Skip to main content

CLP-4 Vector Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 1.1 Derivatives, Velocity, Etc.

This being a Calculus text, one of our main operations is differentiation. We are now interested in parametrizations \(\vr(t)\text{.}\) It is very easy and natural to extend our definition of derivative to \(\vr(t)\) as follows.

Definition 1.1.1.

The derivative of the vector valued function \(\vr(t)\) is defined to be
\begin{equation*} \vr'(t) = \diff{\vr}{t}(t)=\lim_{h\rightarrow 0}\frac{\vr(t+h)-\vr(t)}{h} \end{equation*}
when the limit exists. In particular, if \(\vr(t)=\big(x(t)\,,\,y(t)\,,\,z(t)\big)\text{,}\) then
\begin{equation*} \vr'(t)=\big(x'(t)\,,\,y'(t)\,,\,z'(t)\big) \end{equation*}
That is, to differentiate a vector valued function of \(t\text{,}\) just differentiate each of its components.
And of course differentiation interacts with arithmetic operations, like addition, in the obvious way. Only a little more thought is required to see that differentiation interacts quite nicely with dot and cross products too. Here are some examples.
Let
\begin{align*} \va(t)&= t^2\,\hi + t^4\,\hj + t^6\,\hk\\ \vb(t)&= e^{-t}\,\hi + e^{-3t}\,\hj + e^{-5t}\,\hk\\ \ga(t)&= t^2\\ s(t)&= \sin t \end{align*}
We are about to compute some derivatives. To make it easier to follow what is going on, we'll use some colour. When we apply the product rule
\begin{equation*} \diff{}{t}\big[f(t)\,g(t)\big] ={\color{blue}{f'(t)}}\,g(t) + f(t)\,{\color{blue}{g'(t)}} \end{equation*}
we'll use blue to highlight the factors \(f'(t)\) and \(g'(t)\text{.}\) Here we go.
\begin{align*} \ga(t)\,\vb(t) & = t^2e^{-t}\,\hi + t^2 e^{-3t}\,\hj + t^2 e^{-5t}\,\hk \end{align*}
gives
\begin{align*} \diff{}{t}\big[\ga(t)\vb(t)\big] &=\big[{\color{blue}{2t}} e^{-t}{\color{blue}{-}}t^2{\color{blue}{e^{-t}}}\big]\hi +\big[{\color{blue}{2t}} e^{-3t}{\color{blue}{-3}}t^2{\color{blue}{e^{-3t}}}\big]\hj +\big[{\color{blue}{2t}} e^{-5t}{\color{blue}{-5}}t^2{\color{blue}{e^{-5t}}}\big]\hk\\ &={\color{blue}{2t}}\big\{e^{-t}\,\hi + e^{-3t}\,\hj + e^{-5t}\,\hk\big\} + t^2{\color{blue}{\big\{-e^{-t}\,\hi -3 e^{-3t}\,\hj -5 e^{-5t}\,\hk\big\}}}\\ &={\color{blue}{\ga'(t)}}\vb(t)+\ga(t){\color{blue}{\vb'(t)}} \end{align*}
and
\begin{align*} \va(t)\cdot\vb(t) & = t^2e^{-t} + t^4 e^{-3t} + t^6 e^{-5t} \end{align*}
gives
\begin{align*} \diff{}{t}\big[\va(t)\cdot\vb(t)\big] &=\big[{\color{blue}{2t}} e^{-t}{\color{blue}{-}}t^2{\color{blue}{e^{-t}}}\big] +\big[{\color{blue}{4t^3}} e^{-3t}{\color{blue}{-3}}t^4{\color{blue}{e^{-3t}}}\big] +\big[{\color{blue}{6t^5}} e^{-5t}{\color{blue}{-5}}t^6{\color{blue}{e^{-5t}}}\big]\\ &=\big[{\color{blue}{2t}} e^{-t}+{\color{blue}{4t^3}} e^{-3t}+{\color{blue}{6t^5}} e^{-5t}\big] +\big[{\color{blue}{-}}t^2{\color{blue}{e^{-t}}} {\color{blue}{-3}}t^4{\color{blue}{e^{-3t}}} {\color{blue}{-5}}t^6{\color{blue}{e^{-5t}}}\big]\\ &={\color{blue}{\big\{2t\,\hi+4t^3\,\hj+6t^5\,\hk\big\}}}\cdot \big\{e^{-t}\,\hi + e^{-3t}\,\hj + e^{-5t}\,\hk\big\}\\&\hskip0.5in +\big\{t^2\,\hi + t^4\,\hj + t^6\,\hk\big\}\cdot {\color{blue}{\big\{-e^{-t}\,\hi-3e^{-3t}\,\hj-5e^{-5t}\,\hk\big\}}}\\ &={\color{blue}{\va'(t)}}\cdot\vb(t)+\va(t)\cdot{\color{blue}{\vb'(t)}} \end{align*}
and
\begin{align*} \va(t)\times\vb(t) &=\det\left[\begin{matrix}\hi& \hj &\hk\\ t^2 & t^4 & t^6\\ e^{-t} & e^{-3t} & e^{-5t}\end{matrix}\right]\\ &=\hi\big(t^4 e^{-5t}-t^6 e^{-3t}) -\hj(t^2 e^{-5t}- t^6 e^{-t}) +\hk(t^2 e^{-3t}-t^4 e^{-t}) \end{align*}
gives
