The Calculus of Vector-Valued Functions

The Calculus of Vector-Valued Functions

In our first and second term of calculus, we could handle dynamic situations in only one dimension.

Today we learn how to handle the dynamics of curves in both $2$ and $3$-dimensional space!

Derivatives of Vector-Valued Functions

Let ${\bf r}(t)$ be a vector-valued function in either the plane or space.

Then the derivative of ${\bf r}(t)$ is $$ {\bf r}'(t)=\lim_{\Delta t \rightarrow 0} \frac{{\bf r}(t+\Delta t)-{\bf r}(t)}{\Delta t} $$ provided the limit exists.

Derivatives of Vector-Valued Functions

$\Delta {\bf r}$$=$${\bf r}(t+\Delta t)-{\bf r}(t)$ $,\,\,\,\,$ $\displaystyle \frac{\Delta {\bf r}}{\Delta t}=\frac{{\bf r}(t+\Delta t)-{\bf r}(t)}{\Delta t}$

Derivatives of Vector-Valued Functions

If ${\bf r}(t)$ is a vector-valued function in either the plane or space, then the geometric interpretation of its derivative, ${\bf r}'(t),$ is a tangent vector whose length is the rate of change of ${\bf r}(t)$ at a particular value of $t.$

If ${\bf r}(t)$ describes an object's position with respect to time, then ${\bf r}'(t),$ describes the object's velolcity.

${\bf r}(t)$$, \,\,$ ${\bf r}'(t)$

Example

Use the definition to calculate the derivative of the function $$ {\bf r}(t)=(2t^2+3){\bf i}+(5t-6){\bf j} $$

$$ \begin{array}{lll} \displaystyle {\bf r}'(t) &\displaystyle= \lim_{\Delta t \rightarrow 0} \frac{{\bf r}(t+\Delta t)-{\bf r}(t)}{\Delta t}&\mbox{}\\ \displaystyle &\displaystyle= \lim_{\Delta t \rightarrow 0} \frac{(2(t+\Delta t)^2+3){\bf i}+(5(t+\Delta t)-6){\bf j}-((2t^2+3){\bf i}+(5t-6){\bf j})}{\Delta t}&\mbox{}\\ \displaystyle &\displaystyle= \lim_{\Delta t \rightarrow 0} \frac{(2(t^2+2t\Delta t+\Delta t^2)+3){\bf i}+(5t+5\Delta t-6){\bf j}-(2t^2+3){\bf i}-(5t-6){\bf j}}{\Delta t}&\mbox{}\\ \displaystyle &\displaystyle= \lim_{\Delta t \rightarrow 0} \frac{(2t^2+4t\Delta t+2\Delta t^2+3){\bf i}+(5t+5\Delta t-6){\bf j}-(2t^2+3){\bf i}-(5t-6){\bf j}}{\Delta t}&\mbox{}\\ \displaystyle &\displaystyle= \lim_{\Delta t \rightarrow 0} \frac{(2t^2+4t\Delta t+2\Delta t^2+3-(2t^2+3)){\bf i}+(5t+5\Delta t-6-(5t-6)){\bf j}}{\Delta t}&\mbox{}\\ \displaystyle &\displaystyle= \lim_{\Delta t \rightarrow 0} \frac{(4t\Delta t+2\Delta t^2){\bf i}+(5\Delta t){\bf j}}{\Delta t}&\mbox{}\\ \displaystyle &\displaystyle= \lim_{\Delta t \rightarrow 0} \left[\frac{4t\Delta t+2\Delta t^2}{\Delta t}{\bf i}+\frac{5\Delta t}{\Delta t}{\bf j}\right]&\mbox{}\\ \displaystyle &\displaystyle= \lim_{\Delta t \rightarrow 0} \left[(4t+2\Delta t){\bf i}+5{\bf j}\right]&\mbox{}\\ \displaystyle &\displaystyle= 4t\,{\bf i}+5\,{\bf j}&\mbox{}\\ \end{array} $$

The above example strongly hints at the following result...

