Saturday

January 31, 2015

January 31, 2015

Posted by **COFFEE** on Wednesday, June 13, 2007 at 5:58pm.

x = e^t + e^-t, y = 5 - 2t, from 0 to 3

Length = Integral from 0 to 3 of:

Sqrt[(dx/dt)^2 + (dy/dt)^2]

dx/dt = e^t - e^-t, correct?

dy/dt = -t^2 - 5t, correct?

So: Integral from 0 to 3 of

Sqrt[(e^t - e^-t)^2 + (-t^2 - 5t)^2]

Then what do I do? Thanks.

ok on dx/dt

dy/dt= -2

INT sqrt(4+e^t-e^-t)dt

sqrt (4x + (e^t + e^-t)/loge)

dy/dt = -2 as Bob pointed out.

Next, use that:

Sqrt[(e^t - e^-t)^2 + 4] =

Sqrt[4 + 4 Sinh^2(t)] = =

2 Sqrt[1 + Sinh^2(t)] =

2 Cosh(t)

thanks!!!

**Answer this Question**

**Related Questions**

Calculus - Find a curve through the point (1,1) whose length integral is given ...

calculus - find the exact length of the curve y = ln(1-x^2), 0 <= x <= (1/...

calc: arc length - find the exact length of this curve: y = ( x^3/6 ) + ( 1/2x...

calculus - Consider a plane curve which is described in polar coordinates (r...

Calculus - Consider a plane curve which is described in polar coordinates (r...

calculus - Consider a plane curve which is described in polar coordinates (r...

calc check: curve length - Find the length of the curve y=(1/(x^2)) from ( 1, 1...

Calculus - Find the definite integral that represents the arc length of the ...

calc: arc length - Posted by COFFEE on Monday, June 11, 2007 at 11:48pm. find ...

CALCULUS 2!!! PLEASE HELP!! - I'm having trouble with this question on arc ...