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. If your answer is less than 1, place a leading "0" before the decimal point (ex: 0.48)
So I was trying to do it:
integral from 0 to 1 √[2-cos(t)+ln(t^2)] dt
then: integral from 0 to 1 √[(sint)^2+(2/t)^2] dt
and I am stock, please help
To find the arc length of the curve, you can use the formula for arc length in parametric form. The formula is given by:
L = ∫ √[(dx/dt)^2 + (dy/dt)^2] dt
Let's substitute the given derivatives:
dy/dt = ln(t^2)
dx/dt = 2 - cos(t)
So, the formula becomes:
L = ∫ √[(2 - cos(t))^2 + (ln(t^2))^2] dt
To evaluate this integral, you can simplify the expression inside the square root using trigonometric identities and properties of natural logarithm.
L = ∫ √[4 - 4cos(t) + cos^2(t) + ln^2(t^2)] dt
Simplifying further:
L = ∫ √[4 - 4cos(t) + cos^2(t) + ln^2(t^2)] dt
= ∫ √[4 - 4cos(t) + 1 - sin^2(t) + ln^2(t^2)] dt
= ∫ √[5 - 4cos(t) - sin^2(t) + ln^2(t^2)] dt
Now, you can split this integral into two parts:
L = ∫ √[5 - 4cos(t) - sin^2(t)] dt + ∫ ln(t^2) dt
The first integral involves trigonometric functions, and the second integral is a simple logarithmic integral.
The value of the first integral can be found using various techniques such as substitution or trigonometric identities. Once you have the value of each integral, you can add them together to get the final value of L, which is the arc length of the curve.
Note: Finding an exact solution for this integral might be challenging, and numerical methods such as numerical integration or approximation techniques can be used to get an approximate value for the arc length.
recall that for parametric equations,
ds^2 = (dx/dt)^2 + (dy/dt)^2
So, for your curves, that would be (recalling that ln(t^2) = 2lnt
ds^2 = (2-cost)^2 + (2lnt)^2
So the arc length would be
∫[0,1] √((2-cost)^2 + (2lnt)^2) dt
good luck with that. Better go numeric.