Vector-Valued Funtion-Proof of Constant Angle between

[jaredld]

if r(t)=e^t*cos(t)i+e^t*sin(t)j How can I show that the angle between "r" and "a" never changes. What is the angle?

My first thought was to convert the vector-valued function to a rectangular function. However, I am not sure what to do with the e.

I was thinking that once I convert it, I can show visually that r is always perpendicular to the tangent vector acceleration. I thought it would be relatively easy to draw because I expect r(t) to be a circle.

Is my thinking wrong? Any ideas? Thanks for your time in advance.

 

[pka]

\(\displaystyle \L
\begin{array}{l}
r(t) = \left[ {e^t \cos (t)} \right]i + \left[ {e^t \sin (t)} \right]j \\
a(t) = r''(t) = \left[ { - 2e^t \sin (t)} \right]i + \left[ {2e^t \cos (t)} \right]j \\
\left( {r(t)} \right) \cdot \left( {a(t)} \right) = ? \\
\end{array}\)

 

[jaredld]

Thanks PKA....I do feel stupid now for not having seen that.

 

[soroban]

Hello, jaredld!

Don't feel stupid . . . your reasoning was correct.

If \(\displaystyle {\bf r}(t)\:=\:e^t\sin(t){\bf i}\,+\.i\cdot e^t\cos(t){\bf j}\)
show that the angle between $\displaystyle {\bf r}$ and $\displaystyle {\bf a}$ never changes.
What is the angle?

My first thought was to convert the vector-valued function to a rectangular function.
However, I am not sure what to do with the $\displaystyle e$.

Good thinking . . . I think we're stuck with the $\displaystyle e$.

We have: $\displaystyle \:\begin{Bmatrix}\;x\:=\:e^{^t}\sin(t)\; \\ \;y\:=\:e^{^t}\cos(t)\;\end{Bmatrix}$

Square the equations: \(\displaystyle \:\begin{Bmatrix}\;x^2\:=\:e^{^{2t}}\sin^2(t)\;\\ \;y^2\:=\:e^{^{2t}}\cos^2(t)\;\end{array}\)

Add: $\displaystyle \:x^2\,+\,y^2\:=\:e^{^{2t}}\left[\sin^2(t)\,+\,\cos^2(t)\right] \:=\:e^{^{2t}}\cdot1$

Hence: $\displaystyle \:x^2\,+\,y^2\;=\;e^{2t}$ . . . a "variable" circle.

I thought it would be relatively easy to draw because I expect $\displaystyle {\bf r}(t)$ to be a circle.

Also good thinking . . .

Although the radius of the circle is increasing (the curve is a spiral),
$\displaystyle \;\;$at any given value of $\displaystyle t$, the point is on a circle.

So its vectors will behave accordingly . . . $\displaystyle {\bf a}\:=\:-{\bf r}$ and the angle is $\displaystyle \pi.$

 

[pka]

jaredld, I hope that you have found out that $\displaystyle a \not= -r$ and the angle is $\displaystyle \frac{\pi} {2}$

 

[M98Ranger]

yep...the dotP was equal to zero so that gives me 90 degrees....thanks a lot for your help Soroban and PKA.