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)

The derivative of the function in terms of t is,
2(4x)(dx / dt) + 2(9y)(dy / dt) = 0
Substituting the known values from the given above gives,
2(4)(2)(dx / dt) + 2(9)(2/3)(sqrt of 5)(1 /3) = 0
Solving the equation gives a value of dx/dt equal to -0.559.

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abidemiokin abidemiokin · Answer 2 · 2021-10-22T13:04:12+02:00

Suppose 4x^2+9y^2=36, where x and y are functions of t. If dy/dt=1/3, the value if dx/dt will be -√5/8

Given the expression 4x^2+9y^2=36

If y = 2/3 √5, get the value of x as shown below:

4x^2 + 9( 2/3 √5,)^2 = 36

4x^2 + 9(20/9) = 36

4x^2 = 36 - 20

4x^2 = 16

x^2 = 4

x = 2

Differentiate the function implicitly with respect to t to have;

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

8(4) dx/dy + 18 (2/3 √5)(1/3) = 0

32 dx/dt + 4√5 = 0

d/dt = -4√5/32

dx/dt = -√5/8

Hence dx/dt from the given equation is -√5/8

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