\begin{align*} &\diff{}{t}\big[\va(t)\times\vb(t)\big]\\ &=\ \hi\big(\ {\color{blue}{4t^3}} e^{-5t}\ \ -\ {\color{blue}{6t^5}} e^{-3t}) \ -\ \hj(\ {\color{blue}{2t}} e^{-5t}\ -\ {\color{blue}{6t^5}} e^{-t}) +\hk(\ {\color{blue}{2t}} e^{-3t}\ -\ {\color{blue}{4t^3}} e^{-t}) \\&\hskip0.1in +\hi\big({\color{blue}{-5}}t^4 {\color{blue}{e^{-5t}}}{\color{blue}{+3}}t^6 {\color{blue}{e^{-3t}}}) -\hj({\color{blue}{-5}}t^2 {\color{blue}{e^{-5t}}}{\color{blue}{+}} t^6 {\color{blue}{e^{-t}}}) +\hk({\color{blue}{-3}}t^2 {\color{blue}{e^{-3t}}}{\color{blue}{+}}t^4 {\color{blue}{e^{-t}}})\\ &={\color{blue}{\big\{2t\,\hi+4t^3\,\hj+6t^5\,\hk\big\}}}\times \big\{e^{-t}\,\hi + e^{-3t}\,\hj + e^{-5t}\,\hk\big\}\\&\hskip0.5in +\big\{t^2\,\hi + t^4\,\hj + t^6\,\hk\big\}\times {\color{blue}{\big\{-e^{-t}\,\hi-3e^{-3t}\,\hj-5e^{-5t}\,\hk\big\}}}\\ &={\color{blue}{\va'(t)}}\times\vb(t)+\va(t)\times{\color{blue}{\vb'(t)}} \end{align*}
and
\begin{align*} \va\big(s(t)\big) &=(\sin t)^2\,\hi +(\sin t)^4\,\hj + (\sin t)^6\,\hk\\ \implies \diff{}{t}\big[\va\big(s(t)\big)\big] &=2(\sin t)\cos t\,\hi +4(\sin t)^3\cos t\,\hj + 6(\sin t)^5\cos t\,\hk\\ &=\big\{2(\sin t)\,\hi +4(\sin t)^3\hj + 6(\sin t)^5\hk\big\}\cos t \\ &=\va'\big(s(t)\big)\,s'(t) \end{align*}
Of course these examples extend to general (differentiable) \(\va(t)\text{,}\) \(\vb(t)\text{,}\) \(\ga(t)\) and \(s(t)\) and give us (most of) the following theorem.
Let's think about the geometric significance of \(\vr'(t)\text{.}\) In particular, let's think about the relationship between \(\vr'(t)\) and distances along the curve. The derivative \(\vr'(t)\) is the limit of \(\frac{\vr(t+h)-\vr(t)}{h}\) as \(h\rightarrow 0\text{.}\) The numerator, \(\vr(t+h)-\vr(t)\text{,}\) is the vector with head at \(\vr(t+h)\) and tail at \(\vr(t)\text{.}\)
When \(h\) is very small this vector
  • has the essentially the same direction as the tangent vector to the curve at \(\vr(t)\) and
  • has length being essentially the length of the part of the curve between \(\vr(t)\) and \(\vr(t+h)\text{.}\)
Taking the limit as \(h\rightarrow 0\) yields that
  • \(\vr'(t)\) is a tangent vector to the curve at \(\vr(t)\) that points in the direction of increasing \(t\) and
  • if \(s(t)\) is the length of the part of the curve between \(\vr(0)\) and \(\vr(t)\text{,}\) then \(\diff{s}{t}(t)=\big|\diff{\vr}{t}(t)\big|\text{.}\)
This is worth stating formally.
As an application, we have the
Note that the velocity \(\vv(t) = \vr'(t)\) is a vector quantity while the speed \(\diff{s}{t}(t)=|\vr'(t)|\) is a scalar quantity.
In general it can be quite difficult to compute arc lengths. So, as an easy warmup example, we will compute the circumference of the circle \(x^2+y^2=a^2\text{.}\) We'll also find a unit tangent to the circle at any point on the circle. We'll use the parametrization
\begin{equation*} \vr(\theta) = \big(a\cos\theta\,,\,a\sin\theta\big)\qquad 0\le \theta\le 2\pi \end{equation*}
of Example 1.0.1. Using Lemma 1.1.4, but with the parameter \(t\) renamed to \(\theta\)
\begin{align*} \vr'(\theta) &= a\big(-\sin\theta\,,\cos\theta\big)\\ \hat\vT(\theta) &= \frac{\vr'(\theta)}{|\vr'(\theta)|} = \big(-\sin\theta\,,\cos\theta\big)\\ \diff{s}{\theta}(\theta)&=\big|\vr'(\theta)\big| = a\\ s(\Theta)-s(0)&= \int_{0}^\Theta \big|\vr'(\theta)\big|\,\dee{\theta} =a\Theta \end{align*}
As 1  \(s(\Theta)\) is the arc length of the part of the circle with \(0\le\theta\le\Theta\text{,}\) the circumference of the whole circle is
\begin{equation*} s(2\pi) = 2\pi a \end{equation*}
which is reassuring, since this formula has been known 2  for thousands of years.
The formula \(s(\Theta)-s(0)=a\Theta\) also makes sense — the part of the circle with \(0\le\theta\le\Theta\) is the fraction \(\frac{\Theta}{2\pi}\) of the whole circle, and so should have length \(\frac{\Theta}{2\pi}\times 2\pi a\text{.}\) Also note that
\begin{equation*} \vr(\theta)\cdot\hat\vT(\theta) = \big(a\cos\theta\,,\,a\sin\theta\big) \cdot \big(-\sin\theta\,,\cos\theta\big) =0 \end{equation*}
so that the tangent to the circle at any point is perpendicular to the radius vector of the circle at that point. This is another geometric fact that has been known 3  for thousands of years.
