defined to be the radius of its equator) as noted by the Dutch physicist Christiaan Huygens. The Gaussian curvature of a regular surface can be computed by the Gauss’ formula. The pseudosphere of radius 1 as a parametric surface is represented by the equation $$\varphi(t,s)=(t-\tanh t,\mathrm{sech}\ t\cos s,\mathrm{sech}\ t\sin s)$$ As seen in the figure above, the pseudosphere is not regular along the equator (at $t=0$). It has a constant negative Gaussian curvature $K=-1$ anywhere else. A common misunderstanding (usually by non-mathematicians) is that a surface with constant negative Gaussian curvature is a hyperbolic surface. In order for a surface to be hyperbolic, in addition to having a constant negative curvature, it is required to be complete and regular, so the pseudosphere is not hyperbolic although it was introduced by Eugenio Beltrami as a model of hyperbolic geometry. In fact, Hilbert’s theorem states that there exist no complete regular surface of constant negative Gaussian curvature immersed in $\mathbb{R}^3$. This means that one cannot obtain a model of two-dimensional hyperbolic geometry in $\mathbb{R}^3$. However, one can obtain a model of two-dimensional hyperbolic geoemtry in $\mathbb{R^{2+1}}$, a 3-dimensional Minkowski space-time. There is another interesting surface with constant negative Gaussian curvature called Dini’s surface. It is described by the parametric equations \begin{align*}x&=a\cos u\sin v\\y&=a\sin u\sin v\\z&=a\left(\cos v+\ln\tan\frac{v}{2}\right)+bu\end{align*}

The Gaussian curvature of Dini’s surface is computed to be $K=-\frac{1}{a^2+b^2}$.

Related rates problems often involve (context-wise) real-life applications of the chain rule/implicit differentiation. Here are some of the examples that are commonly seen in calculus textbooks.

Example. Car A is traveling west at 50mi/h and car B is traveling north at 60mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?

Solution.

Denote by $x$ and $y$ the distances from the intersection to car A and to car B, respectively. Then we have $\frac{dx}{dt}=-50$mi/h and $\frac{dy}{dt}=-60$mi/h. Let us denote $z$ the distance between $A$ and $B$. Then by Pythagorean law we have $$z^2=x^2+y^2$$ Differentiating this with respect to $t$, we obtain $$z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$$ and thus \begin{align*}\frac{dz}{dt}&=\frac{1}{z}\left[x\frac{dx}{dt}+y\frac{dy}{dt}\right]\\&=\frac{1}{0.5}[0.3(-50)+0.4(-60)]=-78\mathrm{mi/h}\end{align*}

Example. Air is being pumped into a spherical balloon so that its volume increases at a rate of $100\mathrm{cm}^3/\mathrm{s}$. How fast is the radius of the balloon increasing when the diameter is 50 cm?

Solution. Let $V$ and $r$ denote the volume and the radius of the spherical balloon. Then $V=\frac{4}{3}\pi r^3$. Differentiating this with respect to $t$, we obtain $$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$ So, \begin{align*}\frac{dr}{dt}&=\frac{1}{4\pi r^2}\frac{dV}{dt}\\&=\frac{1}{4\pi(25)^2}100\\&=\frac{1}{25\pi}\mathrm{cm/s}\end{align*}

Example. Gravel is being dumped from a conveyor belt at a rate of $30 \mathrm{ft}^3/\mathrm{min}$ and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are the same. How fast is the height of the pile increasing when the pile is 10 ft high?

Solution. The cross section of the gravel pile is shown in the figure below.

The amount of gravel dumped is the same as the volume of the cone. Let us denote the volume by $V$, its base radius by $r$, and its height by $h$. Then $V=\frac{1}{3}\pi r^2h$. Since $h=2r$, $V$ can be written as $$V=\frac{1}{12}\pi h^3$$ Differentiating this with respect to $t$, we obtain $$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$$ So, we have \begin{align*}\frac{dh}{dt}&=\frac{4}{\pi h^2}\frac{dV}{dt}\\&=\frac{4}{\pi(10)^2}(30)=\frac{1.2}{\pi}\mathrm{ft/min}\approx 0.38\mathrm{ft/min}\end{align*}

Example. A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?

