ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Martoman on July 12, 2010, 12:21:09 am

Title: shadow rate
Post by: Martoman on July 12, 2010, 12:21:09 am: A person 2 meters tall is walking away from a streetlight 8 meters high at a rate of 50 cm per second.

At what rate is the length of the person's shadow changing?

So I set up the triangles. i know $\frac{dx}{dt} = 50$

i'm calling the length of the shadow L and the distance from the man and the pole x.

Similar triangles being: $\frac{height \; of \; man}{L} = \frac{height \; of\;streetlamp}{x+l}$

Doing the mathy thing: $x+ L = 4L$

so $3L = x \rightarrow L = \frac{x}{3}$

I want $\frac{dL}{dx} * \frac{dx}{dt} = \frac{1}{3} *50 = 50/3\; cm/second$ and -ve as its getting smaller

So $-\frac{50}{3}$ ?!?

Or is this just pain wrong?
Title: Re: shadow rate
Post by: moekamo on July 12, 2010, 02:16:10 am: the rate of change of distance is in cm, but u have the heights of the man and pole in metres, i think it has something to do with that...
Title: Re: shadow rate
Post by: Martoman on July 12, 2010, 02:19:12 am: ah god.damn.it is the method right though?
Title: Re: shadow rate
Post by: moekamo on July 12, 2010, 02:42:25 am: yea i think so, do you have the answer?
Title: Re: shadow rate
Post by: Martoman on July 12, 2010, 02:44:19 am: nah which is why i'm posting it to verify if what i've done is right. So long as the method stands up to scrutiny. I don't care about the numbers, its the thought processes that matter. :smitten:
Title: Re: shadow rate
Post by: kakar0t on July 12, 2010, 03:48:26 am: thats what she said :D

ily similar triangles!
Title: Re: shadow rate
Post by: Chavi on July 12, 2010, 10:34:33 am: Firstly, you need to ensure all figures are in the same unit. you can't have dx/dt = 50 in cm, and Dl/dx in meters.

So, $\frac{dx}{dt} = \frac{1}{2}$
and the rest is correct:
$\frac{dL}{dt} = \frac{dL}{dx} * \frac{dx}{dt} = \frac{1}{3} * \frac{1}{2}= 1/6; m/second$