Consider the curve
\begin{equation*} \vr(t) = 6\sin(2t)\hi + 6\cos(2t)\hj +5t\hk \end{equation*}
where the standard basis vectors \(\hi = (1,0,0)\text{,}\) \(\hj=(0,1,0)\) and \(\hk =(0,0,1)\text{.}\) We'll first sketch it, by observing that
  • \(x(t)=6\sin(2t)\) and \(y(t) =6\cos(2t)\) obey \(x(t)^2+y(t)^2 = 36 \sin^2(2t) + 36\cos^2(2t) = 36\text{.}\) So all points of the curve lie on the cylinder \(x^2+y^2=36\) and
  • as \(t\) increases, \(\big(x(t),y(t)\big)\) runs clockwise around the circle \(x^2+y^2=36\) and at the same time \(z(t) = 5t\) just increases linearly.
Our curve is the helix
We have marked three points of the curve on the above sketch. The first has \(t=0\) and is \(0\hi+6\hj+0\hk\text{.}\) The second has \(t=\frac{\pi}{2}\) and is \(0\hi-6\hj+\frac{5\pi}{2}\hk\text{,}\) and the third has \(t=\pi\) and is \(0\hi+6\hj+5\pi\hk\text{.}\) We'll now use Lemma 1.1.4 to find a unit tangent \(\hat\vT(t)\) to the curve at \(\vr(t)\) and also the arclength of the part of curve between \(t=0\) and \(t=\pi\text{.}\)
\begin{align*} \vr(t) &= 6\sin(2t)\hi + 6\cos(2t)\hj +5t\hk\\ \vr'(t) &= 12\cos(2t)\hi -12\sin(2t)\hj +5\hk\\ \diff{s}{t}(t)&=\big|\vr'(t)\big| =\sqrt{12^2\cos^2(2t) +12^2\sin^2(2t)+5^2} = \sqrt{12^2+5^2}\\ &= 13\\ \hat\vT(t) &= \frac{\vr'(t)}{|\vr'(t))|} = \frac{12}{13}\cos(2t)\hi -\frac{12}{13}\sin(2t)\hj +\frac{5}{13}\hk\\ s(\pi)-s(0)&= \int_{0}^\pi \big|\vr'(t)\big|\,\dee{t} =13\pi \end{align*}
Imagine that, at time \(t\text{,}\) a particle is at
\begin{equation*} \vr(t) = \left[h+a\cos\left(2\pi\frac{t}{T}\right)\right]\hi +\left[k+a\sin\left(2\pi\frac{t}{T}\right)\right]\hj \end{equation*}
As \(|\vr(t) -h\,\hi-k\,\hj| = a\text{,}\) the particle is running around the circle of radius \(a\) centred on \((h,k)\text{.}\) When \(t\) increases by \(T\text{,}\) the argument, \(2\pi\frac{t}{T}\text{,}\) of \(\cos\left(2\pi\tfrac{t}{T}\right)\) and \(\sin\left(2\pi\tfrac{t}{T}\right)\) increases by exactly \(2\pi\) and the particle runs exactly once around the circle. In particular, it travels a distance \(2\pi a\text{.}\) So it is moving at speed \(\frac{2\pi a}{T}\text{.}\) According to Lemma 1.1.5, it has
\begin{align*} \text{velocity } =\vr'(t) &=-\frac{2\pi a}{T}\sin\left(2\pi\frac{t}{T}\right)\hi +\frac{2\pi a}{T}\cos\left(2\pi\frac{t}{T}\right)\hj\\ \text{speed} = \diff{s}{t}(t)&=|\vr'(t)|=\frac{2\pi a}{T}\\ \text{acceleration} =\vr''(t) &=-\frac{4\pi^2 a}{T^2}\cos\left(2\pi\frac{t}{T}\right)\hi -\frac{4\pi^2 a}{T^2}\sin\left(2\pi\frac{t}{T}\right)\hj \\ &= - \frac{4\pi^2}{T^2}\big[\vr(t) -h\,\hi-k\,\hj\big] \end{align*}
Here are some observations.
  • The velocity \(\vr'(t)\) has dot product zero with \(\vr(t) -h\,\hi-k\,\hj\text{,}\) which is the radius vector from the centre of the circle to the particle. So the velocity is perpendicular to the radius vector, and hence parallel to the tangent vector of the circle at \(\vr(t)\text{.}\)
  • The speed given by Lemma 1.1.5 is exactly the speed we found above, just before we started applying Lemma 1.1.5.
  • The acceleration \(\vr''(t)\) points in the direction opposite to the radius vector.
In this example, we find the perimeter of the astroid 4 
\begin{equation*} x^{2/3}+y^{2/3} = a^{2/3} \end{equation*}
A geometric construction of this curve, as well as a derivation of its equation is given in the optional section 1.11 later. We'll start by finding a convenient parametrization.
  • To do so, notice that \(x^{2/3}+y^{2/3} = a^{2/3}\) looks somewhat like the equation of the circle \(x^2+y^2=a^2\text{.}\)
  • The standard parametrization of the circle, namely \(x=a\cos t\text{,}\) \(y=a \sin t\) works because of the elementary trig identity \(\cos^2t+\sin^2t=1\text{.}\)
  • If we can arrange that \(x(t)^{2/3} = a^{2/3}\cos^2 t\) and \(y(t)^{2/3}=a^{2/3}\sin^2 t\text{,}\) then the same elementary trig identity will give \(x(t)^{2/3}+y(t)^{2/3} = a^{2/3}\text{,}\) as desired.