Solution.

Let us denote by $x$ and $y$ the distance from the wall to the bottom of the ladder and the distance from the top of the ladder to the floor, respectively. By Pythagorean law, we have $x^2+y^2=100$. Differentiating this with respect to $t$, we obtain $$x\frac{dx}{dt}+y\frac{dy}{dt}=0$$ Hence, we have \begin{align*}\frac{dy}{dt}&=-\frac{x}{y}\frac{dx}{dt}\\&=-\frac{6}{8}(1)=-\frac{3}{4}\mathrm{ft/s}\end{align*}

Example. A water tank has the shape of an inverted circular cone with base radius 2m and heigh 4 m. If water is being pumped into the tank at a rate of $2 \mathrm{m}^3/\mathrm{min}$, find the rate at which the water level is rising when the water is 3 m deep.

Solution. The cross section of the water tank is shown in the figure below.

The amount of water $V$ when the water level is $h$ and the surface radius is $r$ is $V=\frac{1}{3}\pi r^2h$. From the above figure we have the following ratio holds $$\frac{2}{4}=\frac{r}{h}$$ i.e. $r=\frac{h}{2}$. SO $V$ can be written as $$V=\frac{1}{12}\pi h^3$$ Differentiating this with respect to $t$, we obtain $$\frac{dV}{dt}=\frac{1}{4}\pi h^2\frac{dh}{dt}$$ Hence, \begin{align*}\frac{dh}{dt}&=\frac{4}{\pi h^2}\frac{dV}{dt}\\&=\frac{4}{\pi(3)^2}(2)\\&=\frac{8}{9\pi}\mathrm{m/min}\approx 0.28\mathrm{m/min}\end{align*}

Example. A street light is at the top of a 15 feet tall pole. A 6 feet tall woman walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 45 feet from the base of the pole?

Solution. Let $x$ be the distance from the light pole to the woman and $y$ be the distance from the light pole to the tip of her shadow as shown in the figure below.

By similar triangles, we have $\frac{15}{y}=\frac{6}{y-x}$. Solving this equation for $y$, we obtain $y=\frac{5}{3}x$. Differentiating this with respect to $t$, we find how fast the tip of her shadow is moving: $$\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}=\frac{25}{3}\mathrm{ft/s}$$ As seen regardless of how far the woman is from the pole, the speed of the tip is constant.

Example. The fish population, $N$, in a small pond depends on the amount of algae, $a$ (measured in pounds), in it. The equation modeling the fish population is given by $N=(3a^2-20a+26)^4$. If the amount of algae is increasing at a rate of 2 lb/week, at what rate is the fish population changing when the pond contains 5 lb of algae?

Solution. By the chain rule, we obtain \begin{align*}\frac{dN}{dt}&=4(3a^2-20a+26)^3\left(6a\frac{da}{dt}-20\frac{da}{dt}\right)\\&=4(3a^2-20a+26)^3(6a-20)\frac{da}{dt}\end{align*} $\frac{da}{dt}=2$lb/week, so when $a=5$lb, $\frac{dN}{dt}$ is $$\frac{dN}{dt}=4(3(5)^2-20(5)+26)^3(6(5)-20)(2)=-80\ \mathrm{lb/week}$$ What this means is that the fish population is decreasing by 80 lb/week at the instant when the pond contains 5 lb of algae.

Example. The retail price per gallon of gasoline is increasing at \$0.02 per week. The demand equation is given by $$10p-\sqrt{356-x^2}=0$$ where $p$ is the price per gallon (in dollars), when $x$ million gallons are demanded. At what rate is the revenue changing when 10 million gallons are demanded?