Derivatives of Vector-Valued Functions

If ${\bf r}(t)=f(t)\,{\bf i}+g(t)\,{\bf j}=\langle f(t),g(t)\rangle,$ then $${\bf r}'(t)=f'(t)\,{\bf i}+g'(t)\,{\bf j}=\langle f'(t),g'(t)\rangle$$

If ${\bf r}(t)=f(t)\,{\bf i}+g(t)\,{\bf j}+h(t)\,{\bf k}=\langle f(t),g(t),h(t)\rangle,$ then $${\bf r}'(t)=f'(t)\,{\bf i}+g'(t)\,{\bf j}+h'(t)\,{\bf k}=\langle f'(t),g'(t),h'(t)\rangle$$

Example

Use the above result to calculate the derivative of the vector-valued function $$ {\bf r}(t)=3\cos t\,{\bf i}+2\sin t\,{\bf j} $$

$$ {\bf r}(t)=-3\sin t\,{\bf i}+2\cos t\,{\bf j} $$

${\bf r}(t)=3\cos t\,{\bf i}+2\sin t\,{\bf j}$$, \,\,$ ${\bf r}'(t)=-3\sin t\,{\bf i}+2\cos t\,{\bf j}$

Example

Calculate the derivative of the vector-valued function $$ {\bf r}(t)=e^t\sin t\,{\bf i}+e^t\cos t\,{\bf j}-e^{2t}\,{\bf k} $$

$$ \begin{array}{lll} \displaystyle {\bf r}'(t)&\displaystyle= \left(\frac{d}{dt}e^t\sin t\right)\,{\bf i}+ \left(\frac{d}{dt}e^t\cos t\right)\,{\bf j}- \left(\frac{d}{dt}e^{2t}\right)\,{\bf k}&\mbox{}\\ \displaystyle &\displaystyle= (e^t\sin t+e^t\cos t)\,{\bf i}+(e^t\cos t-e^t\sin t)\,{\bf j}-2e^{2t}\,{\bf k}&\mbox{}\\ \displaystyle &\displaystyle= e^t(\sin t+\cos t)\,{\bf i}+e^t(\cos t-\sin t)\,{\bf j}-2e^{2t}\,{\bf k}&\mbox{}\\ \end{array} $$

Properties of the Derivative of Vector-Valued Functions

$$ \begin{array}{ll} \displaystyle \frac{d}{dt}[c{\bf r}(t)]=c{\bf r}'(t)&\\ \displaystyle \frac{d}{dt}[{\bf r}(t)\pm{\bf s}(t)]={\bf r}'(t)\pm{\bf s}'(t)&\\ \displaystyle \frac{d}{dt}[f(t){\bf r}(t)]=f(t){\bf r}'(t)+f'(t){\bf r}(t)&\mbox{Derivative of Scalar Product}\\ \displaystyle \frac{d}{dt}[{\bf r}(t)\cdot{\bf s}(t)]={\bf r}'(t)\cdot{\bf s}(t)+{\bf r}(t)\cdot{\bf s}'(t)&\mbox{Derivative of Dot Product}\\ \displaystyle \frac{d}{dt}[{\bf r}(t)\times{\bf s}(t)]={\bf r}'(t)\times{\bf s}(t)+{\bf r}(t)\times{\bf s}'(t)&\mbox{Derivative of Cross Product}\\ \displaystyle \frac{d}{dt}[{\bf r}(f(t))]={\bf r}'(f(t))f'(t)&\mbox{Chain Rule}\\ \end{array} $$

Corollary: If $\Vert{\bf r}(t)\Vert=c$ for all $t,$ then ${\bf r}(t)\cdot {\bf r}'(t)=0$

Proof:

$$ \begin{array}{lrll} &\displaystyle \Vert {\bf r}(t)\Vert&\displaystyle=c &\mbox{}\\ \implies&\displaystyle \Vert {\bf r}(t) \Vert^2&\displaystyle= c^2 &\mbox{}\\ \implies&\displaystyle {\bf r}(t)\cdot {\bf r}(t)&\displaystyle= c^2 &\mbox{}\\ \implies&\displaystyle \frac{d}{dt}[{\bf r}(t)\cdot {\bf r}(t)]&\displaystyle= \frac{d}{dt}c^2 &\mbox{}\\ \implies&\displaystyle {\bf r}'(t)\cdot{\bf r}(t)+{\bf r}(t)\cdot{\bf r}'(t)&\displaystyle=0 &\mbox{}\\ \implies&\displaystyle 2{\bf r}'(t)\cdot{\bf r}(t)&\displaystyle=0 &\mbox{}\\ \implies&\displaystyle {\bf r}'(t)\cdot{\bf r}(t)&\displaystyle=0 &\mbox{}\\ \end{array} $$ Notice that this says that if ${\bf r}(t)$ is a circle in $\mathbb{R}^2$ or sphere in $\mathbb{R}^3,$ then ${\bf r}(t)$ and ${\bf r}'(t)$ are orthogonal!

