I have another question (studying for finals, haha)
Let ~r(t) = et cost
~i + et sin t~j be parametric equations of a curve C. Find the length of
C from t = 0 to t = π.
so the equation I am using is the integral from a to b (in thsi case 0 to pi) of the magnitude of r'(t) dt
I get 0 because I get 2e^tsint as my integrated magnitude and when sint is 0 or pi it is just 0.
but the answer is sqrt(2) (e^pi -1)
ok so now i get that i did the integral wrong,. it should be 2e^t(sint+cost) but I get 2(e^pi -1) for sqrt2??
To find the length of curve C from t = 0 to t = π, you correctly applied the formula for arc length using parametric equations. The formula is:
L = ∫[a,b] √[(dx/dt)² + (dy/dt)²] dt
In this case, you have ~r(t) = et cos(t)~i + et sin(t)~j. To find the magnitude of the derivative of ~r(t), we need to find dx/dt and dy/dt.
dx/dt = d(et cos(t))/dt = et cos(t) - et sin(t)
dy/dt = d(et sin(t))/dt = et sin(t) + et cos(t)
Now, let's calculate the magnitude of r'(t):
|r'(t)| = √[(dx/dt)² + (dy/dt)²]
= √[(et cos(t) - et sin(t))² + (et sin(t) + et cos(t))²]
= √[e²t²(cos²(t) - 2cos(t)sin(t) + sin²(t)) + e²t²(sin²(t) + 2cos(t)sin(t) + cos²(t))]
= √[e²t² + e²t²]
= √(2e²t²)
= √2et
Next, integrate |r'(t)| from t = 0 to t = π:
L = ∫[0,π] √2et dt
Integrating √2et with respect to t gives you:
L = [√2/2 e^(2t/2)] from 0 to π
L = [√2/2 e^t] from 0 to π
L = (√2/2)(e^π - e^0)
L = (√2/2)(e^π - 1)
L = sqrt(2)(e^π - 1)
Therefore, the length of curve C from t = 0 to t = π is sqrt(2)(e^π - 1).