Solution. The total revenue $R$ is given by the equation $$R=xp$$ Differentiating this equation with respect to $t$, we obtain $$\frac{dR}{dt}=\frac{dx}{dt}p+x\frac{dp}{dt}$$ The only quantity we don’t have to calculate $\frac{dR}{dt}$ is $\frac{dx}{dt}$. To find it, let us differentiate the demand function with respect to $x$. By the chain rule, we obtain $$10\frac{dp}{dt}+\frac{x}{\sqrt{351-x^2}}\frac{dx}{dt}=0$$ When $x=10$ million gallons, with $\frac{dp}{dt}=0.02$, we find from this equation that $$\frac{dx}{dt}=-0.02\sqrt{256}=-0.02\cdot 16= -0.32\ \mbox{million gallons/week}$$ When $x=10$, from the demand function, we find $p$ as $$p=\frac{\sqrt{256}}{10}=\frac{16}{10}=1.6$$ Therefore, the rate of change of revenue when 10 million gallons of gasoline is demanded is $$\frac{dR}{dt}=-0.32(1.6)+10(0.02)=-0.312$$ What this means is that the revenue is decreasing by about 0.31 million dollars per week when the price increases 0.02 dollars per week (consequently the demand decreases by 0.32 million gallons per week as we saw earlier).

Example. A plane flying with a constant speed of 29 km/min passes over a ground radar station at an altitude of 13 km and climbs at an angle of 20 degrees. At what rate is the distance from the plane to the radar station increasing 3 minutes later?

Solution. First, take a look at the following picture

The law of cosine says that the sides $a$, $b$, $c$ and the angle $\theta$ are related by $$a^2=b^2+c^2-2bc\cos\theta$$

The question above can be pictorially represented as the following sketch

Using the law of cosine with $b=13$ km, we have $$a^2=13^2+c^2-2\cdot 13\cdot c\cdot\cos\left(\frac{11}{18}\pi\right)$$ Here, the angle $\theta$ is given by $\theta=90^\circ+20^\circ=110^\circ=\frac{11}{18}\pi$$ (Note: for this question, you can use degree instead of radian but in that case make sure that your calculator is set to use degree angle measurement.) Since we are to find $\frac{da}{dt}$, differentiate the above equation with respect to $t$: $$2a\frac{da}{dt}=2c\frac{dc}{dt}-2\cdot 13\frac{dc}{dt}\cos\left(\frac{11}{18}\pi\right)$$ Since the airplane is flying along the side $c$ at the constant speed 29 km/min, it would have traveled $c=29\cdot 3=87$ km in three minutes. Thus, $$a=\sqrt{13^2+87^2-2\cdot 13\cdot 87\cos\left(\frac{11}{18}\pi\right)}=92.2586$$ and hence $\frac{da}{dt}$ in three minutes is \begin{align*}\frac{da}{dt}&=\frac{c}{a}\frac{dc}{dt}-\frac{13}{a}\frac{dc}{dt}\cos\left(\frac{11}{18}\pi\right)\\&=\frac{1}{a}\frac{dc}{dt}\left(c-13\cos\left(\frac{11}{18}\pi\right)\right)\\&=\frac{1}{92.2586}29\left(87-13\cos\left(\frac{11}{18}\pi\right)\right)\\&=28.7447\ \mathrm{km/min}\end{align*}

We have already discussed the length of a plane curve represented by the parametric equation ${\bf r}(t)=(x(t),y(t))$, $a\leq t\leq b$ here. The same goes for a space curve. Namely, if ${\bf r}(t)=(x(t),y(t),z(t))$, $a\leq t\leq b$, then its arc length $L$ is given by \begin{equation}\begin{aligned}L&=\int_a^b|{\bf r}'(t)|dt\\&=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}dt\end{aligned}\label{eq:spacearclength}\end{equation}

Example. Find the length of the arc of the circular helix $${\bf r}(t)=\cos t{\bf i}+\sin t{\bf j}+t{\bf k}$$ from the point $(1,0,0)$ to the point $(1,0,2\pi)$.

Solution. ${\bf r}'(t)=-\sin t{\bf i}+\cos t{\bf j}+{\bf k}$ so we have $$|{\bf r}'(t)|=\sqrt{(-\sin t)^2+(\cos t)^2+1^2}=\sqrt{2}$$ The arc is going from $(1,0,0)$ to $(1,0,2\pi)$ and the $z$-component of ${\bf r}(t)$ is $t$, so $0\leq t\leq 2\pi$. Now, using \eqref{eq:spacearclength}, we obtain $$L=\int_0^{2\pi}|{\bf r}'(t)|dt=\int_0^{2\pi}\sqrt{2}dt=2\sqrt{2}\pi$$ Figure 1 shows the circular helix from $t=0$ to $t=2\pi$.

