Asked by Anon
Let C be the intersection of x^2+y^2=16 and x+y+z=5. Find the curvature at (0,4,1).
I don't know how to find the intersection between the given equations.
I don't know how to find the intersection between the given equations.
Answers
Answered by
Steve
The parametric equations for the circle are
x = 4cos t
y = 4sin t
so, the curve of the intersection also includes
z = 5-x-y = 5-4cost-4sint
Now, the curvature of the curve described is
k = |dT/ds| = |dT/dt| ÷ |dr/dt|
at (0,4,1), t = π/2
r = <4cost,4sint,5-4cost-4sint>
r' = <-4sint,4cost,4sint-4cost>
at t = π/2, r' = <-4,0,4>
|r'| = 4√2
T = r'/|r| = 1/√2 <-4sint,4cost,4sint-4cost>
dT/dt = 1/√2 <-4cost,-4sint,4cost+4sint>
so, dT/dt = 1/√2 <0,-4,4>
|dT/dt| = 4
so, k = (4)/(4√2) = 1/√2
As always, double-check my math.
x = 4cos t
y = 4sin t
so, the curve of the intersection also includes
z = 5-x-y = 5-4cost-4sint
Now, the curvature of the curve described is
k = |dT/ds| = |dT/dt| ÷ |dr/dt|
at (0,4,1), t = π/2
r = <4cost,4sint,5-4cost-4sint>
r' = <-4sint,4cost,4sint-4cost>
at t = π/2, r' = <-4,0,4>
|r'| = 4√2
T = r'/|r| = 1/√2 <-4sint,4cost,4sint-4cost>
dT/dt = 1/√2 <-4cost,-4sint,4cost+4sint>
so, dT/dt = 1/√2 <0,-4,4>
|dT/dt| = 4
so, k = (4)/(4√2) = 1/√2
As always, double-check my math.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.