Calc Problem - find dy/dt when y^2 = x^2 + 1?

x=2 ---> y = ...

and y(dy/dt) = x(dx/dt) ----> dy/dt = .,.

First we find dy/dx using implicit differentiation:

2y dy/dx = 2x

dy/dx = x/y

When x = 2

y^2 = 2^2 + 1 = 5

y = ± √5

Using chain rule:

dy/dt = dy/dx * dx/dt

dy/dt = x/y dx/dt

dy/dt = 2/± √5 * 500

dy/dt = ± 1000/√5 = ± 200√5

NOTE: It's not possible to know if dy/dt = 200√5 or -200√5 without more information.

Mαthmφm

y² = x² + 1

2y(dy/dt) = 2x(dx/dt) + 0

Solve for y first:

y² = 2² + 1

y = ±√5

±2√5(dy/dt) = 2(2)(500)

dy/dt = ±1000/√5 ≈ ±447.214

2y(dy/dt) = 2x(dx/dt)

dy/dt = x/y * (dx/dt)

When x = 2, y = +/- √5

dy/dt when x = 2: +/- 2/√5 * (500)