Given a curve ${\bf r}(t)$, $a\leq t\leq b$, sometimes we need to reparametrize it by another parameter $s$ for various reasons. Imagine that the curve represents the path of a particle moving in space. A reparametrization does not change the path of the particle (hence nor the distance it traveled) but it changes the particle’s speed! To see this, let $t=t(s)$, $\alpha\leq s\leq\beta$, $a=t(\alpha)$, $b=t(\beta)$ be an increasing and differentiable function. Since $t=t(s)$ is one-to-one and onto, ${\bf r}(t)$ and ${\bf r}(t(s))$, its reparametrization by the parameter $s$, represent the same path. By the chain rule, \begin{equation}\label{eq:reparametrization}\frac{d}{ds}{\bf r}(t(s))=\frac{d}{dt}{\bf r}(t)\frac{dt}{ds}\end{equation} Thus we see that the speed of the reparametrization $\left|{\bf r}(t(s))\right|$ differs from that of ${\bf r}(t)$ by a factor of $\left|\frac{dt}{ds}\right|=\frac{dt}{ds}$ (since $\frac{dt}{ds}>0$). However, the arc length of the reparametrization is \begin{align*}\int_{\alpha}^{\beta}\left|\frac{d}{ds}{\bf r}(t(s))\right|ds&=\int_{\alpha}^{\beta}\left|\frac{d}{dt}{\bf r}(t)\right|\frac{dt}{ds}ds\\&=\int_a^b\left|\frac{d}{dt}{\bf r}(t)\right|dt=L\end{align*} That is, no change of the distance.

There is a particular reparametrization we are interested. To discuss that, suppose ${\bf r}(t)$, $a\leq t\leq b$ be a differentiable curve in space such that ${\bf r}'(t)\ne 0$ for all $t$. Such a curve is said to be regular or smooth. Let us now define the arc length function \begin{equation}\label{eq:arclengthfunction}s(t)=\int_a^t|{\bf r}'(u)|du\end{equation} By the Fundamental Theorem of Calculus, we have \begin{equation}\label{eq:arclengthfunction2}\frac{ds}{dt}=|{\bf r}'(t)|>0\end{equation} and so the arc length function $s=s(t)$ is increasing. This means that $s(t)$ is one-to-one and onto, so it is invertible. It’s inverse function can be written as $t=t(s)$ and ${\bf r}(t(s))$ is called the reparamerization by arc length. The reason we are interested in this particular reparametrization is that it results in the unit speed: From \eqref{eq:reparametrization} and \eqref{eq:arclengthfunction2}, $$\left|\frac{d}{ds}{\bf r}(t(s))\right|=|{\bf r}'(t)|\left|\frac{dt}{ds}\right|=|{\bf r}'(t)|\frac{1}{|{\bf r}'(t)|}=1$$ So it is also called the unit-speed reparametrization. The reparametrization by arc length plays an important role in defining the curvature of a curve. This will be discussed elsewhere.

Example. Reparametrize the helix ${\bf r}(t)=\cos t{\bf i}+\sin t{\bf j}+t{\bf k}$ by arc length measured from $(1,0,0)$in the direction of increasing $t$.

Solution. The initial point $(1,0,0)$ corresponds to $t=0$. From the previous example, we know that the helix has the constant speed $|{\bf r}'(t)|=\sqrt{2}$. Thus, $$s(t)=\int_0^t|{\bf r}'(u)|du=\sqrt{2}t$$ Hence, we obtain $t=\frac{s}{\sqrt{2}}$. The reparametrization is then given by $${\bf r}(t(s))=\cos\left(\frac{s}{\sqrt{2}}\right){\bf i}+\sin\left(\frac{s}{\sqrt{2}}\right){\bf j}+\frac{s}{\sqrt{2}}{\bf k}$$

Examples in this note have been taken from [1].

References.

