when x = 3
9+y^2 = 25
y^2 = 16
y = ±4
2x dx/dt + 2y dy/dt = 0
dy/dt = -x dx/dt/y
= -3(-8)/±4
= ± 6
If x^2+y^2=25, find dy/dt when x=3 and dx/dt= -8.
2 answers
You mean
on this circle find vertical component of velocity.
2 x dx/dx + 2 y dy/dx = 0
y dy/dx = -x
y dy/dx * dx/dt = -x * dx/dt
dy/dt = -(x/y) dx/dt
here x = 3
then y = +4 or -4
if y is +, dy/dt is + for -dx/dt
if y is -, dy/dt is - for -dx/dt
so
dy/dt = -(3/4)(-8) = 6 in first quadrant
dy/dt = -(3/-4)(-8) = -6 in quadrant 4
on this circle find vertical component of velocity.
2 x dx/dx + 2 y dy/dx = 0
y dy/dx = -x
y dy/dx * dx/dt = -x * dx/dt
dy/dt = -(x/y) dx/dt
here x = 3
then y = +4 or -4
if y is +, dy/dt is + for -dx/dt
if y is -, dy/dt is - for -dx/dt
so
dy/dt = -(3/4)(-8) = 6 in first quadrant
dy/dt = -(3/-4)(-8) = -6 in quadrant 4