If #x^2+y^2=25# and #dy/dt=6#, how do you find #dx/dt# when #y=4# ?

Question

Christopher Agresta · Accepted Answer

There are two values depending on the point.

#{dx}/{dt}={(-8 " at " (3,4)),(8 " at "(-3,4)):}#

Let us look at some details.

First, let us find the values of #x#.

By plugging in #y=4# into #x^2+y^2=25#,

#x^2+16=25 Rightarrow x^2=9 Rightarrow x=pm3#

Now, let us find some derivatives.

By differentiating with respect to #t#,

#d/{dt}(x^2+y^2)=d/{dt}(25) Rightarrow 2x{dx}/{dt}+2y{dy}/{dt}=0#

by dividing by #2x#,

#Rightarrow {dx}/{dx}+y/x{dy}/{dt}=0#

by subtracting #y/x{dy}/{dt}#,

#Rightarrow {dx}/{dt}=-y/x{dy}/{dt}#

Since #y=4#, #x=pm3#, and #{dy}/{dt}=6#,

#{dx}/{dt}=-{4}/{pm3}(6)=pm8#

Paisley Johnson · Answer

undefined To find $ \frac{dx}{dt} $ when $ y = 4 $, differentiate the equation $ x^2 + y^2 = 25 $ implicitly with respect to $ t $, then solve for $ \frac{dx}{dt} $:

$ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 $

Given that $ \frac{dy}{dt} = 6 $ and $ y = 4 $:

$ 2x \frac{dx}{dt} + 2(4)(6) = 0 $

Solve for $ \frac{dx}{dt} $:

$ 2x \frac{dx}{dt} = -48 $

$ \frac{dx}{dt} = \frac{-48}{2x} $

Substitute $ y = 4 $ into the given equation $ x^2 + y^2 = 25 $ to find $ x $:

$ x^2 + 4^2 = 25 $

$ x^2 = 25 - 16 $

$ x^2 = 9 $

$ x = \pm 3 $

Since we are looking for $ \frac{dx}{dt} $ when $ y = 4 $, we take the positive value:

$ x = 3 $

Now, substitute $ x = 3 $ into $ \frac{dx}{dt} = \frac{-48}{2x} $:

$ \frac{dx}{dt} = \frac{-48}{2(3)} $

$ \frac{dx}{dt} = -8 $

Therefore, when $ y = 4 $, $ \frac{dx}{dt} = -8 $.