  • But of course its easy to arrange that: just solve \(x(t)^{2/3} = a^{2/3}\cos^2 t\) for \(x(t)\text{,}\) namely \(x(t) = a\cos^3t\text{,}\) and solve \(y(t)^{2/3}=a^{2/3}\sin^2 t\) for \(y(t)\text{,}\) namely \(y(t)=a\sin^3 t\text{.}\)
Our parametrization is
\begin{equation*} \vr(t) = a\cos^3t\,\hi + a\sin^3 t\,\hj \end{equation*}
By Lemma 1.1.4
\begin{align*} \vr(t) &= a\cos^3t\,\hi + a\sin^3 t\,\hj\\ \vr'(t) &= -3a\sin t\cos^2t\,\hi + 3a\sin^2 t\cos t\,\hj\\ \diff{s}{t}(t)&=\big|\vr'(t)\big| = \sqrt{9a^2\sin^2t\cos^4t + 9a^2\sin^4t\cos^2t}\\ & = 3a \sqrt{\sin^2t\cos^2t(\cos^2t+\sin^2t)}\\ &=3a\big|\sin t\cos t\big|\\ \hat\vT(t) &= \frac{\vr'(t)}{|\vr'(t))|} = \frac{\sin t\cos t}{|\sin t\cos t|}\ \big(-\cos t\,\hi+\sin t\,\hj\big)\\ &= \sgn\!\big(\!\sin t\cos t\big)\ \big(-\cos t\,\hi+\sin t\,\hj\big) \end{align*}
Here \(\sgn\!\big(\!\sin t\cos t\big)\) means “the sign of \(\sin t\cos t\)”, i.e \(+1\) when \(\sin t\cos t \gt 0\) and \(-1\) when \(\sin t\cos t \lt 0\text{.}\) So
\begin{align*} &\hat\vT(t)\\ &=\left.\begin{cases}1&\text{if } \sin t \gt 0,\ \cos t \gt 0 \text{ or } \sin t \lt 0,\ \cos t \lt 0\\ -1&\text{if } \sin t \gt 0,\ \cos t \lt 0 \text{ or } \sin t \lt 0,\ \cos t \gt 0 \end{cases}\right\}\big(-\cos t\,\hi+\sin t\,\hj\big)\\ &=\left.\begin{cases}1&\text{if } 0 \lt t \lt \frac{\pi}{2} \text{ or } \pi \lt t \lt \frac{3\pi}{2}\\ -1&\text{if } \frac{\pi}{2} \lt t \lt \pi \text{ or } \frac{3\pi}{2} \lt t \lt 2\pi \end{cases}\right\}\big(-\cos t\,\hi+\sin t\,\hj\big) \end{align*}
Before we go on to sketch the astroid and compute its perimeter, we can make a few observations that will simplify our lives.
  • The signs of both components of \(\vr(t)\) are the same as the signs of the components of \(\cos t\,\hi +\sin t\,\hj\text{;}\) and the signs of both components of \(\vr'(t)\) are the same as the signs of the components of \(-\sin t\,\hi +\cos t\,\hj\text{.}\) Consequently the astroid looks somewhat like a circle in that
    • when \(0\le t\le \frac{\pi}{2}\text{,}\) \(\vr(t)\) lies in the first quadrant and moves upward and to the left as \(t\) increases and
    • when \(\frac{\pi}{2}\le t\le \pi\text{,}\) \(\vr(t)\) lies in the second quadrant and moves downward and to the left as \(t\) increases and
    • when \(\pi\le t\le \frac{3\pi}{2}\text{,}\) \(\vr(t)\) lies in the third quadrant and moves downward and to the right as \(t\) increases and
    • when \(\frac{3\pi}{2}\le t\le 2\pi\text{,}\) \(\vr(t)\) lies in the fourth quadrant and moves upward and to the right as \(t\) increases and
    • \(\vr(2\pi)=\vr(0)\) so that the astroid is a closed curve that circumnavigates the origin exactly once as \(t\) runs from \(0\) to \(2\pi\text{.}\)
  • Something weird happens at those values of \(t\) where \(\sin t\cos t\) changes sign 5 , i.e. at \(t=0\text{,}\) \(\frac{\pi}{2}\text{,}\) \(\pi\text{,}\) \(\frac{3\pi}{2}\text{,}\) etc. Namely \(\hat T(t)\) flips. To be precise
    \begin{align*} \lim_{t\rightarrow 0-}\hat T(t) &= \lim_{t\rightarrow 0-} \sgn\!\big(\!\sin t\cos t\big)\ \lim_{t\rightarrow 0-} \big(-\cos t\,\hi+\sin t\,\hj\big) =\hi \\ \lim_{t\rightarrow 0+}\hat T(t) &= \lim_{t\rightarrow 0+}\sgn\!\big(\!\sin t\cos t\big)\ \lim_{t\rightarrow 0+}\big(-\cos t\,\hi+\sin t\,\hj\big) =-\hi \end{align*}
    and
    \begin{align*} \lim_{t\rightarrow \pi/2-}\hat T(t) &= \lim_{t\rightarrow \pi/2-}\sgn\!\big(\!\sin t\cos t\big)\ \lim_{t\rightarrow \pi/2-}\big(-\cos t\,\hi+\sin t\,\hj\big) =\hj \\ \lim_{t\rightarrow \pi/2+}\hat T(t) &= \lim_{t\rightarrow \pi/2+}\sgn\!\big(\!\sin t\cos t\big)\ \lim_{t\rightarrow \pi/2+}\big(-\cos t\,\hi+\sin t\,\hj\big) =-\hj \end{align*}
    and so on. This signals cusps in the curve at \(t=0\text{,}\) i.e. at \(\vr(0) = a\hi\text{,}\) and at \(t=\frac{\pi}{2}\text{,}\) i.e. at \(\vr(\frac{\pi}{2}) = a\hj\text{,}\) and so on. So while the astroid looks somewhat like a circle, it has cusps at \(\pm a\hi\) and \(\pm a\hj\text{.}\) Here is the sketch.
