# Calculus

posted by .

Graph the curve and find its exact length.

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!!!

## Similar Questions

I'm having trouble with this question on arc length: y=lnx, (squareroot)3/3 greater than or equal to x less than or equal to 1 It sounds as if you want the length of the y = ln x curve from x = sqrt(3)/3 (0.57735..) to 1. The formula …
2. ### calc: arc length

find the exact length of this curve: y = ( x^3/6 ) + ( 1/2x ) 1/2 <or= x <or= 1 im looking over my notes, but i'm getting stuck. here's my work so far: A ( 1 , 2/3 ) B ( 1/2 , 49/48 ) y' = [1/6 (3x^2)] + [1/2 (-1x^-2)] y' = ( …
3. ### calc: arc length

Posted by COFFEE on Monday, June 11, 2007 at 11:48pm. find the exact length of this curve: y = ( x^3/6 ) + ( 1/2x ) 1/2 <or= x <or= 1 im looking over my notes, but i'm getting stuck. here's my work so far: A ( 1 , 2/3 ) B ( 1/2 …
4. ### calc check: curve length

Find the length of the curve y=(1/(x^2)) from ( 1, 1 ) to ( 2, 1/4 ) [set up the problem only, don't integrate/evaluate] this is what i did.. let me know asap if i did it right.. y = (1/(x^2)) dy/dx = (-2/(x^3)) L = integral from a …
5. ### Calculus

Find the volume of the solid whose base is the region in the xy-plane bounded by the given curves and whose cross-sections perpendicular to the x-axis are (a) squares, (b) semicircles, and (c) equilateral triangles. for y=x^2, x=0, …
6. ### Calculus

Find the definite integral that represents the arc length of the curve y=sqrt(x) over the interval [0, 3]
7. ### Calculus

Find a curve through the point (1,1) whose length integral is given below. L= integral from 1 to 4 sqrt(1+(1/4x))dx Let the curve be y=f(x). Determine (dy/dx)^2 PLease help I m nt sure even how to start this
8. ### Calculus

I know how to do this problem, but I'm stuck at the arc length differential. Set up an integral for the arc length of the curve. (Do not evaluate the integral) x=y^2ln(y), 1<y<2 dx/dy = 2yln(y) + y ds= sqrt (1 + (2yln(y)+y)^2 …
9. ### calculus

find the exact length of the curve y = ln(1-x^2), 0 <= x <= (1/2) So by doing the work I eventually get down to integral from 0 to 1/2 (1 + x^2)/(1-x^2) dx but I keep getting -1 + 1/(x-1) - 1/(x+1) after partial fractions which …
10. ### calculus

Find the length of the curve given by the equation y= intergral from -pi to x of sqrt(cos(t)) dt for x between -pi and pi. I think I know to do this- at least part of it. I am using the fundamental theorem of calculus and the arc length …

More Similar Questions