[1] Calculus, Early Transcendentals, James Stewart, 6th Edition, Thompson Brooks/Cole

The derivative $\frac{d{\bf r}}{dt}={\bf r}'(t)$ of a vector-valued function ${\bf r}(t)=(x(t),y(t),z(t))$ is defined by \begin{equation}\label{eq:vectorderivative}\frac{d{\bf r}}{dt}=\lim_{\Delta t\to 0}\frac{{\bf r}(t+\Delta t)-{\bf r}(t)}{\Delta t}\end{equation} In case of a scalar-valued function or a real-valued function, the geometric meaning of derivative is that it is the slope of tangent line. In case of a vector-valued function, the geometric meaning of derivative is that it is the tangent vector. It can be easily seen from Figure 1. As $\Delta t$ gets smaller and smaller, $\frac{{\bf r}(t+\Delta t)-{\bf r}(t)}{\Delta t}$ gets closer to a line tangent to ${\bf r}(t)$.

From the definition \eqref{eq:vectorderivative}, it is straightforward to show \begin{equation}\label{eq:vectorderivative2}{\bf r}'(t)=(x'(t),y'(t),z'(t))\end{equation}

If ${\bf r}'(t)\ne 0$, the unit tangent vector ${\bf T}(t)$ of ${\bf r}(t)$ is given by \begin{equation}\label{eq:unittangent}{\bf T}(t)=\frac{{\bf r}'(t)}{|{\bf r}'(t)|}\end{equation}

Example.

1. Find the derivative of ${\bf r}(t)=(1+t^3){\bf i}+te^{-t}{\bf j}+\sin 2t{\bf k}$.
2. Find the unit tangent vector when $t=0$.

Solution.

1. Using \eqref{eq:vectorderivative2}, we have $${\bf r}'(t)=3t^2{\bf i}+(1-t)e^{-t}{\bf j}+2\cos 2t{\bf k}$$
2. ${\bf r}'(0)={\bf j}+2{\bf k}$ and $|{\bf r}'(0)|=\sqrt{5}$. So by \eqref{eq:unittangent}, we have $${\bf T}(0)=\frac{{\bf r}'(0)}{|{\bf r}'(0)|}=\frac{1}{\sqrt{5}}{\bf j}+\frac{2}{\sqrt{5}}{\bf k}$$

The following theorem is a summary of differentiation rules for vector-valued functions. We omit the proofs of these rules. They can be proved straightforwardly from differentiation rules for real-valued functions. Note that there are three different types of the product rule or the Leibniz rule for vector-valued functions (rules 3, 4, and 5).

Theorem. Let ${\bf u}(t)$ and ${\bf v}(t)$ be differentiable vector-valued functions, $f(t)$ a scalar function, and $c$ a scalar. Then

1. $\frac{d}{t}[{\bf u}(t)+{\bf v}(t)]={\bf u}'(t)+{\bf v}'(t)$
2. $\frac{d}{dt}[c{\bf u}(t)]=c{\bf u}'(t)$
3. $\frac{d}{dt}[f(t){\bf u}(t)]=f'(t){\bf u}(t)+f(t){\bf u}'(t)$
4. $\frac{d}{dt}[{\bf u}(t)\cdot{\bf v}(t)]={\bf u}'(t)\cdot{\bf v}(t)+{\bf u}(t)\cdot{\bf v}'(t)$
5. $\frac{d}{dt}[{\bf u}(t)\times{\bf v}(t)]={\bf u}'(t)\times{\bf v}(t)+{\bf u}(t)\times{\bf v}'(t)$
6. $\frac{d}{dt}[{\bf u}(f(t))]=f'(t){\bf u}'(f(t))$ (Chain Rule)

Example. Show that if $|{\bf r}(t)|=c$ (a constant), then ${\bf r}'(t)$ is orthogonal to ${\bf r}(t)$ for all $t$.

Proof. Differentiating ${\bf r}(t)\cdot{\bf r}(t)=|{\bf r}(t)|^2=c^2$ using the Leibniz rule 5, we obtain $$0=\frac{d}{dt}[{\bf r}(t)\cdot{\bf r}(t)]={\bf r}'(t){\bf r}(t)+{\bf r}(t){\bf r}'(t)=2{\bf r}'(t)\cdot{\bf r}(t)$$ This implies that ${\bf r}'(t)$ is orthogonal to ${\bf r}(t)$.

