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Homework Help: Changing rates

  1. Mar 18, 2005 #1
    In a right triangle with sides x,y,z, the theta between leg z and leg y is increasing at a constant rate of 3 rad/min. What is the rate at which x is increasing in units per minute when x equals 3 units and z is 5 units.

    so the triangle is basically a 3,4,5 triangle. The theta is between the leg 5 and 4. Leg 4 does not chage since it's the base. To set up a changing rates problem, I did:

    [tex]tan\theta = \frac{O}{A}[/tex]
    [tex]4tan \frac{d \theta}{dt} = \frac{dx}{dt}[/tex]
    [tex]\theta=-0.570 rad/min[/tex]

    I think there is something wrong with this but I'm not sure why the theta is negative.
  2. jcsd
  3. Mar 18, 2005 #2


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    Are you sure the derivative of [tex]tan(\theta)[/tex] is [tex]tan (d \theta/dt)[/tex]?
  4. Mar 18, 2005 #3
    from your post it seems that you are dealing with a right triangle where z is the hypotenuse, y is the adjacent, and x is the opposite, with z=5, y=4, x=3.

    your derivative is incorrect.

    tan is a function of theta, and theta is a function of time, so tan is actually a composite function of time.

    You have to use the chairule when differentiating tan wrt t ie:

    if [tex] f(\theta) = tan \theta [/tex] then

    [tex] \frac{df}{dt} = \frac{df}{d \theta} \frac{d \theta}{dt} [/tex]

    just an added note: you can assume that either y or z is constant, and they both work out the same.

    edit: spacetiger beat me to it
  5. Mar 18, 2005 #4

    [tex]tan\theta = \frac{O}{A}[/tex]
    [tex]4sec^2 \frac{d \theta}{dt} = \frac{dx}{dt}[/tex]
    [tex]\theta=196.855 rad/min[/tex]

    that doesnt seem correct though
  6. Mar 18, 2005 #5
    your solving for the wrong quantity... the question asks for the rate at which x increases. The rate of change of theta is already given.

    also your derivative is still incorrect,, there is a small error, do you see it?
  7. Mar 18, 2005 #6
    sorry, that is in units/minutes

    196.855 is the rate at which x is increasing.

    [tex]tan\theta = \frac{O}{A}[/tex]
    [tex]4sec^2 \frac{d \theta}{dt} = \frac{dx}{dt}[/tex]
    [tex]4 \frac{1}{tan(3)^2}= \frac{dx}{dt}[/tex]
    [tex]\frac{dx}{dt}=196.855 units/min[/tex]
  8. Mar 18, 2005 #7
    [tex]tan\theta = \frac{O}{A}[/tex]
    [tex]4sec^2 \theta \frac{d \theta}{dt} = \frac{dx}{dt}[/tex]

    is that what you mean my error?
  9. Mar 18, 2005 #8
    yes, that was it...

    whats the answer you get now?
    Last edited: Mar 18, 2005
  10. Mar 18, 2005 #9
    21.333 units/min
  11. Mar 18, 2005 #10
    how did you come up with that answer?

    hint: your solving for [tex] \frac{dx}{dt} [/itex], so you need both the values of [tex] \frac{d \theta}{dt} [/tex] and [tex] sec\theta [/tex]
    Last edited: Mar 18, 2005
  12. Mar 18, 2005 #11
    by the way... did the problem tell you that y doesnt change or is that your assumption? You get a different answer if z is constant.
  13. Mar 19, 2005 #12
    yeah, the problem tells me that y doesnt change so I have to use tan(theta)

    [tex]4sec^2\theta \frac{d\theta}{dt}=\frac{dx}{dt}[/tex]
    [tex]\frac{4}{cos^236.87} 3rad/min =\frac{dx}{dt}[/tex]

    sorry, I made a mistake, this is my final answer
    Last edited: Mar 19, 2005
  14. Mar 19, 2005 #13
    Thats the right answer.

    Just to let you know, there is another way of solving this that lets you avoid having to find what theta is, by using the definitions of sec(theta) for a right triangle.

    [tex] cos \theta = \frac{A}{H} [/tex]

    [tex] sec \theta = \frac{1}{cos \thet} = \frac{H}{A}[/tex]

    [tex] \Rightarrow sec^2 \theta = \frac{H^2}{A^2} [/tex]

    Therefore the problem boils down to simple arithmetic...

    [tex]\frac{dx}{dt} = 4sec^2\theta \frac{d\theta}{dt} [/tex]

    [tex] \Rightarrow \frac{dx}{dt} = 4 (\frac{5^2}{4^2})(3) [/tex]
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