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
# x^2 + 6^2 =100 #
# :. x^2 = 100-36 = 64#
# x =+-8#
Differentiating implicitly wrt
# 2x dx/dt + 2ydy/dt = 0 #
# :. x dx/dt + ydy/dt = 0 #
So when
# (+-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
graph{ x^2 + y^2 =100 [-22, 22, -11, 11]}