[b][size=150]Differentiation of a vector-valued function[/size][/b][br][br][u]Definition[/u]: A vector-valued function [math]\vec{r}(t)=\langle f(t),g(t),h(t)\rangle[/math] for [math]a < t < b[/math] is [u]differentiable[/u] if [math]f(t),g(t),h(t)[/math] are differentiable on [math](a,b)[/math]. Then the [b]derivative[/b] of [math]\vec{r}(t)[/math] is defined as follows:[br][br][math]\vec{r'}(t)=\langle f'(t),g'(t),h'(t)\rangle[/math][br][br]Equivalently, we can write the derivative in term of limit:[br][br][math]\vec{r'}(t)=\lim_{\Delta t\to 0}\frac 1{\Delta t} (\vec{r}(t+\Delta t)-\vec{r}(t))[/math][br][br]In the applet below, we visualize [math]\vec{r'}(t)[/math] as [b]a tangent vector[/b] in the positive orientation of the curve (if it is not a zero vector).[br]
[u]Exercise[/u]: Let [math]\vec{r}(t)=\langle t^2, 4t, 4\ln t\rangle[/math] for [math]t>0[/math]. Then compute the unit tangent vector of the curve parametrized by [math]\vec{r}(t)[/math] in terms of [math]t[/math].[br]
[b]Answer[/b]:[br][br]Tangent vector [math]\vec{r'}(t)=\left\langle 2t, 4, \frac 4t\right\rangle[/math].[br][br]Therefore, the unit tangent vector [math]=\frac 1{\sqrt{(2t)^2+4^2+\left(\frac 4t\right)^2}}\left\langle 2t, 4, \frac 4t\right\rangle=\frac 1{2t+\frac 4t}\left\langle 2t, 4, \frac 4t\right\rangle=\left\langle \frac{t^2}{t^2+2},\frac{2t}{t^2+2},\frac 2{t^2+2}\right\rangle[/math].
[u]Properties of differentation[/u][br][br]Let [math]\vec{u}(t)[/math] and [math]\vec{v}(t)[/math] be differentiable vector-valued function, [math]p(t)[/math] be a differentiable real-valued function, and [math]\vec{c}[/math] be a constant vector.[br][br][list=1][*][math]\frac{d}{dt}(\vec{c})=\vec{0} [/math][/*][br][*][math]\frac{d}{dt}(\vec{u}(t)+\vec{v}(t))=\vec{u'}(t)+\vec{v'}(t)[/math][/*][br][*][math]\frac{d}{dt}(p(t)\vec{u}(t))=p'(t)\vec{u}(t)+p(t)\vec{u'}(t)[/math][/*][br][*][math]\frac{d}{dt}\vec{u}(p(t))=p'(t)\vec{u'}(p(t))[/math][/*][br][*][math]\frac{d}{dt}(\vec{u}(t)\cdot\vec{v}(t))=\vec{u'}(t)\cdot\vec{v}(t)+\vec{u}(t)\cdot\vec{v'}(t)[/math][/*][br][*][math]\frac{d}{dt}(\vec{u}(t)\times\vec{v}(t))=\vec{u'}(t)\times\vec{v}(t)+\vec{u}(t)\times\vec{v'}(t)[/math][/*][br][/list][br]All the above properties can be easily proved by writing down both sides in terms of the vector components.[br][br]
[u]Exercise[/u]: Let [math]\vec{u}(t)=\langle \sin t, 2\cos t, \cos t\rangle[/math] and [math]\vec{v}(t)=\langle -t,t^2,2t^3\rangle[/math]. Compute[br][list=1][*][math]\frac{d}{dt}(\vec{u}(t^2))[/math][/*][br][*][math]\frac{d}{dt}(\vec{u}\cdot\vec{v})[/math][/*][br][/list]
[b]Answer[/b]:[br][br]1. [math]\frac{d}{dt}(\vec{u}(t^2))=2t\langle \cos (t^2), -2\sin (t^2), -\sin (t^2) \rangle=\langle 2t\cos (t^2), -4t\sin (t^2), -2t\sin (t^2) \rangle[/math][br][br]2. [math]\frac{d}{dt}(\vec{u}\cdot\vec{v})=\langle \cos t, -2\sin t,[br]-\sin t\rangle \cdot \langle -t,t^2,2t^3\rangle+\langle \sin t,2\cos t,\cos t\rangle\cdot \langle -1,2t,6t^2\rangle[/math][br][math] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -t\cos t-2t^2\sin t-2t^3\sin t-\sin t+4t\cos t+6t^2\cos t[/math][br][math] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (-2t^3-2t^2-1)\sin t+(6t^2+4t-t)\cos t[/math]
[u]Exercise[/u]: Let [math]k[/math] be a positive constant and [math]\vec{r}(t)[/math] be a vector-valued function such that [math]|\vec{r}(t)|=k[/math] for any [math]t[/math]. Prove that [math]\vec{r}(t)[/math] is orthogonal to [math]\vec{r'}(t)[/math] for all [math]t[/math].[br][br]Can you explain this result from a geometry point of view?[br]
[b]Answer[/b]:[br][br]Since [math]\vec{r}(t)\cdot \vec{r}(t)=|\vec{r}(t)|^2=k^2[/math], [math]\frac{d}{dt}(\vec{r}(t)\cdot \vec{r}(t))=0[/math].[br][br]Hence, for all [math]t[/math], we have [math]\vec{r'}(t)\cdot \vec{r}(t)+\vec{r}\cdot\vec{r'}(t)=0[/math][br][math]\implies 2\vec{r'}(t)\cdot\vec{r}(t)=0[/math][br][math]\implies \vec{r'}(t)\cdot\vec{r}(t)=0[/math] i.e. [math]\vec{r'}(t)[/math] is orthogonal to [math]\vec{r}(t)[/math] for all [math]t[/math].
