Find the length of the following curve?

Question

r(t) = ((2^(1/2))t)i + (e^t)j + (e^-t)k is a given verctor-valued function with domain ranging from 0 to 6... I have to find the length of arc... formula for it is...

s = int ( norm ( deri ( r(t)))) with respect to t... upper and lower bounds are given 6 and 0...

r'(t) = (2^(1/2))i + (e^t)j - (e^t)k

||r'(t)|| = (2+(e^4t)+(e^-4t))^(1/2)

when I simplify it, I get...

||r'(t)|| = ((e^2t)+1)/(e^2t)

I am sure I have not made mistakes in simplifying, right???

once I integrate the norm of derivative of r(t), I get...

(t+((e^-2t)/-2))... once I apply limits, my answer should be...

13/2 - (e^-12/2)...

so my basic question is, anyway i can simplify this???

btw my given options are following...

1. e^6 +e^-6
2. 2e^-6
3. 2e^6
4. e^6-e^-6
5. e^6

which one am i choose?

Answer 1

Your curve is described by the vector function
r(t) = ((sqrt 2) t, e^t, e^(-t))

The derivative is
dr/dt = (sqrt 2, e^t, -e^(-t))

The square norm of the derivative is
|dr/dt|^2 = 2 + e^(2t) + e^(-2t)

and the length of the arc between t and t + dt is the square root of this. It is not hard to see that
2 + e^(2t) + e^(-2t) = (e^t + e^(-t))^2
so you need the integral

INT (e^t + e^(-t)) dt ... from 0 to 6

This is equal to
(e^6 - e^(-6)) - (e^0 - e^(-0)) = e^6 - e^(-6)

[If you like working with hyperbolic functions,
|dr/dt|^2 = 2 + 2 cosh 2t = (2 cosh t)^2

so you need the integral of 2 cosh t dt, which is 2 sinh t, so the answer is 2 sinh 6.

Answer 2

First of all, in your k-component of r'(t), you should have -e^-t, not -e^t.

Second, when you compute ||r'(t)||, you should have (2 + (e^2t) + (e^-2t))^(1/2); you had 4t where you should have had 2t.

To simplify, use the fact that (a+b)^2 = a^2+2ab+b^2.

Answer 3

In physics and geometry, the catenary is the theoretical shape of a putting versatile chain or cable whilst supported at its ends and acted upon via a uniform gravitational tension (its very own weight) and in equilibrium. The chain is steepest close to the factors of suspension because of the fact this area of the chain has the main weight knocking down on it. in direction of the backside, the slope of the chain decreases because of the fact the chain helps much less weight. The formula is merely too complicated to kind in this container. stick to the link under for information.