As seen in \eqref{eq:vectorderivative2}, the derivative of a vector-valued functions is obtained by differentiating component-wise. The indefinite and definite integral of a vector-valued function are done similarly by integrating component-wise, namely\begin{equation}\label{eq:vectorintegral}\int{\bf r}(t)dt=\left(\int x(t)dt\right){\bf i}+\left(\int y(t)dt\right){\bf j}+\left(\int z(t)dt\right){\bf k}\end{equation} and \begin{equation}\label{eq:vectorintegral2}\int_a^b{\bf r}(t)dt=\left(\int_a^b x(t)dt\right){\bf i}+\left(\int_a^b y(t)dt\right){\bf j}+\left(\int_a^b z(t)dt\right){\bf k}\end{equation}, respectively. When evaluate the definite integral of a vector-valued function, one can use \eqref{eq:vectorintegral2} but it would be easier to first find the indefinite integral using \eqref{eq:vectorintegral} and then evaluate the definite integral the Fundamental Theorem of Calculus (and yes, the Fundamental Theorem of Calculus still works for vector-valued functions).

Example. Find $\int_0^{\frac{\pi}{2}}{\bf r}(t)dt$ if ${\bf r}(t)=2\cos t{\bf i}+\sin t{\bf j}+2t{\bf k}$.

Solution. \begin{align*}\int{\bf r}(t)dt&=\left(\int 2\cos tdt\right){\bf i}+\left(\int \sin tdt\right){\bf j}+\left(\int 2tdt\right){\bf k}\\&=2\sin t{\bf i}-\cos t{\bf j}+t^2{\bf k}+{\bf C}\end{align*} where ${\bf C}$ is a vector-valued constant of integration. Now, $$\int_0^{\frac{\pi}{2}}{\bf r}(t)dt=[2\sin t{\bf i}-\cos t{\bf j}+t^2{\bf k}]_0^{\frac{\pi}{2}}=2{\bf i}+{\bf j}+\frac{\pi^2}{4}{\bf k}$$

Examples in this note have been taken from [1].

References.

[1] Calculus, Early Transcendentals, James Stewart, 6th Edition, Thompson Brooks/Cole

You remember from algebra that a line in the plane can be determined by its slope and a point on the line or two points on the line (in which case those two points determine the slope). For a space line, slope is not a suitable ingredient. It’s alternative ingredient is a vector parallel to the line. As shown in Figure 1, with a point ${\bf r}_0=(x_0,y_0,z_0)$ on the line $L$ and a vector ${\bf v}=(a,b,c)$ parallel to $L$, we can determine any point ${\bf r}=(x,y,z)$ on $L$ by vector addition of ${\bf r}_0$ and $t{\bf v}$, a dilation of ${\bf v}$: \begin{equation}\label{eq:spaceline}{\bf r}={\bf r}_0+t{\bf v}\end{equation} where $-\infty<t<\infty$. The equation \eqref{eq:spaceline} is called a vector equation of $L$.

In terms of the components, \eqref{eq:spaceline} can be written as \begin{equation}\begin{aligned}x&=x_0+at\\y&=y_0+bt\\z&=z_0+ct\end{aligned}\label{eq:spaceline2}\end{equation} where $-\infty<t<\infty$. The equations in \eqref{eq:spaceline2} are called parametric equations of $L$. Solving the parametric equations in \eqref{eq:spaceline2} for $t$, we also obtain \begin{equation}\label{eq:spaceline3}\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}\end{equation} The equations in \eqref{eq:spaceline3} are called symmetric equations of $L$. Often we need to work with a line segment. The vector equation \eqref{eq:spaceline} can be used to figure out an equation of a line segment. Consider the line segment from ${\bf r}_0$ to ${\bf r}_1$. Then the difference ${\bf r}_1-{\bf r}_0$ is a vector parallel to the line segment and by \eqref{eq:spaceline}, we have \begin{align*}{\bf r}(t)&={\bf r}_0+t({\bf r}_1-{\bf r}_0)\\&=(1-t){\bf r}_0+t{\bf r}_1\end{align*} This still represents an infinite line through ${\bf r}_0$ and ${\bf r}_1$. By limiting the range of $t$ to represent only the line segment from ${\bf r}_0$ to ${\bf r}_1$, we obtain an equation of the line segment \begin{equation}\label{eq:linesegment}{\bf r}(t)=(1-t){\bf r}_0+t{\bf r}_1\end{equation} where $0\leq t\leq 1$.

