Question

How to solve this using partial derivatives?

Answer 1

both x and y are function of t

2x x' -36 y y'=0

2x dx/dt -36 y dy/dt=0

dy/dt=9

Answer 2

x=9

Answer 3

2x dx/dt -36 y dy/dt=0

18 dx/dt -36 y=0

dx/dt=36y/18

dx/dt= 2y

Answer 4

We have
\[x^2-18y^2=9\]
We are given
\[\frac{dy}{dt}= 9\ units/s\]
We have to find \[\frac{dx}{dt}\ at\ x= 9\]
At x=9
\[9^2-18y^2=9\]
\[81-9=18y^2\]
\[90=18y^2=>y=\pm \sqrt{5}\]
\[x^2-18y^2=9\]
Let's differentiate this with respect to t
\[2x\frac {dx}{dt}-36y\frac{dy}{dt}=0\]
or
\[x\frac {dx}{dt}=36y\frac{dy}{dt}\]
substitute x=9
\(y=\sqrt {5}\) since we want the rate of increase
and \(\frac{dy}{dt}=9\)
Find \(\frac{dx}{dt}\) that's your rate of increase in x

Answer 5

@doncatch Did you understand?

Answer 6

@ash2326 little mistake there... y=2or -2

Answer 7

@doncatch Sorry \(y=\pm 2\)
substitute y=2 in the last equation