Arclength parameter for a path with exponential and trigonometric functions

[yango_17]

Homework Statement

Find the arclength parameter s=s(t) for the path
x(t)=e^(at)cos(bt)i + e^(at)sin(bt)j + e^(at)k

Homework Equations

$s(t)=\int_{a}^{t}\left | \mathbf{x}(\tau ) \right |d\tau$

The Attempt at a Solution

I took the derivative and squared it, arriving at the equation $(-be\sin bt)^{2}+(ae\cos bt))^{2}+(be\cos bt)^{2}+(ae\sin bt)^{2}+(ae^{at})^{2}$. How would I go about simplifying this in order to take the square root and integrate it? It seems extremely complex to be able to integrate at the moment. Thanks.

 

[SteamKing]

Homework Statement

Find the arclength parameter s=s(t) for the path
x(t)=e^(at)cos(bt)i + e^(at)sin(bt)j + e^(at)k

Homework Equations

$s(t)=\int_{a}^{t}\left | \mathbf{x}(\tau ) \right |d\tau$

The Attempt at a Solution

I took the derivative and squared it, arriving at the equation $(-be\sin bt)^{2}+(ae\cos bt))^{2}+(be\cos bt)^{2}+(ae\sin bt)^{2}+(ae^{at})^{2}$. How would I go about simplifying this in order to take the square root and integrate it? It seems extremely complex to be able to integrate at the moment. Thanks.

Well, what have you tried? With a lot of sines and cosines being squared and all, are there any trig identities you think which might be useful in simplifying this integrand?

 

[yango_17]

The cos and sin for the ae terms cancel to 1, but then I am left with a -be and a be, which would let me factor out $-\sin ^{2}(bt)+cos^{2}(bt)$, which I am unsure how to simplify. Also, not too sure what to do with the (ae^ae)^2 term.

 

[SteamKing]

The cos and sin for the ae terms cancel to 1, but then I am left with a -be and a be, which would let me factor out $-\sin ^{2}(bt)+cos^{2}(bt)$, which I am unsure how to simplify. Also, not too sure what to do with the (ae^ae)^2 term.

You're squaring everything. How can you have any negative quantities in the integrand?

 

[yango_17]

Ah, I see. So, I'd be left with $2ae^{at^{2}}+be^{2}$ under the square root?

 

[SteamKing]

Ah, I see. So, I'd be left with $2ae^{at^{2}}+be^{2}$ under the square root?

What is (e^at)² ? Remember the laws of exponents.

 

[yango_17]

It would be $2ae^{2at}$. So, under the square root we have $2ae^{2at}+be^{2}$?

 

[SteamKing]

It would be $2ae^{2at}$. So, under the square root we have $2ae^{2at}+be^{2}$?

You seem to have lost some terms along the way. How did you start with e^at and e^bt and wind up with just plain ae and be ?

Remember, many a calculus student's career was derailed not by the calculus, but by algebra or arithmetic.

 

[yango_17]

Going back, it seems like I should have $be^{2at}+2ae^{2at}$. Correct?

 

[SteamKing]

Going back, it seems like I should have $be^{2at}+2ae^{2at}$. Correct?

That's better. How about a little factoring, now, when you put this expression under the radical?

 

[yango_17]

Factoring? What do you mean, exactly?

 

[SteamKing]

Factoring? What do you mean, exactly?

Don't you notice any common terms in your expression under the radical? You know, things which could be factored out from under the square root sign?

 

[yango_17]

e^(at) could be factored out, right?

 

[SteamKing]

e^(at) could be factored out, right?

Yes, yes it could.

 

[yango_17]

So, in the end we're left with $e^{at}\sqrt{2a^{2}+b^{2}}$. Am I correct in assuming that the things under the radical can no longer be simplified?

 

[SteamKing]

So, in the end we're left with $e^{at}\sqrt{2a^{2}+b^{2}}$. Am I correct in assuming that the things under the radical can no longer be simplified?

You are correct. The stuff left under the radical is just another constant anyway.

Now does this expression look like it can be integrated?

 

[yango_17]

Yes! I'm assuming $\sqrt{2a^{2}+b^{2}}$ can be pulled out of the integral and we just have to integrate e^{at}. Doing this integral from 0 to t, I obtained $\frac{1}{a}e^{at}\sqrt{2a^{2}+b^{2}}-\frac{1}{a}\sqrt{2a^{2}+b^{2}}$. Does this seem reasonable?

 

[SteamKing]

Yes! I'm assuming $\sqrt{2a^{2}+b^{2}}$ can be pulled out of the integral and we just have to integrate e^{at}. Doing this integral from 0 to t, I obtained $\frac{1}{a}e^{at}\sqrt{2a^{2}+b^{2}}-\frac{1}{a}\sqrt{2a^{2}+b^{2}}$. Does this seem reasonable?

Yes, but I believe this expression can be simplified.

 

[yango_17]

I've arrived at $\sqrt{2a^{2}+b^{2}}(\frac{1}{a}e^{at}-\frac{1}{a})$ by factoring out the 1/a

 

[yango_17]

Factoring out the radical i mean

 

[SteamKing]

Factoring out the radical i mean

You can still take out one more minor factor.