Find the arc length of the curve from t = 0 to t = 1 whose derivatives in parametric form are dx/dt=2-cos(t) and dy/dt=ln(t^2). Type your answer in the space below and give 2 decimal places
My work:
2-cost(t) = sin(t) = sin^2(t)
ln(t^2)= 2/t= 4/t^2
integral from 0 to 1 √(sin^2(t)+4/t^2) dt
I got stuck, please finish this. Show me your work
part of your problem is that you didn't use what they gave you.
You don't have to take the derivative, since they gave you the derivatives.
That is, you already know dx/dt and dy/dt. Just plug them into the formula for ds.
And your sloppy use of the "=" signs makes things even more confusing.
That said, good luck with the integral; you will have to use some numeric method to evaluate it.
I got 1.98 oobleck
Hmmm. wolframalpha got 2.52
https://www.wolframalpha.com/input/?i=%E2%88%AB%5B0,1%5D+sqrt((2-cost)%5E2%2Bln(t%5E2)%5E2)+dt
To find the arc length of the curve from t = 0 to t = 1, we can use the formula for arc length:
L = ∫[a to b] √[(dx/dt)^2 + (dy/dt)^2] dt
Given the derivatives in parametric form, we have:
dx/dt = 2 - cos(t)
dy/dt = ln(t^2)
Substituting these derivatives into the formula, we get:
L = ∫[0 to 1] √[(2 - cos(t))^2 + (ln(t^2))^2] dt
Now let's simplify the expression inside the square root:
(2 - cos(t))^2 + (ln(t^2))^2
= (2 - cos(t))^2 + (2 * ln(t))^2
= 4 - 4cos(t) + cos^2(t) + 4ln^2(t)
Notice the trigonometric identity cos^2(t) + sin^2(t) = 1. We can use this identity and rearrange the expression:
4 - 4cos(t) + cos^2(t) + 4ln^2(t)
= 5 - 4cos(t) + sin^2(t) + 4ln^2(t)
= 5 - 4cos(t) + sin^2(t) + (2ln(t))^2
Now our arc length formula becomes:
L = ∫[0 to 1] √[5 - 4cos(t) + sin^2(t) + (2ln(t))^2] dt
To find the integral, we can approximate it numerically using a computational tool or software. Once we evaluate the integral, we get the arc length of the curve over the given interval.