For example...

Example

If $ {\bf r}(t)=\cos t\,{\bf i}+\sin t \,{\bf j}-e^{2t}\,{\bf k} $ and $ {\bf s}(t)=t\,{\bf i}+\sin t\,{\bf j}+\cos t\,{\bf k}, $ find

(a) $\displaystyle \frac{d}{dt}[{\bf r}(t)\cdot {\bf r}'(t)]$

$$ \begin{array}{lll} \displaystyle \frac{d}{dt}[{\bf r}(t)\cdot {\bf r}'(t)]&\displaystyle={\bf r}'(t)\cdot{\bf r}'(t)+{\bf r}(t)\cdot{\bf r}''(t) &\mbox{}\\ \displaystyle &\displaystyle=\langle-\sin t,\cos t,-2e^{2t}\rangle\cdot\langle-\sin t,\cos t,-2e^{2t}\rangle+\langle\cos t, \sin t,-e^{2t}\rangle\cdot\langle-\cos t,-\sin t,-4e^{2t}\rangle &\mbox{}\\ \displaystyle &\displaystyle=\sin^2 t+\cos^2t+4e^{4t}-\cos^2 t-\sin^2 t+4e^{4t} &\mbox{}\\ \displaystyle &\displaystyle=8e^{4t} &\mbox{}\\ \end{array} $$

(b) $\displaystyle \frac{d}{dt}[{\bf s}(t)\times {\bf r}(t)]$

$$ \begin{array}{lll} \displaystyle \frac{d}{dt}[{\bf s}(t)\times {\bf r}(t)]&\displaystyle= {\bf s}'(t)\times{\bf r}(t)+{\bf s}(t)\times{\bf r}'(t)&\mbox{}\\ \displaystyle &\displaystyle= \langle 1,\cos t,-\sin t \rangle\times\langle\cos t, \sin t,-e^{2t}\rangle+\langle t,\sin t,\cos t \rangle\times\langle -\sin t, \cos t,-2e^{2t}\rangle&\mbox{}\\ \displaystyle &\displaystyle=\left|\begin{array}{ccc}{\bf i}&{\bf j}&{\bf k}\\ 1&\cos t&-\sin t \\ \cos t&\sin t&-e^{2t} \\ \end{array}\right| + \left|\begin{array}{ccc}{\bf i}&{\bf j}&{\bf k}\\ t&\sin t&\cos t \\ -\sin t&\cos t&-2e^{2t} \\ \end{array}\right| &\mbox{}\\ \displaystyle &\displaystyle=(-e^{2t}\cos t+\sin^2 t){\bf i}-(-e^{2t}+\sin t\cos t){\bf j}+(\sin t -\cos^2 t){\bf k}+(-2e^{2t}\sin t-\cos^2 t){\bf i}-(-2te^{2t}+\sin t \cos t){\bf j}+(t\cos t+\sin^2 t){\bf k} &\mbox{}\\ \displaystyle &\displaystyle=\langle-e^{2t}\cos t+\sin^2 t,-(-e^{2t}+\sin t\cos t),\sin t -\cos^2 t\rangle+\langle-2e^{2t}\sin t-\cos^2 t,-(-2te^{2t}+\sin t \cos t),t\cos t+\sin^2 t\rangle &\mbox{}\\ \displaystyle &\displaystyle=\langle-e^{2t}\cos t+\sin^2 t,e^{2t}-\sin t\cos t,\sin t -\cos^2 t\rangle+\langle-2e^{2t}\sin t-\cos^2 t,2te^{2t}-\sin t \cos t,t\cos t+\sin^2 t\rangle &\mbox{}\\ \displaystyle &\displaystyle=\langle-e^{2t}\cos t+\sin^2 t-2e^{2t}\sin t-\cos^2 t,e^{2t}-\sin t\cos t+2te^{2t}-\sin t \cos t,\sin t -\cos^2 t+t\cos t+\sin^2 t\rangle &\mbox{}\\ \displaystyle &\displaystyle=\langle -e^{2t}(\cos t+2\sin t)-\cos^2 t+\sin^2 t,(2t+1)e^{2t}-2\sin t \cos t,\sin t +t\cos t-\cos^2 t+\sin^2 t\rangle &\mbox{}\\ \displaystyle &\displaystyle=\langle -e^{2t}(\cos t+2\sin t)-\cos(2t),(2t+1)e^{2t}-2\sin (2t),\sin t +t\cos t-\cos(2t)\rangle &\mbox{}\\ \displaystyle &\displaystyle= (-e^{2t}(\cos t+2\sin t)-\cos(2t)){\bf i}+((2t+1)e^{2t}-2\sin (2t)){\bf j}+(\sin t +t\cos t-\cos(2t)){\bf k} &\mbox{}\\ \displaystyle &\displaystyle= -(e^{2t}(\cos t+2\sin t)+\cos(2t)){\bf i}+((2t+1)e^{2t}-2\sin (2t)){\bf j}+(\sin t +t\cos t-\cos(2t)){\bf k} &\mbox{}\\ \end{array} $$ Alternatively, we could have calculated ${\bf s}(t)\times {\bf r}(t)$ first, and then taken the derivative.

