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Thread: Word problem help

  1. #1
    Junior Member
    Oct 2012

    Word problem help

    The brightness of a lamp is L(s) = 1500s^-2 lumens where s is the distance in metres
    from the observer to the lamp. An observer is walking at 1 metre per second along a
    straight line that comes within 3 metres of the lamp at its closest point. How fast (in
    lumens per second) is the brightness changing when the observer has passed the lamp and
    is 5 metres away from it?

    Please help
    Thanks in advance
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  2. #2
    MHF Contributor MarkFL's Avatar
    Dec 2011
    St. Augustine, FL.

    Re: Word problem help

    I would orient the coordinate axes such that the lamp is at the origin and the observer moves along the line $\displaystyle y=3$ from left to right. So, put the observer at the point $\displaystyle (x,3)$ and then the distance between the observer and the lamp is:

    $\displaystyle s=\sqrt{x^2+3^2}$ and we have:

    $\displaystyle L(x)=\frac{1500}{x^2+9}$

    Now, we want to find $\displaystyle \frac{dL}{dt}$, so differentiate with respect to time $\displaystyle t$, and use the given:

    $\displaystyle \frac{dx}{dt}=1\frac{\text{m}}{\text{s}}$

    $\displaystyle s=5\text{ m}$
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  3. #3
    MHF Contributor
    Sep 2012

    Re: Word problem help

    Hey nubshat.

    A you familiar with the chain rule? Recall that if you have x(t) = f(g(t)) then dx/dt = dg/dt * f'(g(t)) where ' means the derivative. In this case your f is 1500s^-2 and your g will be how the distance changes as a function of time where s will be the distance from the lamp as a function of t.

    Since the closest point is 3m away then intuitively it will look like a triangle where the distance away is the hypotenuse of the triangle where s^2 = 3^2 + (1*t)^2 since the person is moving perpendicular to the line where the shortest distance is (you might want to draw a right angled triangle to see what is going on).

    So you have L as a function of s and s as a function of t and you want to find L'(t) when s^2 = 25.
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