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Rate of Change in theta

  1. Jan 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Ship A is traveling due West towards a lighthouse at a speed of 15km/hr. Ship B is traveling due north away from the lighthouse at a speed of 10 km/hr. Let x be the distance between ship A and the lighthouse at time t, and let y be the distance between ship B and the light house at time t.

    Since I cannot show a picture, the lighthouse is directly below ship B which is headed north (So, the light house is south of ship B) and ship A is directly east of the lighthouse headed toward the lighthouse to directly west.

    a)I figured this one out. It wasn't that bad to me.
    b) Let θ be the angle in reference to ship A between ship B and the lighthouse. Find the rate of change of θ, in radians per hour, when x=4km and y=3km

    2. Relevant equations

    θ=(taninverse y/x)
    θ'=((1)/((y/x)^2 + 1)))(x(dy/dx))+(y)/(x^2)

    3. The attempt at a solution

    θ'=((1)/((3km/4km)^2 +1)))(4km(dy/dx))+(3km)/(4km^2)

    but how would you find dy/dx
    do I assume it is y/x?
  2. jcsd
  3. Jan 5, 2012 #2
    Remember that
    [tex]\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{dy}{dx}[/tex]
  4. Jan 5, 2012 #3


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    Remember that both x and y are functions of time, x(t) and y(t). The components of velocity are the time derivatives of the coordinates. Ship A moves towards west with constant velocity of 15 km/h. Its position is x(t)=x(0)-15t. Ship B moves to south with 10 km/h. its position is y(t)=y(0)+10t.

    The angle is also function of time through x and y.
    You wrote it correctly that [tex]\theta= \tan^{-1}(y(t)/x(t))[/tex]
    Calculate dθ/dt applying the chain rule. Now both x and y are function of t. And remember the derivative of (f/g).


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    Last edited: Jan 5, 2012
  5. Jan 5, 2012 #4
    WHOA! Thats it...man wow! Thanks again!
  6. Jan 5, 2012 #5
    I meant JHamm. Thank you for showing me the error I had which lead me to a false conclusion.
  7. Jan 5, 2012 #6
    @ehild oh yea if you apply the chain rule to functions of time, you get Leibniz notations of functions of time in terms of both dy/dt or dx/dt. OH!!!! thanks.
  8. Jan 5, 2012 #7


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    y is not function of x in this problem, but theta is function of both y and x, and both depend on time.

    so [tex]\frac{d\theta}{dt}=\frac{d\tan^{-1}(y/x)}{dt}=\frac{1}{1+(y/x)^2}\frac{y'x -x'y}{x^2}[/tex]
    y'=10 km/h, x'=-15 km/h.

    Last edited: Jan 5, 2012
  9. Jan 5, 2012 #8
    So, Yes, I have a lot of these corrections from old tests written on my paper, but as far as this problem specifically, I didn't do it until just a few minutes ago. I felt compelled to make a correction. When I was going through the corrections, I noticed that you have no other choice but to use the function of theta in terms of a function of x and a function of y. After you write the function out and compute the derivative, it is much much much different than my original calculation. I feel like I oversimplify things too much. Good thing I seek help. I'll get this stuff down one of these days. I mean, I'm not even in calculus 1 yet. These are just old tests I had from high school AP calculus AB that I'm using to practice for this upcoming semester. Technically, I could move on to calculus 2, but I want to make sure I can get an "A" in calculus 1 and have a boost on my gpa. I really want to go to UC Berkeley to get a B.S. in civil engineering. I have accepted the challenge. I feel confident that I could do it from the study skills I learned throughout high school and this past semester at Chabot Community College, in Hayward.

    So, Here's my correction

    since Theta(t)=taninverse(y(t)/x(t))

    then Theta'(t)=((1)/((1+y(t)/x(t))^2)(x(dy/dt) - y(dx/dt))/(x^2) like what you wrote out.


    x=4km, y=3km
    dtheta/dt[4km, 3km]=((1)/((1)+(3km/4km)^2))(4km(10km/hr) - 3km(-15km/hr))/(4km^2)

    dtheta/dt[4km, 3km]=+3.4rad/hr
  10. Jan 6, 2012 #9


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    Well done!

    You must understand the principles of Calculus before stepping further. High-school calculus is not of the same level as that at a university. Better to start from the beginning.

  11. Jan 6, 2012 #10
    Well, the AP calculus system was meant to contain the same classes as a university leveled calculus course, which explains why I never even passed one exam and why I totally failed the class until my teacher was nice enough to give me a c- since I not only shaped up on the material near the end of the class and aced 4 retake test along with the completion of a video project I did on various topics throughout the course, but also because I got a 3 on the AP exam which means I passed the AP exam being eligible for college credit at certain colleges for calculus 1 and a seat in calculus 2. The only difference is the pace really. The AP calculus class I took was meant for the first semester university leveled calculus spread out into one year.
  12. Jan 6, 2012 #11
    So, in so many words, getting an "A" in calculus 1 will redeem my 'completely fill to the brim' inferiority I once had for AP calculus AB with a witch teacher and will give me confidence and skill to do well in calculus 2. Now I know for myself that the best way to do things is to be 80 percent prepared for anything you will encounter in such a difficult or comprehensive class such as calculus, chemistry, physics, biology, or any engineering class. Of course, for the engineering classes, I don't feel that I would need to go through this 1 semester preparation process of being 80 percent prepared, because physics and other sciences will take care of that I'm sure. I haven't taken a class you need for engineering such as a thermodynamics course or statics course, but I feel that if I already spend enough time thinking about this stuff and trying to expand upon the material, rather than what's there, I will be fine with those other classes not having to spend really long hours a semester before the course has started to be prepared to Ace the class or perhaps to get a B since this will be at UC Berkeley hopefully.
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