Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve.
r(t) = (t)i + (2/3)t^(3/2)k,
0 less than or equal to t less than or equal to 8
All but this one of your 22 posts have been removed.
Once you write up YOUR THOUGHTS about how to solve all these, please re-post, and someone here will be happy to comment.
To find the unit tangent vector of a curve, you need to find the first derivative of the curve vector and then normalize it. Here's how you can do it step by step:
Step 1: Find the first derivative of the curve vector r(t) with respect to t. Differentiate each component separately:
r'(t) = i + (2/3)(3/2)t^(1/2)k
= i + tk^(1/2)
Step 2: Normalize the derivative vector to get the unit tangent vector. To normalize a vector, divide it by its length:
T(t) = r'(t) / ||r'(t)||
To find the length of the curve, you need to calculate the arc length integral. Here's how to do it:
Step 1: Find the derivative of the curve vector r(t) with respect to t, which we've already done in the previous step.
Step 2: Calculate the magnitude of the derivative vector ||r'(t)||:
||r'(t)|| = √(1^2 + (t^(1/2))^2)
= √(1 + t)
Step 3: Write the integral for the arc length of the curve:
L = ∫[a,b] ||r'(t)|| dt, where a = 0 and b = 8
Step 4: Integrate the magnitude of the derivative vector with respect to t over the given interval [0, 8]:
L = ∫[0,8] √(1 + t) dt
Step 5: Evaluate the integral to find the length of the indicated portion of the curve:
L = ∫[0,8] √(1 + t) dt = [2/3(1 + t)^(3/2)] from 0 to 8
Step 6: Evaluate the integral limits at t = 8 and t = 0:
L = [2/3(1 + 8)^(3/2)] - [2/3(1 + 0)^(3/2)]
= [2/3(9)^(3/2)] - [2/3(1)^(3/2)]
= [2/3(9^(3/2))] - [2/3(1)]
= [2/3(27)] - [2/3]
= 18 - 2/3
= 54/3 - 2/3
= 52/3
So, the unit tangent vector T(t) = (i + tk^(1/2)) / √(1 + t), and the length of the indicated portion of the curve is 52/3.