Hey ahmedb.
You might want to draw a diagram of a circle and draw the point where the runner is and then find the derivative of the function at that point (you will have to use implicit differentiation) with the equation being y^2 + x^2 = r^2.
draw a velocity vector to represent the following (1cm represents 10km/h):
a) a girl running around a circular track travelling at 15km/h heading due east.
I checked the answer in the back and it shows:



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i don't understand why is it facing south, shouldn't it be : > ??
Hey ahmedb.
You might want to draw a diagram of a circle and draw the point where the runner is and then find the derivative of the function at that point (you will have to use implicit differentiation) with the equation being y^2 + x^2 = r^2.
Basically the equation of a circle which represents the position of the running going counterclockwise is that of a circle with center (0,0) and radius r and is given by x^2 + y^2 = r^2 or y^2 = r^2  x^2.
Velocity is always the derivative of position with respect to time and implicitly differentiating gives us:
d/dx(y^2) = d/dx(r^2  x^2) gives
2y*dy/dx = 2x which gives
dy/dx = x/y.
Now its better for this application to work in terms of representing circle with respect to a time parameter and this is given as
x = r*cos(a*t)
y = r*sin(a*t)
where a is parameter affecting the velocity.
The velocity with respect to x and y is given by
dx/dt = r*a*sin(at)
dy/dt = r*a*cos(at)
The rate of change is a combination of the rates of change with respect to x and y velocities which is given by:
ds^2 = dx^2 + dy^2 which gives
ds/dt = SQRT([dx/dt]^2 + [dy/dt]^2) or
ds/dt = r*a.
So you have a velocity in terms of x and y components and a velocity length which is a vector sum of the two with magnitude r*a.