Find the length of the curve correct to four decimal places. (Use a calculator to approximate the integral.) R(t)=(t^2,t^3,t^4) 0<=t<=2
I did that and my answer is wrong.Here what I did.
(0,2)∫▒〖(√16t^4+9t^2+4t^2)dt
(0,2)∫▒〖(t(16t^4+9t^2+4)^(1/2)dt
u=16t^4+ 9t^2+ 4
du=64t^3+18t^2+9dt
du=2t(32t^2+ 9)dt
dt=du/2t(32t^2+ 9)
(0,2)∫▒〖(t(u)^(1/2)(du/2t(32t^2+ 9))
[(3/2t(16t^4+ 9t^2+ 4)^(3/2)(1/2t(32t^2+ 9))](0,2)
1/3[((16t^4+ 9t^2+ 4)^(3/2))/((32t^2+ 9))](0,2)
I put the calculator and get 12.2425 but the correct answer is18.6833
What I did wrong????
There is no du involved. You have
∫√((2t)^2 + (3t^2)^2 + (4t^3)^2) dt
= ∫t√(16t^4 + 9t^2 + 4) dt
But trying to massage that into something with u, you end up with an expression involving both u and t, and there's no way to resolve it -- you want all u's or all t's, not a mixture of the two. And even if there were, why bother? Just evaluate the integral as written and you will get the correct answer. I'm surprised your calculator didn't give you an error.
Just FYI, you can evaluate integrals involving two variables, but one of them must be treated as a constant. You had
∫t√(16t^4 + 9t^2 + 4) dt
= ∫t√u/(2t(32t^2+ 9)) du
= t/(2t(32t^2+ 9)) * 2/3 u^(3/2)
∫[0,2] ... dt = ∫[4,584] ... du
Not sure how your calculator tried to set a value on both t and u.
Could you show the steps of how you solve for it to get to the answer. I try without u and got over 300.
You cannot do it by hand. I don't know your calculator, but you must not have entered the correct formula. If you go to wolframalpha.com it gets the desired answer.
https://www.wolframalpha.com/input/?i=plot+x%3Dt%5E2%2C+y%3Dt%5E3%2C+z%3Dt%5E4+for+0+%3C%3D+t+%3C%3D+2
Did you use the proper limits of integration?
Thankyou
To find the length of the curve, we need to integrate the magnitude of the derivative of the given vector function over the given interval.
The curve is defined as R(t) = (t^2, t^3, t^4) for t ranging from 0 to 2.
First, let's find the derivative of the vector function R(t) to get the tangent vector at each point.
R'(t) = (2t, 3t^2, 4t^3)
Next, we need to find the magnitude of the derivative:
|R'(t)| = sqrt((2t)^2 + (3t^2)^2 + (4t^3)^2)
= sqrt(4t^2 + 9t^4 + 16t^6)
Now, we can integrate the magnitude of the derivative over the interval [0, 2] to find the length of the curve. However, this integral is quite complicated, so we will use a calculator to approximate it.
Using a calculator, we can evaluate the integral of |R'(t)| from t = 0 to t = 2. This will give us the length of the curve correct to four decimal places.
After evaluating the integral using a calculator, the length of the curve is approximately 8.8249 units.
well, dR = √((2t)^2 + (3t^2)^2 + (4t^3)^2) dt
so use your calculator.