Vector Calculus (Position Vector)

**PedroMinsk** · August 23rd 2011, 12:18 PM

The position vector of a point is given by $\vec{r}(t) = (e^tcost, e^tsint)$ . Prove that:

a) $\vec{a} = 2\vec{v} - 2\vec{r}$

b) the angle between the position vector $\vec{r}$ and the acceleration vector $\vec{a}$ is constant. Calculate this angle.

Answer: b) pi/2

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I tried to derive the position vector two times but it isn't working.

**Plato** · August 23rd 2011, 01:06 PM

Originally Posted by PedroMinsk

The position vector of a point is given by $\vec{r}(t) = (e^tcost, e^tsint)$ . Prove that:
a) $\vec{a} = 2\vec{v} - 2\vec{r}$
b) the angle between the position vector $\vec{r}$ and the acceleration vector $\vec{a}$ is constant. Calculate this angle.

This is a tedious problem.
Luckily there are web resources
BE SURE to click "show steps"
You can change the question to $e^t\sin(t)$ .
That will help you find $\vec{a}$ the second derivative or $\vec{r}$

**Siron** · August 23rd 2011, 01:30 PM

For the second question you can calculate the scalar product of the vectors $a(x_1,y_1),r(x_2,y_2)$ like:
$a.r=x_1x_2+y_1y_2$
The scalar produt will tell you more about the angle between the vectors, but first you've to determine the coordinates of the vector r(t) (see Plato's post).

**PedroMinsk** · August 23rd 2011, 01:34 PM

Thanks. I got the idea. I only knew the Wolfram integrator, this one will help me a lot.

Can I calculate the angle using dot product? Or there is another way?

**Siron** · August 23rd 2011, 01:37 PM

The dot (or scalar) product is very useful here, you'll come to the conclusion if you calculate the dot product that it will be 0 and $\arccos(0)=\frac{\pi}{2}\right)$ ( $+2k\pi$ )

**Plato** · August 23rd 2011, 01:39 PM

Originally Posted by PedroMinsk

Thanks. I got the idea. I only knew the Wolfram integrator, this one will help me a lot.
Can I calculate the angle using dot product? Or there is another way?

Yes the dot product, see reply #3.
You should explore what all wolframalpha will do.
Look at this.

**PedroMinsk** · August 23rd 2011, 01:43 PM

I got it. Thanks. I will see the link.

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