[b][size=150]Integration of vector-valued functions[/size][/b][br][br]An [b]antiderivative[/b] of a vector-valued function [math]\vec{r}(t)[/math] is [math]\vec{R}(t)[/math] if [math]\vec{R'}(t)=\vec{r}(t)[/math]. Explicitly, if [math]\vec{r}(t)=\langle f(t),g(t),h(t)\rangle[/math] and [math]\vec{R}(t)=\langle F(t),G(t),H(t)\rangle[/math], then [math]F'(t)=f(t), G'(t)=g(t), H'(t)=h(t)[/math], which means that [math]F,G,H[/math] are antiderivative of [math]f,g,h[/math] respectively. Moreover, the indefinite integral of [math]\vec{r}(t)[/math] is[br][br][math]\int \vec{r}(t) \ dt=\vec{R}(t)+\vec{C}[/math][br][br]where [math]\vec{C}[/math] is an arbitrary constant vector.[br][br][u]Example[/u][br][br]Find [math]\vec{r}(t)[/math] with [math]\vec{r'}(t)=\langle \sin t, 3t, e^t\rangle[/math] and [math]\vec{r}(0)=\vec{i}[/math][br][br][u]Answer[/u]:[br][br][math]\vec{r}(t)=\int \vec{r'}(t) \ dt = \langle -\cos t, \frac{3t^2}2, e^t\rangle + \vec{C}[/math][br][br][math]\vec{r}(0)=\langle -1, 0, 1\rangle+\vec{C}=\langle 1,0,0\rangle[/math][br][br][math]\implies \vec{C}=\langle 2,0,-1\rangle[/math][br][br]Therefore, [math]\vec{r}(t)=\langle -\cos t+2, \frac{3t^2}2, e^t-1\rangle[/math].[br][br][br]The definite integral of [math]\vec{r}(t)[/math] is defined as follows:[br][br][math]\int_a^b \vec{r}(t) \ dt = \left\langle \int_a^b f(t) \ dt, \int_a^b g(t) \ dt, \int_a^b h(t) \ dt\right\rangle[/math][br][br]By Fundamental Theorem of Calculus, we have[br][br][math]\int_a^b \vec{r}(t) \ dt = \langle F(b)-F(a),G(b)-G(a),H(b)-H(a)\rangle = \vec{R}(b)-\vec{R}(a)[/math][br][br]where [math]\vec{R}(t)[/math] is an antiderivative of [math]\vec{r}(t)[/math].[br]
[u]Exercise[/u]: Suppose the tangent vector of a curve parametrized by [math]\vec{r}(t)[/math] for [math]0\leq t \leq \pi[/math] is [math]\left\langle 1, 3\cos \left(\frac t2\right),-3t\right\rangle[/math]. Find the vector from the point on the curve when [math]t=0[/math] to the point on the curve when [math]t=\pi[/math].[br]
[b]Answer[/b]:[br][br]It is given that [math]\vec{r'}(t)=\left\langle 1, 3\cos \left(\frac t2\right),-3t\right\rangle[/math]. Then by the Fundatmental Theorem of Calculus, we have[br][br][math]\vec{r}(\pi)-\vec{r}(0)=\int_0^{\pi} \vec{r'}(t) \ dt = \left. \left\langle t, 6\sin\left(\frac t2\right),-\frac{3t^2}2\right\rangle \right|_0^{\pi}=\left\langle \pi,6,-\frac{3\pi^2}2\right\rangle-\langle 0,0,0\rangle=\left\langle \pi,6,-\frac{3\pi^2}2\right\rangle[/math][br][br]Hence, the required vector is [math]\left\langle \pi,6,-\frac{3\pi^2}2\right\rangle[/math][br]