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Math Help - Calc: Related rates of change question

  1. #1
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    Calc: Related rates of change question

    Chris, who is 6 feet tall, is walking away from a street light pole 30 deet high at a rate of 2 feet per second.

    a. How fast is his shadow increasing in length when chris is 24ft from the pole? 30 feet? -- I already found that to be 1/2 ft per sec.

    b. How fast is the tip of his shadow moving?

    c. To follow the tip of his shadow, at what angular rate must Chris be lifting his eyes when his shadow is 6ft long?



    Please help with b and c!
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  2. #2
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    Quote Originally Posted by cedwards08 View Post
    Chris, who is 6 feet tall, is walking away from a street light pole 30 deet high at a rate of 2 feet per second.

    a. How fast is his shadow increasing in length when chris is 24ft from the pole? 30 feet? -- I already found that to be 1/2 ft per sec.

    b. How fast is the tip of his shadow moving?

    c. To follow the tip of his shadow, at what angular rate must Chris be lifting his eyes when his shadow is 6ft long?



    Please help with b and c!
    b. Let the shadow length be l, and the distance walked be w, then the tip of the shadow is l+w from the pole, and the tip is moving at:

     <br />
v=\frac{d}{dt}(l+w)=\frac{dl}{dt}+\frac{dw}{dt} <br />

    and both of the speeds on the right you should already know.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by cedwards08 View Post
    Chris, who is 6 feet tall, is walking away from a street light pole 30 deet high at a rate of 2 feet per second.

    a. How fast is his shadow increasing in length when chris is 24ft from the pole? 30 feet? -- I already found that to be 1/2 ft per sec.

    b. How fast is the tip of his shadow moving?

    c. To follow the tip of his shadow, at what angular rate must Chris be lifting his eyes when his shadow is 6ft long?



    Please help with b and c!
    c. Let l be the length of the shadow, then the look down angle of Chris' eyes is \theta where:

    \tan(\theta)=\frac{6}{l}

    So the rate that he raises his eyes is -\frac{d\theta}{dt} , and:

    \frac{d}{dt}\tan(\theta)=\sec^2(\theta)\frac{d \theta}{dt}=-\ \frac{6}{l^2}\frac{dl}{dt}

    RonL
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  4. #4
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    Quote Originally Posted by cedwards08 View Post
    Chris, who is 6 feet tall, is walking away from a street light pole 30 deet high at a rate of 2 feet per second.

    a. How fast is his shadow increasing in length when chris is 24ft from the pole? 30 feet? -- I already found that to be 1/2 ft per sec.

    b. How fast is the tip of his shadow moving?

    c. To follow the tip of his shadow, at what angular rate must Chris be lifting his eyes when his shadow is 6ft long?



    Please help with b and c!
    Let's go back to part (a) anyway because it is connected to parts (b) and (c).

    Draw the figure.
    It is a large right triangle whose vertical leg, the pole, is 30 ft and whose horizontal leg is L.
    L = x +y
    where x = distance of Chris from the pole.
    And y = length of Chris' shadow.

    Thus, a smaller right triangle, similar to the larger one, is inside the larger one.
    This smaller one has a vertical leg, Chris, that is 6 ft and a horizontal leg that is y ft.

    By proportion,
    30/(x+y) = 6/y
    Cross multiply,
    30y = 6x +6y
    24y = 6x
    4y = x ------(i)

    Differentiate both sides with respect to time t,
    4(dy/dt) = dx/dt

    It is given in the question that dx/dt = 2 ft/sec, so,
    4(dy/dt) = 2
    dy/dt = 2/4 = 1/2 ft/sec ----------the rate of increase of the length of the shadow.

    When x = 24 ft? .....dy/dt = 1/2 ft/sec
    When x = 30 ft? .....dy/dt = 1/2 ft/sec
    The rate of increase of Chris's shadow's length has nothing to do on how far away Chris is from the pole. Meaningg dy/dt = 1/2 ft/sec always.

    ------------------------
    b. How fast is the tip of his shadow moving?

    Here, we call the tip of the shadow to be L away from the pole.
    In the figure, Chris is still x ft away from the pole, so, the the previous y becomes (L -x) ft.
    By the same proportion,
    30/L = 6/(L -x)
    30(L -x) = 6L
    30L -30x = 6L
    24L = 30x
    4L = 5x
    Differentiate both sides with respect to time t,
    4(dL/dt) = 5(dx/dt)
    4(dL/dt) = 5(2)
    dL/dt = 10/4 = 2.5 ft/sec ---------the rate the tip of the shadow is moving.

    ---------------------------
    c. To follow the tip of his shadow, at what angular rate must Chris be lifting his eyes when his shadow is 6ft long?

    In the original figure, see the smaller right triangle.
    Vertical leg = 6 ft
    Horizontal leg = y = Chris' shadow.

    If Chris is looking at the tip of his shadow, the angle of his eyesight is that one between his eyes and the line of sight....the angle between the vertical leg and the hypotenuse.
    Let's call this angle, theta.
    tan(theta) = y/6
    Differentiate both sides with respect to time t,
    sec^2(theta) *dtheta/dt = (1/6)dy/dt

    We found out in part (a) that dy/dt = 1/2 ft/sec always, so,
    sec^2(theta) *dtheta/dt = (1/6)(1/2) = 1/12 ----------**

    We don't know sec(theta) so we find it.
    When the shadow is 6 ft long, the y is 6 ft.
    Since the vertical leg is 6 ft and the horizontal leg is 6 ft also, then theta = 45 degrees.
    And, sec(45deg) = sqrt(2), so,
    [sqrt(2)]^2 *dtheta/dt = 1/12
    dtheta/dt = (1/12)/2
    dtheta/dt = 1/24 rad/sec

    Therefore, to follow the tip of his shadow, Chris must be lifting his eyesight at the rate of 1/24 radian per second at the instant that the his shadow is 6 ft long.
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