Question

Suppose 4x^2+9y^2=36, where x and y are functions of t.
If dy/dt=1/3, find dx/dt when x=2 and y=2/3*sqrt(5)

Answer 1

Diffferentiate implitly, you get:

8x dx/dt + 18y dy/dt = 0

Now, substitute, you get

8(2) dx/dt + 18(1/3)(2/3) = 0

16 dx/dt + 4 = 0

16 dx/dt = -4

dx/dt = -4/16 or -1/4

Answer 2

implicitly*

Answer 3

The 2/3 actually \[2/3 \sqrt{5}\] so it would be \[-4\sqrt{5}/16\] right?

Answer 4

Yes..I didnt see that 3sqrt(5)...you're absolutely correct.

Answer 5

okay thank you!!