Unit Tangent Vectors

Let ${\bf r}(t)$ be a vector-valued function in either the plane or space.

Then the principal unit tangent vector, ${\bf T}(t)$ is defined to be $$ {\bf T}(t)=\frac{{\bf r}'(t)}{\Vert{\bf r}'(t)\Vert} $$ provided $\Vert{\bf r}'(t)\Vert \gt 0$

Example

For $${\bf r}(t)=3\cos t\,{\bf i}+2\sin t\,{\bf j},$$ find $${\bf T}(t)=\displaystyle \frac{{\bf r}'(t)}{\Vert{\bf r}'(t)\Vert}.$$

Since ${\bf r}'(t)=-3\sin t\,{\bf i}+2\cos t\,{\bf j},$ we have $$ \begin{array}{lll} \displaystyle {\bf T}(t)&\displaystyle= \frac{{\bf r}'(t)}{\Vert{\bf r}'(t)\Vert}&\mbox{}\\ \displaystyle &\displaystyle=\frac{-3\sin t\,{\bf i}+2\cos t\,{\bf j}}{\sqrt{(-3\sin t)^2+(2\cos t)^2}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-3\sin t\,{\bf i}+2\cos t\,{\bf j}}{\sqrt{9\sin^2 t+4\cos^2 t}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-3\sin t\,{\bf i}+2\cos t\,{\bf j}}{\sqrt{5\sin^2 t+4\sin^2 t+4\cos^2 t}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-3\sin t\,{\bf i}+2\cos t\,{\bf j}}{\sqrt{5\sin^2 t+4}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{-3\sin t}{\sqrt{5\sin^2 t+4}}\,{\bf i}+\frac{2\cos t}{\sqrt{5\sin^2 t+4}}\,{\bf j} &\mbox{}\\ \end{array} $$

${\bf r}(t)=3\cos t\,{\bf i}+2\sin t\,{\bf j}$$, \,\,$ ${\bf T}(t) %=\displaystyle \frac{-3\sin t\,{\bf i}+2\cos t\,{\bf j}}{\Vert -3\sin t\,{\bf i}+2\cos t\,{\bf j} \Vert} %=\displaystyle \frac{-3\sin t\,{\bf i}+2\cos t\,{\bf j}}{\sqrt{5\sin^2 t+4}} =\displaystyle \frac{-3\sin t}{\sqrt{5\sin^2 t+4}}\,{\bf i}+\frac{2\cos t}{\sqrt{5\sin^2 t+4}}\,{\bf j} $

Example

Find the unit tangent vector for the vector-valued function $$ {\bf r}(t)=(t^2-3)\,{\bf i}+(2t+1)\,{\bf j}+(t-2)\,{\bf k} $$

Since ${\bf r},(t)=2t\,{\bf i}+2\,{\bf j}+\,{\bf k}$ $$ \begin{array}{lll} \displaystyle {\bf T}(t)&\displaystyle= \frac{{\bf r}'(t)}{\Vert{\bf r}'(t)\Vert}&\mbox{}\\ \displaystyle &\displaystyle=\frac{2t\,{\bf i}+2\,{\bf j}+\,{\bf k}}{\sqrt{(2t)^2+(2)^2+(1)^2}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{2t\,{\bf i}+2\,{\bf j}+\,{\bf k}}{\sqrt{4t^2+5}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{2t}{\sqrt{4t^2+5}}\,{\bf i}+\frac{2}{\sqrt{4t^2+5}}\,{\bf j}+\frac{1}{\sqrt{4t^2+5}}\,{\bf k} &\mbox{}\\ \end{array} $$

Integration of Vector-Valued Functions

We now turn our attention to what it means to integrate vector-valued functions.