  • The astroid is invariant under reflections in the \(x\)-axis and in the \(y\)-axis. That is, \(x^{2/3}+y^{2/3} = a^{2/3}\) is invariant under \(x\rightarrow -x\) and also under \(y\rightarrow -y\text{.}\) So to find the whole perimeter, it suffices to find the arc length of the part of the astroid in the first quadrant, and then multiply by \(4\text{.}\)
    \begin{align*} \text{perimeter} &=4\int_0^{\pi/2}\diff{s}{t}\ \dee{t} = 4\int_0^{\pi/2}3a\sin t\cos t\ \dee{t}\\ & = 6a\int_0^{\pi/2}\sin(2t)\ \dee{t} =6a\Big[-\frac{\cos(2t)}{2}\Big]_0^{\pi/2}\\ &=6a \end{align*}
In the last example, we found that the astroid had cusps at those points \(\vr(t)\) where the velocity \(\vr'(t)\) vanished. In this example, we will explore a little further what can happen when \(\vr'(t)=\vZero\text{.}\)
Suppose that you are out for a walk and that your position at time \(t\) is \(\vr(t)\text{.}\) If at some time you have nonzero velocity, it is very hard for you to change your direction of motion discontinuously 6 . On the other hand, when \(\vr'(t)=0\text{,}\) you are not moving at all and it is easy for you to turn and leave in any direction you choose. You could reverse direction completely, or make a sharp left turn, or not change direction at all. Here are examples of all of these. They all have \(\vr'(t)=0\text{.}\) They are sketched below.
\begin{alignat*}{4} \vr_1(t) &= (t^5, t^2) & \vr_1'(t)&= (5t^4, 2t)\\ \vr_2(t) &= \left.\begin{cases} (t^2, 0) & \text{if } t\ge 0 \\ (0 , t^2) & \text{if } t\le 0 \end{cases}\right\}\qquad & \vr_2'(t)&= \left.\begin{cases} (2t, 0) & \text{if } t\ge 0 \\ (0 , 2t) & \text{if } t\le 0 \end{cases} \right\}\\ \vr_3(t) &= (t^3, 0) & \vr_3'(t) &= (3t^2, 0) \end{alignat*}
We'll find the arc length of
\begin{equation*} \vr(t) = t\cos t\,\hi + t\sin t\,\hj +t\,\hk\qquad 0\le t\le\sqrt{2} \end{equation*}
We'll first sketch it, by observing that
  • \(x(t)=t\cos t \text{,}\) \(y(t) =t\sin t \) and \(z(t)=t\) obey \(x(t)^2+y(t)^2 = t^2 = z(t)^2\text{.}\) So all points of the curve lie on the cone \(x^2+y^2=z^2\) and
  • as \(t\) increases, \(\big(x(t),y(t)\big)\) runs counterclockwise around a “circle” whose radius increases linearly with \(t\) and at the same time \(z(t)\) also increases linearly.
Our curve is the “corkscrew”
By Lemma 1.1.4
\begin{align*} \vr(t) &= t\cos t\,\hi + t\sin t\,\hj +t\,\hk\\ \vr'(t) &= [\cos t -t\sin t ]\hi +[\sin t+t\cos t]\hj + \hk \end{align*}
so that
\begin{align*} &\diff{s}{t}(t)=\big|\vr'(t)\big|\\ &= \sqrt{\big(\cos^2t\!-\!2 t\sin t\cos t\!+\!t^2\sin^2 t\big) \!+\!\big(\sin^2t\!+\!2 t\sin t\cos t \!+\!t^2\cos^2 t\big)\!+\!1}\\ &=\sqrt{2+t^2} \end{align*}
Our goal, stated at the beginning of this example, was to compute
\begin{equation*} s(\sqrt{2})-s(0)= \int_{0}^{\sqrt{2}} \big|\vr'(t)\big|\,\dee{t} = \int_{0}^{\sqrt{2}} \sqrt{2+t^2}\,\dee{t} \end{equation*}
To evaluate the integral, we'll use three techniques that you learned in your first integral calculus course. First, motivated by the \(\sqrt{2+t^2}\text{,}\) we'll use the trigonometric substitution
\begin{equation*} t=\sqrt{2}\tan u\qquad \dee{t} =\sqrt{2}\sec^2 u\,\dee{u}\qquad 2+t^2=2\big[1+\tan^2u\big]=2\sec^2 u \end{equation*}
When \(t=0\text{,}\) \(u=0\) and when \(t=\sqrt{2}\text{,}\) \(\tan u=1\) so that \(u=\frac{\pi}{4}\) and
\begin{align*} s(\sqrt{2})-s(0)&= \int_{0}^{\pi/4} \sqrt{2\sec^2u}\,\sqrt{2}\sec^2 u\,\dee{u} = 2\int_{0}^{\pi/4} \sec^3 u\,\dee{u} \end{align*}
You may have evaluated this integral in first year. There are several ways of doing so. Perhaps the most straight forward, but also most tedious, method is to rewrite the integral as
\begin{equation*} s(\sqrt{2})-s(0)= 2\int_{0}^{\pi/4} \frac{\cos u}{\cos^4 u}\,\dee{u} \end{equation*}
We recognize that this is a trigonometric integral that contains an odd power of \(\cos u\text{,}\) so we substitute \(w=\sin u\text{,}\) \(\dee{w}=\cos u\,\dee{u}\text{,}\) \(\cos^2 u= 1-w^2\text{.}\) When \(u=0\text{,}\) \(w=0\) and when \(u=\frac{\pi}{4}\text{,}\) \(w=\frac{1}{\sqrt{2}}\) so that
\begin{equation*} s(\sqrt{2})-s(0)= 2\int_{0}^{1/\sqrt{2}} \frac{\dee{w}}{{(1-w^2)}^2} \end{equation*}
The integrand is now a rational function, i.e. a ratio of polynomials. So we apply partial fractions.
