Use your calculator to find the length of the arc from t = 0 to t = 1 of x = t^3 + 1, y = 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)
ds^2 = dx^2 + dy^2
so, the arc length is
∫[0,1] √((3t^2)^2 + (2t)^2) dt
= ∫[0,1] t√(9t^2 + 4) dt
If u = 9t^2+4
du = 18t
and so we have
1/18 ∫[4,13] √u du
= 1/18 (2/3 u^(3/2)) [4,13]
= 1/27 (13√13 - 8)
To find the length of the arc from t = 0 to t = 1, we can use the formula for arc length in parametric equations:
L = ∫(a to b) √(dx/dt)^2 + (dy/dt)^2 dt
In this case, we have x = t^3 + 1 and y = t^2. Let's differentiate x and y with respect to t to find dx/dt and dy/dt:
dx/dt = 3t^2
dy/dt = 2t
Now, we can plug these derivatives into the formula and integrate:
L = ∫(0 to 1) √((3t^2)^2 + (2t)^2) dt
Simplifying the expression inside the square root:
L = ∫(0 to 1) √(9t^4 + 4t^2) dt
To solve this integral, we can employ numerical methods or use a calculator with an integral function. Since you mentioned using a calculator, we can proceed with that method.
1. Open your calculator and go to the integration function.
2. Input the function inside the square root: √(9t^4 + 4t^2).
3. Set the lower limit of integration (a) to 0 and the upper limit (b) to 1.
4. Compute the integral using the calculator's integration function.
5. Round the result to two decimal places.
The calculated length will then be the answer to the question. If the length is less than 1, make sure to add a leading "0" before the decimal point.