PLEASE HELP! Use your calculator to find the length of the arc from t = 0 to t = 3 of x = 2t + 1, y = t2 - 1.
Type your answer in the space below and give 3 decimal places. If your answer is less than 1, place a leading "0" before the decimal point (ex: 0.482). (5 points)
Which one or ones of the following integrals produces the area that lies inside the circle r = 3 sin θ and outside the cardioid r = 1 + sin θ? (10 points)
I. 1/2 the integral from pi/6 to 5pi/6 of 9*sin^2 theta minus (1 + sin theta)^2 d theta
II. the integral from pi/ 6 to pi/2 of 9*sin^2 theta minus (1 + sin theta)^2 d theta
III. 1/2 the integral from 0 to pi/6 of 9*sin^2 theta minus (1 + sin theta)^2, d theta plus 1/2 the integral from 5pi/6 to pi of 9sin^2 theta minus (1 + sin theta)^2, d theta
A) I only
B) I and II only
C) II only
D) III only
didn't we already do these?
∫[0,3]√((2)^2 + (2t)^2) dt = ∫[0,3] 2√(1+t^2) dt = ____
since r=3sinθ intersects r=1+sinθ at θ = π/6 and 5π/6, the area is
∫[π/6,5π/6] 1/2 (R^2-r^2) dθ
where R = 3sinθ and r = 1+cosθ
Looks like B to me, due to the symmetry of the region.
To find the length of the arc from t = 0 to t = 3 of the curve described by the parametric equations x = 2t + 1 and y = t^2 - 1, we can use the arc length formula for parametric curves.
The arc length formula for a parametric curve given by x = f(t) and y = g(t) is given by:
L = ∫[a to b] √[ (dx/dt)^2 + (dy/dt)^2 ] dt
In this case, the range of t is from t = 0 to t = 3, so we need to evaluate the integral from t = 0 to t = 3.
First, let's calculate dx/dt and dy/dt:
dx/dt = 2
dy/dt = 2t
Now, substitute these values into the arc length formula:
L = ∫[0 to 3] √[ (2)^2 + (2t)^2 ] dt
L = ∫[0 to 3] √(4 + 4t^2) dt
To calculate the integral, you can use a calculator or software that can perform definite integrals. Approximating the value with a calculator or software that can perform numerical integration techniques, we get:
L ≈ 7.043
Rounding to three decimal places, the length of the arc from t = 0 to t = 3 is approximately 7.043.