# Thread: Rate of Change, Circle

1. ## Rate of Change, Circle

Point moves anticlockwise around circle x^2 + y^2 =1 at uniform speed of 2 m/s

a) Find expression for rates of change of its x - coordinate in terms of x, when the point is above the x -axis

My attempt:
Since v = 2,
Then dx/dt^2 + dy/dt^2 = v = 2

then dx/dt^2 = 2- dy/dt

Is that correct, cause the book has -2sqrt(1-x^2)

2. Originally Posted by Lukybear
Point moves anticlockwise around circle x^2 + y^2 =1 at uniform speed of 2 m/s

a) Find expression for rates of change of its x - coordinate in terms of x, when the point is above the x -axis

My attempt:
Since v = 2,
Then dx/dt^2 + dy/dt^2 = v = 2

then dx/dt^2 = 2- dy/dt

Is that correct, cause the book has -2sqrt(1-x^2)
The easy way is to work in polar coordinates:

$\displaystyle x = r \cos (\theta) = \cos \theta$

$\displaystyle y = r \sin (\theta) = \sin \theta$

$\displaystyle \displaystyle \frac{dx}{dt} = \frac{dx}{d \theta} \cdot \frac{d \theta}{dt} = -sin (\theta) \cdot \frac{d \theta}{dt}$.

But since r = 1, $\displaystyle \frac{d \theta}{dt} = 2$.

Therefore $\displaystyle \displaystyle \frac{dx}{dt} = - 2 sin (\theta) = -2y = -2 \sqrt{1 - y^2}$.

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The longer way is to note that:

$\displaystyle \displaystyle 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = - \frac{x}{y} \frac{dx}{dt}$ .... (1)

$\displaystyle \displaystyle \left( \frac{dx}{dt}\right)^2 + \left( \frac{dy}{dt}\right)^2 = 4$ .... (2)

Substitute (1) into (2) and solve for $\displaystyle \frac{dx}{dt}$. Reject the positive solution since x is decreasing for points above the x-axis if the point moves anticlockwise.

3. Thanks so much.