find the point (or points) where the ellipse 9x^2+4y^2=36 has maximum curvature. don't even know where to start with this one.
4 answers
just a note i now know how to do it but the method to do it will literally take up like 4 pages. if i solve for y=f(x) and use the equation for the curve, k. please tell me im missing something that will make this simpler.
You want dy/dx to be a maximum
So the 2nd derivative has to be zero
instead of solving for y, differentiatte implicitly
18x + 8y dy/dx = 0
dy/dx = =18x/8y = -9x/4y
second derivative = (4y(-9) - (-9x)(4dy/dx) )/16y^2
= (-36y + 36x dy/dx)/(16y^2)
= 0 for a max/min of dy/dx
36x dy/x = 36y
x dy/dx = y
dy/dx = y/x
but dy/dx = -9x/4y
continue please
So the 2nd derivative has to be zero
instead of solving for y, differentiatte implicitly
18x + 8y dy/dx = 0
dy/dx = =18x/8y = -9x/4y
second derivative = (4y(-9) - (-9x)(4dy/dx) )/16y^2
= (-36y + 36x dy/dx)/(16y^2)
= 0 for a max/min of dy/dx
36x dy/x = 36y
x dy/dx = y
dy/dx = y/x
but dy/dx = -9x/4y
continue please
ok got it. thank you. i also solve for y and plug it in right? so i can solve for x.
reiny is wrong you cannot just use the second derivative to find the maximum curvature.