\begin{align*} s(\sqrt{2})-s(0) &= 2\int_{0}^{1/\sqrt{2}} \frac{\dee{w}}{{[(1-w)(1+w)]}^2}\\ &= \frac{1}{2}\int_{0}^{1/\sqrt{2}} \Big[\frac{1}{1-w}+\frac{1}{1+w}\Big]^2 \dee{w}\\ &= \frac{1}{2}\int_{0}^{1/\sqrt{2}} \Big[\frac{1}{(1-w)^2}+\frac{2}{(1-w)(1+w)}+\frac{1}{(1+w)^2}\Big] \dee{w}\\ &= \frac{1}{2}\int_{0}^{1/\sqrt{2}} \Big[\frac{1}{(1-w)^2}+\frac{1}{1-w}+\frac{1}{1+w} +\frac{1}{(1+w)^2}\Big] \dee{w}\\ &= \frac{1}{2} \Big[\frac{1}{1-w}-\ln|1-w|+\ln|1+w| -\frac{1}{1+w}\Big]_{0}^{1/\sqrt{2}}\\ &= \frac{1}{2} \Big[\frac{2w}{1-w^2}+\ln\frac{1+w}{1-w}\Big]_{0}^{1/\sqrt{2}} = \frac{1}{2} \Big[2\sqrt{2}+\ln\frac{\sqrt{2}+1}{\sqrt{2}-1}\Big]\\ &\approx 2.2956 \end{align*}
Ooof!

Exercises Exercises

Exercise Group.

Exercises — Stage 1
Questions 1.1.1 through 1.1.5 provide practice with curve parametrization. Being comfortable with the algebra and interpretation of these descriptions are essential ingredients in working effectively with parametrizations.
1.
Find the specified parametrization of the first quadrant part of the circle \(x^2+y^2=a^2\text{.}\)
  1. In terms of the \(y\) coordinate.
  2. In terms of the angle between the tangent line and the positive \(x\)-axis.
  3. In terms of the arc length from \((0,a)\text{.}\)
2.
Consider the following time-parametrized curve:
\begin{equation*} \vr(t)=\left( \cos\left(\frac{\pi}{4}t\right), (t-5)^2\right) \end{equation*}
List the three points \((-1/\sqrt{2},0)\text{,}\) \((1,25)\text{,}\) and \((0,25)\) in chronological order.
3.
At what points in the \(xy\)-plane does the curve \((\sin t, t^2)\) cross itself? What is the difference in \(t\) between the first time the curve crosses through a point, and the last?
4.
A circle of radius \(a\) rolls along the \(x\)-axis in the positive direction, starting with its centre at \((a,a)\text{.}\) In that position, we mark the topmost point on the circle \(P\text{.}\) As the circle moves, \(P\) moves with it. Let \(\theta\) be the angle the circle has rolled — see the diagram below.
  1. Give the position of the centre of the circle as a function of \(\theta\text{.}\)
  2. Give the position of \(P\) as a function of \(\theta\text{.}\)
5.
The curve \(C\) is defined to be the intersection of the hyperboloid
\begin{equation*} x^2-\frac{1}{4}y^2+3z^2=1 \end{equation*}
and the plane
\begin{equation*} x+y+z=0. \end{equation*}
When \(y\) is very close to 0, and \(z\) is negative, find an expression giving \(z\) in terms of \(y\text{.}\)
6.
A particle traces out a curve in space, so that its position at time \(t\) is
\begin{equation*} \vr(t)=e^{-t}\,\hi+\frac{1}{t}\,\hj+(t-1)^2(t-3)^2\,\hk \end{equation*}
for \(t \gt 0\text{.}\)
Let the positive \(z\) axis point vertically upwards, as usual. When is the particle moving upwards, and when is it moving downwards? Is it moving faster at time \(t=1\) or at time \(t=3\text{?}\)
7.
Below is the graph of the parametrized function \(\vr(t)\text{.}\) Let \(s(t)\) be the arclength along the curve from \(\vr(0)\) to \(\vr(t)\text{.}\)
Indicate on the graph \(s(t+h)-s(t)\) and \(\vr(t+h)-\vr(t)\text{.}\) Are the quantities scalars or vectors?
8.
What is the relationship between velocity and speed in a vector-valued function of time?
9. (✳).
