Rates of change (1 Viewer)

[NSBHSchoolie]

Yes I suck at rates of change - here is a cambridge 3U q. (ex 7E, Q11)

A point moves around the circle x^2 + y^2 = 1 at a uniform speed of 2m/s.

(a) Find an expression for the rate of change of its x-coordinate in terms of x, when the point is above the x-axis. (The units on the axes are in metres.)

(b) Use your answer to part (a) to find the rate of change of the x-coordinate as it crosses the y-axis at P(0,1). Why should this answer have been obvious without this formula?

If memory serves me right the second part of (b) is because its a circle, at (0,1), dy/dx = undefined and therefore v(x)=0m/s and v(y)=2m/s.

Thanks in advance

 

[ND]

A point moves around the circle x^2 + y^2 = 1 at a uniform speed of 2m/s.

(a) Find an expression for the rate of change of its x-coordinate in terms of x, when the point is above the x-axis. (The units on the axes are in metres.)

Remember that dy/dx = rise/run = tan@. Think about it in terms of horizontal (x) and vertical (y) components.

(b) Use your answer to part (a) to find the rate of change of the x-coordinate as it crosses the y-axis at P(0,1). Why should this answer have been obvious without this formula?

If memory serves me right the second part of (b) is because its a circle, at (0,1), dy/dx = undefined and therefore v(x)=0m/s and v(y)=2m/s.

Actually, at (0,1), dy/dx = 0, so v_x = 2m/s. Alternatively, you could just think about something moving in circular motion; when it's at the top of the circle, it's motion is purely horizontal.

 

[NSBHSchoolie]

But you need dx/dt right? How would you go about that, chain rule?

 

[ND]

Yep use chain rule. Remember that d@/dt=2.