Question

A particle moves in a circular orbit x^2+y^2=16. As it passes through the point (2,2sqrt3), its y-coordinate decreases at a rate of 3 units per second. at what rate is x-coordinate changing?

Answer 1

The rate at which the x-coordinate is changing is 3√3 units per second .

In the question ,

it is given that

the equation of the particle moving in the circular orbit is x²+y²=16

given the y coordinate decreases at 3units per second

So, dy/dt = -3.

to find the rate at which the x-coordinate is changing , we will differentiate the given equation.

Differentiating the equation  x²+y²=16 with respect to t , we get

d/dt(x²+y²) = d/dt(16)

2x(dx/dt) + 2y(dy/dt) = 0

2x(dx/dt) =  -2y(dy/dt)

Dividing both sides by 2 ,

we get ,

x(dx/dt) =  -y(dy/dt)

$\frac{dx}{dt} =\frac{-y(\frac{dy}{dt}) }{x}$

As the particle is passing through the point (2,2√3) , So, the point will satisfy the equation.

Substituting the value of x=2 , y=2√3 and dy/dt= -3 ,

we get ,

dx/dt = ((-2√3)*(-3))/2

= (6√3)/2

= 3√3

Therefore , the rate at which the x-coordinate is changing is 3√3 units per second .

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