As we shall see later, what we cover here is only the beginning of a topic that vastly generalizes the the notion integration.

We begin with a straightforward generalization of the definite integral we covered in second-term calculus.

Integration of Vector-Valued Functions in the Plane

If ${\bf r}(t)=f(t)\,{\bf i}+g(t)\,{\bf j}=\langle f(t),g(t)\rangle,$ then the indefinite integral of ${\bf r}(t)$ is given by $$\int {\bf r}(t)\, dt=\left(\int f(t)\,dt\right){\bf i}+\left(\int g(t)\,dt\right){\bf j}=\left\langle \int f(t)\,dt,\int g(t)\,dt\right\rangle$$ and the definite integral is given by $$\int_{a}^{b} {\bf r}(t)\, dt=\left(\int_{a}^{b} f(t)\,dt\right){\bf i}+\left(\int_{a}^{b} g(t)\,dt\right){\bf j}=\left\langle \int_{a}^{b} f(t)\,dt,\int_{a}^{b} g(t)\,dt\right\rangle$$

Integration of Vector-Valued Functions in Space

If ${\bf r}(t)=f(t)\,{\bf i}+g(t)\,{\bf j}+h(t)\,{\bf k}=\langle f(t),g(t),h(t)\rangle,$ then the indefinite integral of ${\bf r}(t)$ is given by $$\int {\bf r}(t)\, dt=\left(\int f(t)\,dt\right){\bf i}+\left(\int g(t)\,dt\right){\bf j}+\left(\int h(t)\,dt\right){\bf k}=\left\langle \int f(t)\,dt,\int g(t)\,dt,\int h(t)\,dt\right\rangle$$ and the definite integral is given by $$\int_{a}^{b} {\bf r}(t)\, dt=\left(\int_{a}^{b} f(t)\,dt\right){\bf i}+\left(\int_{a}^{b} g(t)\,dt\right){\bf j}+\left(\int_{a}^{b} h(t)\,dt\right){\bf k}=\left\langle \int_{a}^{b} f(t)\,dt,\int_{a}^{b} g(t)\,dt,\int_{a}^{b} h(t)\,dt\right\rangle$$

Example

For ${\bf r}(t)=\sin(2t)\,{\bf i}+\tan t\,{\bf j},$ find $$\int {\bf r}(t)\, dt$$

$$ \begin{array}{lll} \displaystyle \int {\bf r}(t)\, dt&\displaystyle= \int [\sin(2t)\,{\bf i}+\tan t\,{\bf j}]\, dt&\mbox{}\\ \displaystyle &\displaystyle=\left( \int \sin(2t)\,dt\right) \,{\bf i}+\left(\int \tan t\, dt\right)\,{\bf j}&\mbox{}\\ \displaystyle &\displaystyle=\left(-\frac{1}{2}\cos(2t)+C_1\right) \,{\bf i}+\left(\ln|\sec t|+C_2\right)\,{\bf j}&\mbox{}\\ \displaystyle &\displaystyle=\left(-\frac{1}{2}\cos(2t)\right) \,{\bf i}+\left(\ln|\sec t|\right)\,{\bf j}+C_1{\bf i}+C_2{\bf j}&\mbox{}\\ \displaystyle &\displaystyle=\left(-\frac{1}{2}\cos(2t)\right) \,{\bf i}+\left(\ln|\sec t|\right)\,{\bf j}+{\bf C}&\mbox{}\\ \end{array} $$

