Related rates: A train is travelling from left to right....

[Sophie]

A train is travelling from left to right along a long strait horizontal track at 0.9km/min. A movie camera 1km away from the train is focused on the train.

a. How fast is the distance between the camera and the train changing when the train is 2km from the camera?

b. How fast is the camera rotating (in radians/min) at the moment when the train is 2km from the camera.

I have done the following, however I am unsure if I have interprited the question right with my right angled triangle. I found the wording of this question challenging I am not 100% sure where the camera is suppose to be?

a. I drew a right angle triangle with the camera 1km from the tracks and the hyp as 2km.

x = 1km (distance from tracks to camera)
y= 2km distance from train to camera (hyp)
z = train tracks

x^2 + z^2 = y^2

z = 2.24

1^2 + z^2 = y^2

2z dz/dt = 2y dy/dt

dz/dt = 0.9km/hr
z = 2.24
y = 2

dy/dt = 1.006 km/min


b. Using the same triangle

cosx = 1/y

-sinx dx/dt = -1/(y^2) dy/dt

dy/dt = 1.006 km/min
y = 2
sinx = 2.24/2

dx/dt = (-1/(-sinx *y^2)) * dy/dt


dx/dt = (1/((2.24/20)*4)) * 1.006

dx/dt = 0.22 Rad/min

Thanks Sophie

 

[galactus]

I got something a little different for the first part.

As you stated, Pythagoras.

x=distance along track, y=1, z=distance from camera to train(hypoteneuse).

Knowns: z=2, dx/dt=9/10, y=1

\(\displaystyle \L\\z^{2}=x^{2}+y^{2}\).........[1]

Using Pythagoras, we see that \(\displaystyle \L\\\sqrt{2^{2}-1^{2}}=\sqrt{3}\)

Differentiate [1]:

\(\displaystyle \L\\z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\)

\(\displaystyle \L\\2\frac{dz}{dt}=(\sqrt{3})(\frac{9}{10})+(1)(0)\)

\(\displaystyle \L\\\frac{dz}{dt}=\frac{9\sqrt{3}}{20}\approx{0.779} \;\ km/min\)

For the 2nd part.

\(\displaystyle \L\\tan({\theta})=x\)

\(\displaystyle \L\\sec^{2}({\theta})\frac{d{\theta}}{dt}=\frac{dx}{dt}\)

\(\displaystyle \L\\\frac{d{\theta}}{dt}=\frac{1}{sec^{2}({\theta})}\frac{dx}{dt}\)

\(\displaystyle \L\\sec({\theta})=\frac{2}{1}, \;\ sec^{2}({\theta})=4\)

\(\displaystyle \L\\\frac{d{\theta}}{dt}=\frac{1}{4}(\frac{9}{10})=\frac{9}{40}={0.225} \;\ rad/min\)

 

[Sophie]

Thanks

That was a very silly mistake I made during part a...

Sophie