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Rate of change

  1. Mar 11, 2009 #1
    Q) a plane is flying horizontally at an altitude of 1.5 miles and a speed of 400 mph. it is on a linear path that will take it directly over a radar station. find the rate at which the distance from the plane to the radar station is changing when the plane is exactly 2.5 miles from the station.

    I know the answer is -320 mph, but i don't know how to get it.
     
  2. jcsd
  3. Mar 11, 2009 #2
    Have you drawn a diagram yet? If not, do so and figure out what kind of shape you are dealing with. Then you can see what distance you need to find and derive the equations from there.
     
  4. Mar 11, 2009 #3
    Its a triangle, and i got the distance as 2.915? If thats right, I still don't know what to do from there
     
  5. Mar 11, 2009 #4

    AEM

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    Notice that the triangle is a right triangle and two of those sides are changing in length. What equation relates these lengths to each other? What process evaluates the rate of change? Use it on the equation and substitute in the instantaneous values given in your problem.
     
  6. Mar 11, 2009 #5
    Well, I don't think you've solved for the right value. The hypotenuse is the distance from the plane to the station, right? That means that it has the value of 2.5 at that time. In other words, the x-component is not 2.5. Also, think about the equation that you're using and how it relates all of the pieces of info that you are given.
     
  7. Mar 11, 2009 #6
    ok, so the distance is; 2.5^2= 1.5^2 + D^2
    d=2
     
  8. Mar 11, 2009 #7
    Yeah, that's right. Now do what AEM said.
     
  9. Mar 11, 2009 #8
    1.5dy/dt + 2.5dx/dt= 2dr/dt
    1.5(0) +2.5dx/dt= 2(-400)
    2.5(dx/dt)=-800
    dx/dt= -800/2.5 = -320

    ??????
     
  10. Mar 11, 2009 #9
    Well, you got the right answer but your equations are a little backwards. First of all, since y is a constant value of 1.5, not 5, you can plug it in before you differentiate. Second, remember that the velocity of the plane is in the x direction, not along the hypotenuse of the triangle.

    1. x^2 + y^2 = r^2
    2. x^2 + (1.5)^2 = r^2
    3. 2(x)(dx/dt) = 2(r)(dr/dt)
    4. 2(2)(-400) = 2(2.5)(dr/dt)
    5. -1600 = 5(dr/dt)
    6. (dr/dt) = -320
     
    Last edited: Mar 11, 2009
  11. Mar 12, 2009 #10

    AEM

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    Let's start over. You're going about this poorly. We'll put the radar station at our origin, let the vertical direction be y and the horizontal direction be x. Let the direct distance of the plane from the radar station along (the hypotenuse) be s. Then

    [tex] s^2 = x^2 + y^2 [/tex]

    When you differentiate this equation with respect to time, keeping in mind that y is a constant, you get

    [tex] 2 s \frac{ds}{dt} = 2 x \frac{dx}{dt} [/tex]

    Notice that the derivative of y with respect to t is zero. This step must be done before you substitute in numbers. Cancel off the common factor of 2. Figure out what values to substitute in for s, x, and dx/dt and you've got it. Remember: do not substitute values into your equation before you differentiate otherwise you mak make the mistake of treating something as a constant that isn't. You might want to think about why the answer is negative.
     
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