If #x^2+y^2=100# and #(dy)/(dt)=4#, what is #(dx)/(dt)# when #y=6#?

1 Answer
Apr 12, 2018

# dx/dt = +-3 # when #y=6#

Explanation:

We have:

# x^2 + y^2 =100 #

When #y=6#, we have:

# x^2 + 6^2 =100 #

# :. x^2 = 100-36 = 64#
# x =+-8#

Differentiating implicitly wrt #t# we have:

# 2x dx/dt + 2ydy/dt = 0 #

# :. x dx/dt + ydy/dt = 0 #

So when #y=4#, we have:

# (+-8) dx/dt + (6)(4) = 0 #

# :. dx/dt = - (6)(4)/(+-8) #

# :. dx/dt = +-3 #

We gain two possible solutions due to the fact that the original equation is that of a circle of radius #10# centered on the origin:

graph{ x^2 + y^2 =100 [-22, 22, -11, 11]}