How can I simplify sqrt( 4t^2 + 4 + ( 1/t ) ) to (2t^2 + 1)/t?

[Peter5897]

Couldn’t decide if I should put this in the calculus or general math forums but...

I’m studying for a final that’s coming up this Wednesday and I’ve been looking at some past quizzes with the steps to finding the solutions that my instructor has posted online. Given the problem:

1. Compute the length of the curve~r(t) = (t^2, 2t, ln(t)), from t = 1 to t = e.

I understand that I need to take the integral from 1 to e of the sqrt( (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2) but unfortunately I’m having trouble taking the sqrt.

I get down to… *EDITED*sqrt( 4t^2 + 4 + ( 1/t^2 ) ) and I get stuck even with the answer and the steps in front of me. In the next step the problem gets simplified so as to have everything over a t^2 and then the sqrt is done and I’m left with the integral from 1 to e of (2t^2 + 1)/t.

I’m hoping someone could explain to me how sqrt( 4t^2 + 4 + ( 1/t ) ) becomes (2t^2 + 1)/t.

I’m ashamed I don’t know this but I don’t want to get stuck trying to do algebra on a calculus final.

Thanks in advance.

 

[Data]

Well,

$$\sqrt{4t^2+4+\frac{1}{t}}$$

doesn't reduce to

$$\frac{2t^2 +1}{t}.$$

So it's good that you can't figure that out!

Your error was forgetting to square $$\frac{dz}{dt}$$.

 

[Peter5897]

Ack, I wish that was my problem but I actually just mistyped it up there.

I still don't see how sqrt( 4t^2 + 4 + ( 1/t^2 ) ) gets reduced to (2t^2 + 1)/t and I'm sure it's some simple step that I'm missing.

Sorry...

 

[Data]

$$\sqrt{4t^2 + 4 + \frac{1}{t^2}} = \sqrt{\frac{4t^4 + 4t^2 + 1}{t^2}} = \frac{1}{t}\sqrt{4t^4+4t^2+1} = \frac{1}{t}\sqrt{(2t^2+1)^2} = \frac{2t^2+1}{t}$$
(for $t > 0$)

 

[Peter5897]

Thank you, I always got stuck at step 2. In fact at one point I was sitting there with sqrt( ( 2t^2 + 1)^2 / t^2 ) ) and was baffled...

Thanks again, the world makes sense now.

 

[Data]

In fact at one point I was sitting there with sqrt( ( 2t^2 + 1)^2 / t^2 ) ) and was baffled...

:rofl:
happens to everyone sometimes.

Thanks again, the world makes sense now.

Good . Don't be afraid to come back with more questions!