Example.

1. Find a vector equation and parametric equations for the line that passes through the point $(5,1,3)$ and is parallel to the vector ${\bf i}+4{\bf j}-2{\bf k}$.
2. Find two other points on the line.

Solution.

1. ${\bf r}_0=(5,1,3)$ and ${\bf v}=(1,4,-2)$. Hence by \eqref{eq:spaceline}, we have $${\bf r}(t)=(5,1,3)+t(1,4,-2)=(5+t,1+4t,3-2t)$$ Parametric equations are $$x(t)=5+t,\ y(t)=1+4t,\ z(t)=3-2t$$
2. From the parametric equations, for example, $t=1$ gives $(6,5,1)$ and $t=-1$ gives $(4,-3,5)$.

Example.

1. Find parametric equations and symmetric equations of the line that passes through the points $A(2,4,-3)$ and $B(3,-1,1)$.
2. At what point does this line intersect the $xy$-plane?

Solution.

1. Note that the vector ${\bf v}=\overrightarrow{AB}=(1, -5, 4)$ is parallel to the line. So $a=1$, $b=-5$, and $c=4$. By taking ${\bf r}_0=(x_0,y_0,z_0)=(2,4,-3)$, we have the parametric equations $$x=2+t,\ y=4-5t,\ z=-3+4t$$ Symmetric equations are then $$\frac{x-2}{1}=\frac{y-4}{-5}=\frac{z+3}{4}$$
2. The line intersects the $xy$-plane when $z=0$. By setting $z=0$ in the symmetric equations from part 1, we get $$\frac{x-2}{1}=\frac{y-4}{-5}=\frac{3}{4}$$ Solving these equations for $x$ and $y$ respectively, we find $x=\frac{11}{4}$ and $y=\frac{1}{4}$.

As we have seen, a line is determined by a point on the line and a vector parallel to the line. A plane, on the other hand, can be determined by a point ${\bf r}_0$ on the plane and a vector ${\bf n}$ perpendicular to the plane (such a vector is called a normal vector to the plane). See Figure 2.

From Figure 2, we see that \begin{equation}\label{eq:plane}{\bf n}\cdot({\bf r}-{\bf r}_0)=0\end{equation} The equation \eqref{eq:plane} is called a vector equation of the plane. If ${\bf n}=(a,b,c)$, ${\bf r}=(x,y,z)$, and ${\bf r}_0=(x_0,y_0,z_0)$, then the equation \eqref{eq:plane} can be written as \begin{equation}\label{eq:plane2}a(x-x_0)+b(y-y_0)+c(z-z_0)=0\end{equation}

Example. Find an equation of the plane through the point $(2,4,-1)$ with normal vector ${\bf n}=(2,3,4)$. Find the intercepts and sketch the plane.

Solution. $a=2$, $b=3$, $c=4$, $x_0=2$, $y_0=4$, and $z_0=-1$. So by the equation \eqref{eq:plane2}, we have $$2(x-2)+3(y-4)+4(z+1)=0$$ or $$2x+3y+4z=12$$ To find the $x$-intercept, set $y=z=0$ in the equation and we get $x=6$. Similarly, we find the $y$- and $z$-intercepts $y=4$ and $z=3$, respectively. Figure 3 shows the plane.

Example. Find an equation of the plane that passes through the points $P(1,3,2)$, $Q(3,-1,6)$, and $R(5,2,0)$.

