# Calculus rate of change

I'm not sure why I'm not getting the same answer.This is the question.

A spotlight is located 500 feet from a straight road. A truck is driving down the road at the rate of 100 ft/sec (68.2 mi/hour). The spotlight is rotating to remain focused on the truck. How fast is the spotlight rotating to remain focused on the truck when the truck is 1300 feet way from the spotlight?

I used 500 tan θ = y ..and then differentiate it with respect to t

However,if I use 1300 sin θ = y ...and then differentiate it with respect to t,I'm getting different answer.

Can anyone tell me why?

Draw the right triangle. When the truck is 1300ft away from the spotlight, it's 1200 ft away from the point on the road directly opposite the light (500-1200-1300 triangle).

Your equation is basically correct. $\displaystyle 500\tan\theta = x$ (I used x instead of your 'y'). x here represents the distance along the road of the truck from that point directly opposite the light.

Differentiate implicitly wrt t:

$\displaystyle 500\sec^2\theta \frac{d\theta}{dt} = \frac{dx}{dt} = v$, where v is the speed of the truck in ft/s.

So you get $\displaystyle \frac{d\theta}{dt} = \frac{\cos^2\theta}{500}.v = \frac{1}{500}.(\frac{5}{13})^2.100 = \frac{5}{169} rad/s$, which is the answer you require.

• Right exactly.But from the right angle triangle,I can also use the equation 1300 sin θ = x right? (your x)..If I differentiate it implicitly with respect to t,I don't seem to get the same answer.Why is it like this? – Zhi J Teoh May 22 '14 at 5:28
• Because the 1300 doesn't stay constant, whereas the 500 does. Remember that as the truck moves, the hypotenuse changes with time, but the distance of the spotlight to the road is a constant. :) – Deepak May 22 '14 at 5:29
• So that means 1300ft is useless here?I mean we didn't even use the 1300ft. – Zhi J Teoh May 22 '14 at 5:33
• The 1300ft is useful, but only to fix the position of the truck at the instant you want to determine the rate of change of the angle. I used the 1300ft to calculate that at that instant, the "base" of the right triangle is 1200ft using Pythagoras' theorem. – Deepak May 22 '14 at 5:35
• So basically from what I see,when calculating the rate of change of the angle,one of the variable must be a constant,and another will be a changing variable?like 500 tan θ = x ....500 is constant while x changes(since the truck is travelling with respect to time) – Zhi J Teoh May 22 '14 at 5:39