Second Derivative of x^2-y^2=1?

Question

Could you please show how the Second Derivative of x^2-y^2=1 is equal to 1/y-x^2/y^3.

Thanks

Hrm I see its not a clear question, can you find the second derivative of this? My Ans doesn't agree with the ans my calculator gives me. You can do implict dif on your calc using other programs fyi.

Answer 1

Let's see:

x²-y²=1
2x - 2y dy/dx = 0
2 - 2y d²y/dx² - 2 (dy/dx)² = 0

From the second equation:
dy/dx=x/y

So:
2 - 2y d²y/dx² - 2x²/y² = 0
2y d²y/dx² = 2 - 2x²/y²
d²y/dx² = 1/y - x²/y³

Q.E.D.

Answer 2

Set x1=x2 and remedy for t: 3sin t = -3+cos t, 9(sin t)^2 = 9-6cos t+(cos t)^2. 9(a million-(cos t)^2) = 9-6cos t+(cos t)^2, 9-9(cos t)^2=9-6cos t+(cos t)^2 0=-6cos t+10(cos t)^2, (cos t)(-6+10cos t)=0, cos t =0 or cos t = 6/10. watching the graphs of x1 and x2 there are 2 intersection factors: 3pi/2 and 3pi/2+arccos(3/5). Now replace those 2 values into y1 and y2 to locate that: y1=y2 whilst t=3pi/2. A collision will ensue whilst t=3pi/2. area b could be solved further.

Answer 3

Seems to be something wrong in the question...or maybe I'm wrong...