Solution. The vectors $\overrightarrow{PQ}=(2,-4,4)$ and $\overrightarrow{PR}=(4,-1,-2)$ are on the plane, so the cross product $$\overrightarrow{PQ}\times\overrightarrow{PR}=\begin{vmatrix}{\bf i} & {\bf j} & {\bf k}\\2 & -4 & 4\\4 & -1 & -2\end{vmatrix}=12{\bf i}+20{\bf j}+14{\bf k}$$ is a normal vector to the plane. With $(x_0,y_0,z_0)=(1,3,2)$ and $(a,b,c)=(12,20,14)$, we find an equation of the plane $$12(x-1)+20(y-3)+14(z-2)=0$$ or $$6x+10y+7z=50$$

Using basic geometry, we see that the angle between two planes $P_1$ and $P_2$ is the same as the angle between their respective normal vectors ${\bf n}_1$ and ${\bf n}_2$. See Figure 4 where cross sections of two planes $P_1$ and $P_2$ are shown.

Example.

1. Find the angle between the planes $x+y+z=1$ and $x-2y+3z=1$.
2. Find the symmetric equations for the line of intersection $L$ of these two planes

Solution.

1. The normal vectors of these planes are ${\bf n}_1=(1,1,1)$ and ${\bf n}_2=(1,-2,3)$. Let $\theta$ be the angle between ${\bf n}_1$ and ${\bf n}_2$ . Then $$\cos\theta=\frac{{\bf n}_1\cdot{\bf n }_2}{|{\bf n}_1||{\bf n}_2|}=\frac{1(1)+1(-2)+1(3)}{\sqrt{1^2+1^2+1^2}\sqrt{1^2+(-2)^2+3^2}}=\frac{2}{\sqrt{42}}$$ Thus, $$\theta=\cos^{-1}\left(\frac{2}{\sqrt{42}}\right)\approx 72^\circ$$
2. First, let us find a point on $L$. Set $z=0$. Then we have $x+y=1$ and $x-2y=1$. Solving these two equations simultaneously we find $x=1$ and $y=0$. So $(1,1,0)$ is on the line $L$. Now we need a vector parallel to the line $L$. The cross product $${\bf n}_1\times{\bf n}_2=\begin{vmatrix}{\bf i} & {\bf j} & {\bf k}\\1 & 1 & 1\\1 & -2 & 3\end{vmatrix}=5{\bf i}-2{\bf j}-3{\bf k}$$ is perpendicular to both ${\bf n}_1$ and ${\bf n}_2$, hence it is parallel to $L$. Therefore, the symmetric equations of $L$ are $$\frac{x-1}{5}=\frac{y}{-5}=\frac{z}{-3}$$

Let us find the distance $D$ from a point $Q(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$. See Figure 5.

Let $P(x_0,y_0,z_0)$ be a point in the plane and let ${\bf b}=\overrightarrow{PQ}=(x_1-x_0,y_1-y_0,z_1-z_0)$. Then from Figure 5, we see that the distance from $Q$ to the plane is given by the scalar projection of ${\bf b}$ onto the normal vector ${\bf n}$: \begin{align*}D&=|\mathrm{comp}_{\bf n}{\bf b}|=\frac{|{\bf b}\cdot{\bf n}|}{|{\bf n}|}\\&=\frac{|a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)|}{\sqrt{a^2+b^2+c^2}}\\&=\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}\end{align*} The last expression is obtained by the fact that $ax_0+by_0+cz_0=-d$. Therefore, the distance $D$ from a point $Q(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ is \begin{equation}\label{eq:distance2plane}D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}\end{equation}

Example. Find the distance between the parallel planes $10x+2y-2z=5$ and $5x+y-z=1$.

Solution. One can easily see that the two planes are parallel because their respective normal vectors $(10,2,-2)$ and $(5,1,-1)$ are parallel. To find the distance between the planes, first one will have to find a point in one plane and then use the formula \eqref{eq:distance2plane} to find the distance. Let us find a point in the plane $10x+2y-2z=5$. One can, for instance, use the $x$-intercept, so let $y=z=0$. Then $10x=5$ i.e. $x=\frac{1}{2}$. The distance from $\left(\frac{1}{2},0,0\right)$ to the plane $5x+y-z=1$ is $$D=\frac{\left|5\left(\frac{1}{2}\right)+1(0)-(0)\right|}{\sqrt{5^2+1^2+(-1)^2}}=\frac{\sqrt{3}}{6}$$

Examples in this note have been taken from [1].

References.

[1] Calculus, Early Transcendentals, James Stewart, 6th Edition, Thompson Brooks/Cole