10BAntidiff


 * AntiDifferentiation of Parametric Vectors **

To antidifferentiate, simple find the antiderivative of each component seperately. The **constant of antidifferentiation** will be a vector __**c**__ where **__c__** = a__**i**__ + b**__j__** + c**__k__**

Example A light plane is initially 1200m above a point on the ground (the origin) when a wheel falls off the plane. If North is **__j__**, East is __**i**__ and up is **__k__**, The initial velocity of the plane (and therefore the wheel) is __**v**__ = 25**__i__** + 32__**j**__ m/s

The acceleration of the wheel can be described as -9.8**__k__** m/s2 or: math . \qquad \underline{\ddot{r}}=-9.8\underline{k}, \quad t \geq 0 \quad \textit{r is in metres} math

a) Find the velocity vector of the wheel math . \qquad \underline{\dot{r}}=-9.8t\underline{k}+\underline{c} math
 * By antidifferentiation, we get **

math . \qquad \underline{\dot{r}}(0)=25\underline{i}+32\underline{j} math
 * From the question, the initial velocity is **

math . \qquad \underline{c}=25\underline{i}+32\underline{j} math
 * hence **

math . \qquad \underline{\dot{r}}=25\underline{i}+32\underline{j}-9.8t\underline{k} math
 * hence **

b) Find the position vector of the wheel

math . \qquad \underline{r}=25t\underline{i}+32t\underline{j}-4.9t^2\underline{k}+\underline{d} math
 * By antidifferentiation, we get **

math . \qquad \underline{r}(0)=1200\underline{k} math
 * From the question, the initial position is **

math . \qquad \underline{d}=1200\underline{k} math
 * hence **

math \\ . \qquad \underline{r}=25t\underline{i}+32t\underline{j}-4.9t^2\underline{k}+1200\underline{k} \\. \\ . \qquad \underline{r}=25t\underline{i}+32t\underline{j}+(1200-4.9t^2)\underline{k} math
 * hence **

c) FInd the place where the wheel first hits the ground

math \\ . \qquad 1200-4.9t^2=0 \\. \\ . \qquad 4.9t^2=1200 \\. \\ . \qquad t^2=244.90 \\ .\\ . \qquad t=15.6 math
 * {Wheel hits the ground when the k component is zero} **

At t = 15.6, position is: math . \qquad \underline{r}=390\underline{i}+499.2\underline{j}+0\underline{k} math

The distance from the origin is math . \qquad \sqrt{390^2+499.2^2}=633.5 \, m math

At an angle of math . \qquad \tan^{-1}\frac{499.2}{390}=52^\circ \quad \textit{degrees anticlockwise from x-axis which was described as East} math

Which as a bearing is N38°E .