Let \(\vr(t)\) be a vector valued function. Let \(\vr'\text{,}\) \(\vr''\) , and \(\vr'''\) denote \(\diff{\vr}{t}\text{,}\) \(\difftwo{\vr}{t}\) and \(\frac{\mathrm{d^3} \vr }{\mathrm{d}{t}^3}\text{,}\) respectively. Express
\begin{equation*} \diff{ }{t}\big[ (\vr \times \vr')\cdot\vr'' \big] \end{equation*}
in terms of \(\vr\text{,}\) \(\vr'\) , \(\vr''\) , and \(\vr'''\text{.}\) Select the correct answer.
  1. \(\displaystyle (\vr'\times\vr'' )\cdot\vr'''\)
  2. \(\displaystyle (\vr'\times\vr'' )\cdot\vr + (\vr\times\vr' )\cdot\vr'''\)
  3. \(\displaystyle (\vr\times\vr' )\cdot\vr'''\)
  4. \(\displaystyle 0\)
  5. None of the above.
10.
Show that, if the position and velocity vectors of a moving particle are always perpendicular, then the path of the particle lies on a sphere.

Exercise Group.

Exercises — Stage 2
11. (✳).
Find the speed of a particle with the given position function
\begin{equation*} \vr(t) = 5 \sqrt{2}\,t\,\hi + e^{5t}\,\hj - e^{-5t}\,\hk \end{equation*}
Select the correct answer:
  1. \(\displaystyle |\vv(t)| = \big(e^{5t} + e^{-5t}\big)\)
  2. \(\displaystyle |\vv(t)| = \sqrt{10 + 5e^{t} + 5e^{-t}}\)
  3. \(\displaystyle |\vv(t)| = \sqrt{10 + e^{10t} + e^{-10t}}\)
  4. \(\displaystyle |\vv(t)| = 5\big(e^{5t} + e^{-5t}\big)\)
  5. \(\displaystyle |\vv(t)| = 5\big(e^t + e^{-t}\big)\)
12.
Find the velocity, speed and acceleration at time \(t\) of the particle whose position is
\begin{equation*} \vr(t)= a \cos t\,\hi+a\sin t\,\hj+ct\,\hk \end{equation*}
Describe the path of the particle.
13. (✳).
  1. Let
    \begin{equation*} \vr(t) = \left(t^2 , 3, \tfrac{1}{3} t^3 \right) \end{equation*}
    Find the unit tangent vector to this parametrized curve at \(t = 1\text{,}\) pointing in the direction of increasing \(t\text{.}\)
  2. Find the arc length of the curve from (a) between the points \((0, 3, 0)\) and \((1, 3, -\frac{1}{3})\text{.}\)
14.
Using Lemma 1.1.4, find the arclength of \(\vr(t)=\left(t,\sqrt{\frac{3}{2}}t^2,t^3\right)\) from \(t=0\) to \(t=1\text{.}\)
15.
Find the length of the parametric curve
\begin{equation*} x=a\cos t\sin t\qquad y=a\sin^2 t\qquad z=bt \end{equation*}
between \(t=0\) and \(t=T \gt 0\text{.}\)
16.
A particle's position at time \(t\) is given by \(\vr(t)=(t+\sin t, \cos t)\)  7 . What is the magnitude of the acceleration of the particle at time \(t\text{?}\)
17. (✳).
A curve in \(\bbbr^3\) is given by the vector equation \(\vr(t) = \left(2t \cos t, 2t \sin t,\frac{t^3}{3}\right)\)
  1. Find the length of the curve between \(t = 0\) and \(t = 2\text{.}\)
  2. Find the parametric equations of the tangent line to the curve at \(t = \pi\text{.}\)
18. (✳).
Let \(\vr(t) = \big(3 \cos t, 3 \sin t, 4t\big)\) be the position vector of a particle as a function of time \(t \ge 0\text{.}\)
  1. Find the velocity of the particle as a function of time \(t\text{.}\)
  2. Find the arclength of its path between \(t = 1\) and \(t = 2\text{.}\)
19.
The plane \(\ z=2x+3y\ \) intersects the cylinder \(\ x^2+y^2=9\ \) in an ellipse.
  1. Find a parametrization of the ellipse.
  2. Express the circumference of this ellipse as an integral. You need not evaluate the integral 8 .
20. (✳).
Consider the curve
\begin{equation*} \vr(t) = \frac{1}{3}\cos^3 t\,\hi +\frac{1}{3} \sin^3 t\,\hj + \sin^3 t\,\hk \end{equation*}
  1. Compute the arc length of the curve from \(t = 0\) to \(t = \frac{\pi}{2}\text{.}\)
  2. Compute the arc length of the curve from \(t = 0\) to \(t = \pi\text{.}\)
21. (✳).
Let \(\vr(t)=\big(\frac{1}{3}t^3,\frac{1}{2}t^2,\frac{1}{2}t\big)\text{,}\) \(t\ge 0\text{.}\) Compute \(s(t\)), the arclength of the curve at time \(t\text{.}\)
22. (✳).
Find the arc length of the curve \(\vr(t) = \big(t^m\,,\, t^m\,,\, t^{3m/2}\big)\) for \(0 \le a \le t \le b\text{,}\) and where \(m \gt 0\text{.}\) Express your result in terms of \(m\text{,}\) \(a\text{,}\) and \(b\text{.}\)
23.
Let \(C\) be the part of the curve of intersection of the parabolic cylinder \(x = y^2\) and the hyperbolic paraboloid \(3z = 2xy\) with \(y\ge 0\text{.}\)
  1. Write a vector parametric equation for \(C\) using \(x\) as the parameter.
  2. Find the length of the part of \(C\) between the origin and the point \((9, 3, 18)\text{.}\)
  3. A particle moves along \(C\) in the direction for which \(x\) is increasing. If the particle moves with constant speed 9, find its velocity vector when it is at the point \((1, 1, \frac{2}{3})\text{.}\)
  4. Find the acceleration vector of the particle of part (c) when it is at the point \((1, 1, \frac{2}{3})\text{.}\)
24.