$$\int_{0}^{\pi/3} {\bf r}(t)\, dt$$

From the above, we have $$ \begin{array}{lll} \displaystyle \int_{0}^{\pi/3} {\bf r}(t)\, dt&\displaystyle= \int_{0}^{\pi/3} [\sin(2t)\,{\bf i}+\tan t\,{\bf j}]\, dt&\mbox{}\\ \displaystyle &\displaystyle=\left[-\frac{1}{2}\cos(2t)\right]_{0}^{\pi/3} \,{\bf i}+\left[\ln|\sec t|\right]_{0}^{\pi/3}\,{\bf j}&\mbox{}\\ \displaystyle &\displaystyle=\left[-\frac{1}{2}\cos\left(2\cdot \frac{\pi}{3}\right)-\left(-\frac{1}{2}\cos(2\cdot 0)\right)\right] \,{\bf i}+\left[\ln\left|\sec \frac{\pi}{3}\right|-\ln\left|\sec 0\right|\right]\,{\bf j}&\mbox{}\\ \displaystyle &\displaystyle=\left[-\frac{1}{2}\left(-\frac{1}{2}\right)+\frac{1}{2}\right] \,{\bf i}+\left[\ln\left|2\right|-\ln|1|\right]\,{\bf j}&\mbox{}\\ \displaystyle &\displaystyle=\frac{3}{4} \,{\bf i}+\ln 2\,{\bf j}&\mbox{}\\ \end{array} $$

Example

Find $$\int \langle t,t^2,t^3\rangle \times \langle t^3,t^2,t\rangle \, dt$$

First we note that $$ \begin{array}{lll} \displaystyle \langle t,t^2,t^3\rangle \times \langle t^3,t^2,t\rangle &\displaystyle= \left|\begin{array}{ccc}{\bf i}&{\bf j}&{\bf k}\\ t&t^2&t^3 \\ t^3&t^2&t \\ \end{array}\right|&\mbox{}\\ \displaystyle &\displaystyle= (t^3-t^5){\bf i}-(t^2-t^6){\bf j}+(t^3-t^5){\bf k}&\mbox{}\\ \displaystyle &\displaystyle= (t^3-t^5){\bf i}+(t^6-t^2){\bf j}+(t^3-t^5){\bf k}&\mbox{}\\ \end{array} $$ Then $$ \begin{array}{lll} \displaystyle \int \langle t,t^2,t^3\rangle \times \langle t^3,t^2,t\rangle \, dt&\displaystyle= \int \left[(t^3-t^5){\bf i}+(t^6-t^2){\bf j}+(t^3-t^5){\bf k}\right] \, dt&\mbox{}\\ \displaystyle &\displaystyle= \left(\frac{t^4}{4}-\frac{t^6}{6}\right){\bf i}+\left(\frac{t^7}{7}-\frac{t^3}{3} \right){\bf j}+\left(\frac{t^4}{4}-\frac{t^6}{6}\right){\bf k}+{\bf C}&\mbox{}\\ \end{array} $$

Applications of Vector-Valued Integration

We have seen that the definite integral of a velocity function is a displacement.

With the language of vectors we can now track a change in an object's position in $2$-dimensional and $3$-dimensional space.

Application: A low-flying plane is dropping off a supply package to a remote Alaskan village. At the moment the package is released, the plane's altitude is $250$ meters and its speed is $50$ meters per second. Ignoring air resistance, the velocity ${\bf v}(t)$ the package is modelled by the vector-valued function ${\bf v}(t)=50\,{\bf i}-9.8t\,{\bf j}.$ If the initial position of the package is given by ${\bf s}(0)=250\,{\bf j},$ find the position vector ${\bf s}$ of the object after $3$ seconds.

${\bf s}($$t$$)=$$x(t)$$\,{\bf i}$$+y(t)$$\,{\bf j}$

Integrating the velocity function, we obtain the position function ${\bf s}(t).$ $${\bf s}(t)=\int (50\,{\bf i}-9.8t\,{\bf j})\, dt=50t\,{\bf i}-4.9t^2\,{\bf j}+{\bf s}_0$$ where ${\bf s}_0$ is an initial position vector.

From the given information, ${\bf s}_0={\bf s}(0)=250{\bf j}.$ Thus, $${\bf s}(t)=50t\,{\bf i}-4.9t^2\,{\bf j}+250{\bf j}=50t\,{\bf i}+(250-4.9t^2)\,{\bf j}$$ It follows that $${\bf s}(3)=50\cdot 3\,{\bf i}+(250-4.9\cdot 3^2)\,{\bf j}=150{\bf i}+205.9{\bf j}$$