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I Related Rates (+Trig) Question

  1. Apr 10, 2016 #1
    I'm a bit iffy with the whole of the 'related rates' topic of my calculus course. I've tried coming up with a question of my own to see if I can solve it. The question is as follows:

    The distance between a point on the ground and the bottom of a pole is 26m. The angle of inclination from that point to the top of the pole is increasing at 4ºs-1. What is the rate of change of the distance between that point and the top of the pole?

    (Note that the answer would be an expression and not a number.)

    So it took me a lot longer than I was expecting and I got stuck a few times, but here's what I got. So if the triangle is PBT where PT is the side you have to find the rate for, then I get:

    ##{\frac {d(PT)}{dt}} ={\frac {4*\sqrt{1-({\frac {26}{PT}})^2}*PT^2}{26}}##

    However obviously since this question is my own composition I can't compare it against a mark scheme or anything, can anyone confirm that I've done this correctly? Happy to provide working upon request.
     
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  3. Apr 10, 2016 #2

    Simon Bridge

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    Lets just check - The bottom of the pole is at point B on the ground, and it's top is at point T which is not on the ground.
    There is a point P also on the ground, some known distance away |BP|=d
    BTP makes a triangle with unknown (top) apex angle A, and we know dA/dt = w > 0 so the pole is being lowered.
    If we set x=|PT| then we want to know dx/dt given w and d.
    We are not told the length h=|BT| of the pole
    Have I got this right?

    How did you do that without knowing the length of the pole?
    Does it make sense that the rate that x changes is a constant?
    You can test this out by making a scale model - use a straw and a protractor - and measure how x changes with angle.
     
  4. Apr 10, 2016 #3
    It's not a constant, as you can see from my answer for the derivative of PT with respect to time includes PT itself, i.e. the rate is dependent on its current value

    Your understanding of the question is correct, but you're looking at the wrong angle, so the pole is actually being raised, my bad I thought it was implicit in "angle of inclination"

    Perhaps a diagram would help. I've taken the liberty of drawing one in the spoiler below.

    14dkqqh.png

    Thanks for the help
     
  5. Apr 10, 2016 #4

    Simon Bridge

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    Oh yes - I see the extra "PT^2" in your calculation... excuse me.
    Seeing the correct angle, I see that you can eliminate |PT| by differentiating.
    Have you tried making a model? You could also sketch for 10 different angles and see how it compares with your equation.

    What was the method that finally worked?
    Did you start with the cosine rule?
     
    Last edited: Apr 10, 2016
  6. Apr 10, 2016 #5
    Since (presumably) the pole is level it must be perpendicular to the ground so <PBT is 90 degrees, hence right angle triangle and basic trig applies:

    ##\cos\theta = {\frac {26}{PT}}##

    Rearranging for θ gives ##\theta = \cos^{-1}({\frac {26}{PT}})##

    Equating the derivative of that with respect to t (worked out via chain rule and implicitly) to 4 and rearranging for ##{\frac {d(PT)}{dt}}## gave me the expression I had in the original post
     
  7. Apr 10, 2016 #6

    Simon Bridge

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    But you said the pole is being raised - so it will only be perpendicular to the ground at the end of the process.
     
  8. Apr 10, 2016 #7
    Pardon me, I meant extended, not raised

    (edit: i.e. assume it's a straight line getting longer)
     
    Last edited: Apr 10, 2016
  9. Apr 11, 2016 #8

    Simon Bridge

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    Oh I see - the pole is extending straight up out of the ground?
    Well - fair enough ... I think, then, the main lesson for you hear is accurately and clearly communicating the details of a problem.

    Part of the point of these forums is to help each other with this communication:
    So, at point B, a horizontal distance d=|BP| from the observer at point P, a pole is extended vertically so that the angle (∠TPB) of the sight-line from P to the top of the pole T increases at constant rate w. You want to know how the distance x=|PT| changes with time - via dx/dt. Your diagram would have the pole drawn bold at an intermediate position and dotted or greyed out in a more extended position, and an arrow next to it pointing upwards indicating the motion.

    You'll notice that I don't like multi-letter variable names... best practise is to use single-letter variables.

    x=|PT| is the length of the hypotenuse of a rt triangle. ##\dot x =## to be found;
    h=|BT| is the opposite side, is unknown, and changes with time: ##\dot h \neq 0##;
    d=|PB| is the adjacent side, is given, and is constant: ##\dot d = 0##;
    θ=∠TPB so that ##\dot\theta = \omega## is given and is constant so ##\dot\omega = 0##;
    And JIC: the dot-notation means ##\dot x = \frac{dx}{dt}##.

    By trigonometry:
    ##x\cos\theta = d \implies \dot x\cos\theta - x\dot\theta\sin\theta = 0 \implies \dot x =x\omega\tan\theta## (product and chain rules)
    ##\tan\theta = h/d = \sqrt{x^2-d^2}/d## by trig + Pythagoras. So: $$\dot x = \frac{x\omega\sqrt{x^2-d^2}}{d}$$ ... you may want to see if that rearranges to what you got or find a problem with my working. (It's late I can easily make a mistake.)

    I suppose the next logical exercise would be:
    ... initially P is to the left of B and B is to the left of T ... there is a rope going from T to P where there is a winch... you want to know how fast you need the winch to pull the rope in so as to maintain a constant angular speed of ascent.
    We will note that winch-base distance |PB| is constant, as is the height of the pole |BT| ... the pole is fixed to rotate on a pivot at B.
     
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