If a particle has constant mass \(m\text{,}\) position \(\vr\text{,}\) and is moving with velocity \(\vv\text{,}\) then its angular momentum is \(\vL=m(\vr\times\vv)\text{.}\)
For a particle with mass \(m=1\) and position function \(\vr=(\sin t, \cos t, t)\text{,}\) find \(\left|\diff{\vL}{t} \right|\text{.}\)

Exercise Group.

Exercises — Stage 3
25. (✳).
A particle moves along the curve \(\cC\) of intersection of the surfaces \(z^2=12y\) and \(18x=yz\) in the upward direction. When the particle is at \((1,3,6)\) its velocity \(\vv\) and acceleration \(\va\) are given by
\begin{equation*} \vv =6\,\hi+12\,\hj+12\,\hk\qquad \va = 27\,\hi+30\,\hj+6\,\hk \end{equation*}
  1. Write a vector parametric equation for \(\cC\) using \(u=\frac{z}{6}\) as a parameter.
  2. Find the length of \(\cC\) from \((0,0,0)\) to \((1,3,6)\text{.}\)
  3. If \(u=u(t)\) is the parameter value for the particle's position at time \(t\text{,}\) find \(\diff{u}{t}\) when the particle is at \((1,3,6)\text{.}\)
  4. Find \(\difftwo{u}{t}\) when the particle is at \((1,3,6)\text{.}\)
26. (✳).
A particle of mass \(m = 1\) has position \(\vr_0 = \frac{1}{2}\,\hk\) and velocity \(\vv_0 =\frac{\pi^2}{2}\,\hi\) at time \(0\text{.}\) It moves under a force
\begin{equation*} \vF(t) = -3t\,\hi + \sin t\,\hj + 2e^{2t}\,\hk. \end{equation*}
  1. Determine the position \(\vr(t)\) of the particle depending on \(t\text{.}\)
  2. At what time after time \(t = 0\) does the particle cross the plane \(x = 0\) for the first time?
  3. What is the velocity of the particle when it crosses the plane \(x = 0\) in part (b)?
27. (✳).
Let \(C\) be the curve of intersection of the surfaces \(y=x^2\) and \(z=\frac{2}{3}x^3\text{.}\) A particle moves along \(C\) with constant speed such that \(\diff{x}{t} \gt 0\text{.}\) The particle is at \((0,0,0)\) at time \(t=0\) and is at \((3,9,18)\) at time \(t=\frac{7}{2}\text{.}\)
  1. Find the length of the part of \(C\) between \((0,0,0)\) and \((3,9,18)\text{.}\)
  2. Find the constant speed of the particle.
  3. Find the velocity of the particle when it is at \(\big(1,1,\frac{2}{3}\big)\text{.}\)
  4. Find the acceleration of the particle when it is at \(\big(1,1,\frac{2}{3}\big)\text{.}\)
28.
A camera mounted to a pole can swivel around in a full circle. It is tracking an object whose position at time \(t\) seconds is \(x(t)\) metres east of the pole, and \(y(t)\) metres north of the pole.
In order to always be pointing directly at the object, how fast should the camera be programmed to rotate at time \(t\text{?}\) (Give your answer in terms of \(x(t)\) and \(y(t)\) and their derivatives, in the units rad/sec.)
29.
A pipe of radius 3 follows the path of the curve \(\vr(t)=(\frac{2\sqrt2}{3}t^{3/2} , \frac12t^2 , t+2)\text{,}\) for \(0 \le t \le 10\text{.}\)
What is the volume inside the pipe? What is the surface area of the pipe?
30.
A wire of total length 1000cm is formed into a flexible coil that is a circular helix. If there are 10 turns to each centimetre of height and the radius of the helix is 3 cm, how tall is the coil?
31.
A projectile falling under the influence of gravity and slowed by air resistance proportional to its speed has position satisfying
\begin{equation*} \difftwo{\vr}{t}=-g\hk-\alpha\diff{\vr}{t} \end{equation*}
where \(\alpha\) is a positive constant. If \(\vr=\vr_0\) and \(\diff{\vr}{t}=\vv_0\) at time \(t=0\text{,}\) find \(\vr(t)\text{.}\)
You might guess that \(\Theta\) is a capital Greek theta. You'd be right.
The earliest known written approximations of \(\pi\text{,}\) in Egypt and Babylon, date from 1900--1600BC. The first recorded algorithm for rigorously evaluating \(\pi\) was developed by Archimedes around 250 BC. The first use of the symbol \(\pi\text{,}\) for the ratio between the circumference of a circle and its diameter, in print was in 1706 by William Jones.
It is Proposition 18 in Book 3 of Euclid's Elements. It was published around 300BC.
Astroid should not be confused with asteroid, though both words derive from the Greek word for star.
Like a cross-walk sign.
For your velocity to jump discontinuously, your acceleration has to be infinite, which requires an infinite force. You might not look so healthy afterwards
The particle traces out a cycloid — see Question 1.1.4
The indefinite integral involved is one of a class of integrals called elliptic integrals because of their connections to arc lengths of ellipses. In general, elliptic integrals cannot be expressed in terms of elementary functions. You can easily find discussions of elliptic integrals using your favourite search engine.
<Prev^TopNext>
Feedback