Science:Math Exam Resources/Courses/MATH100/December 2012/Question 08/Solution 1

To get started, we draw a diagram and label with variables.

Note that H and h (the height of the lamppost and the person) are constant but x and y (the distance of the person from the lamppost, and the length of the person's shadow) are changing.

We relate all the variables in the above picture by using similar triangles:

${\frac {y}{h}}={\frac {x+y}{H}}$

Cross-multiplying gives

$\displaystyle yH=xh+yh$

If we differentiate with respect to time, we get

$H{\frac {dy}{dt}}=h{\frac {dx}{dt}}+h{\frac {dy}{dt}}$

(Recall that H, h are constants, so they aren't differentiated.)

The question is asking us for the rate at which the person's shadow is changing, which is the same as the rate at which y is changing, or $dy/dt$. So we solve the above expression for $dy/dt$.

$H{\frac {dy}{dt}}-h{\frac {dy}{dt}}=h{\frac {dx}{dt}}$

${\frac {dy}{dt}}(H-h)=h{\frac {dx}{dt}}$

${\frac {dy}{dt}}={\frac {h{\frac {dx}{dt}}}{H-h}}$

The rate v that is given in the question is the speed at which the person is walking, i.e. the rate at which x changes, so:

${\frac {dx}{dt}}=v$

Plugging this into our previous formula, we get

${\frac {dy}{dt}}={\frac {hv}{H-h}}$

which is the final answer.