Need help with this related rate problem

Question

Laura is hitting a golf ball at the driving range. Suppose that the altitude of the ball is given in feet by the equation y = 200t − 40t 2, where t is the time in seconds after she hits the ball. Suppose also that the horizontal velocity of the ball is constant and equal to 80 feet per second. At what rate is the angle of elevation of the ball changing 2 seconds after Laura hits the ball?

Answer 1

Hi Bailey.

 

After 2 seconds, the ball has a vertical altitude of:

 

y(2) = 200(2) - 40(2)² = 400 - 160 = 240ft

 

The horizontal component of velocity is constant throughout the flight. After two seconds, the ball has traveled a horizontal distance of:

 

80(2) = 160ft

 

If we draw an angle to the horizontal from the start point to the ball, we can see that:

 

tan(Θ) = y/x

 

So

 

Θ = arctan(y/x)

 

You want to know how fast the angle is changing. In other words, you want dΘ/dt.

 

If we want to find out what dΘ/dt is, we can implicitly differentiate both sides of the equation.

 

d/dt(Θ = arctan(y/x))

 

dΘ/dt = 1/(1+(y/x)^2) * (x*dy/dt - y*dx/dt)/x²

 

Remember that the derivative of arctan(x) is 1/(1+x²). In this case, we don't have x, we have y/x, so we need to chain rule this quotient.

 

We can clean this expression up a little bit.

 

dΘ/dt = x²/(x² + y²) * (x*dy/dt - y*dx/dt)/x²

 

dΘ/dt = (x*dy/dt - y*dx/dt)/(x² + y²)

 

We know what y is (240 ft).

We know what x is (160 ft).

 

dx/dt is the rate of change of the horizontal component, which we know is a constant 80.

 

What about dy/dt?

 

y = 200t - 40t²

 

So

 

dy/dt = 200 - 80t

 

We want dy/dt when t = 2 seconds.

 

dy/dt(2) = 200 - 80*2 = 200 - 160 = 40

 

Plug in all these values to your dΘ/dt equation.

 

y = 240

x = 160

dx/dt = 80

dy/dt = 40

 

dΘ/dt = (x*dy/dt - y*dx/dt)/(x² + y²)

 

= (160*40 - 240*80)/(160² + 240²)

 

= -12800/83200

 

= -0.15385 rad/ sec

 

= -8.8 deg/ sec

 

Your angle is decreasing at a rate of 8.8 deg per second. Hope this helps (and is correct!).