Compute the length of the curve r(t)=5ti+8tj+(t2+4)k over the interval 0≤t≤6
Hint: use the formula
∫t2+a2‾‾‾‾‾‾‾√dt=12tt2+a2‾‾‾‾‾‾‾√+12a2ln(t+t2+a2‾‾‾‾‾‾‾√)+C
Find Length?
so, did you use the formula? What did you get?
83.73
To find the length of the curve defined by the vector-valued function, r(t) = 5ti + 8tj + (t^2 + 4)k, over the interval 0 ≤ t ≤ 6, we can use the arc length formula.
The arc length formula states that the length of a curve defined by a vector function r(t) = f(t)i + g(t)j + h(t)k, over the interval a ≤ t ≤ b, is given by the integral of the magnitude of the derivative of r(t) with respect to t.
In this case, the derivative of r(t) = 5ti + 8tj + (t^2 + 4)k with respect to t will give us the velocity vector:
r'(t) = 5i + 8j + 2tk
To find the magnitude of r'(t), we use the formula for the magnitude of a vector:
|r'(t)| = √(5^2 + 8^2 + (2t)^2)
= √(25 + 64 + 4t^2)
= √(89 + 4t^2)
Now, we can use the arc length formula:
Length = ∫(0 to 6) √(89 + 4t^2) dt
To evaluate this integral, we can make a substitution using the hint given:
Let u = 4t^2 + 89
Then, du = 8t dt
Substituting these into the integral, we have:
Length = 1/2 ∫(0 to 6) √u du
= 1/2 [1/2 u^(1/2) + 1/4(2/3) u^(3/2)] from 0 to 6
= 1/4 [(1/2 (4t^2 + 89)^(1/2) + 1/3 (4t^2 + 89)^(3/2))] from 0 to 6
Evaluating this expression at 6 and 0, we get:
Length = 1/4 [(1/2 (4(6)^2 + 89)^(1/2) + 1/3 (4(6)^2 + 89)^(3/2))] - [(1/2 (4(0)^2 + 89)^(1/2) + 1/3 (4(0)^2 + 89)^(3/2))]
= 1/4 [(1/2 (144 + 89)^(1/2) + 1/3 (144 + 89)^(3/2)) - (1/2 (89)^(1/2) + 1/3 (89)^(3/2))]
Now, we can simplify and compute the length using a calculator or software.