# What is the arc length of #r(t)=(t^2,2t,4-t)# on #tin [0,5]#?

Arc length is given by:

Simplify:

This is a known integral:

Reverse the substitution:

Hence

By signing up, you agree to our Terms of Service and Privacy Policy

To find the arc length of the curve defined by ( r(t) = (t^2, 2t, 4 - t) ) on the interval ( [0, 5] ), we first need to calculate the derivative of ( r(t) ) with respect to ( t ), which gives us the velocity vector:

[ r'(t) = (2t, 2, -1) ]

Then, we compute the magnitude of this velocity vector:

[ | r'(t) | = \sqrt{(2t)^2 + 2^2 + (-1)^2} = \sqrt{4t^2 + 4 + 1} = \sqrt{4t^2 + 5} ]

Now, we integrate this magnitude over the interval ( [0, 5] ) to find the arc length:

[ \text{Arc Length} = \int_{0}^{5} | r'(t) | , dt = \int_{0}^{5} \sqrt{4t^2 + 5} , dt ]

This integral may require techniques such as trigonometric substitution or integration by parts to evaluate, depending on the preferred method of integration. Once evaluated, it will give us the arc length of the curve on the interval ( [0, 5] ).

By signing up, you agree to our Terms of Service and Privacy Policy

- What is the area bounded by the parametric equations? : # x=acos theta # and # y=bsin theta #
- What is the arclength of #(t-2t^3,4t+1)# on #t in [-2,4]#?
- How do you find the (shortest) distance from the point P(1, 1, 5) to the line whose parametric equations are x = 1 + t, y = 3 - t, and z = 2t?
- How do you differentiate the following parametric equation: # x(t)=lnt, y(t)= tcost #?
- How do you differentiate the following parametric equation: # x(t)=-3e^t-2t, y(t)= -5t^2-e^(4t) #?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7