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Math Help - Implicit differentiation problem

  1. #1
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    Implicit differentiation problem

    A weather balloon V, rising vertically from a point P on the ground, is observed from a point O on the ground, 90 m from P. Let θ be the angle between OP and OV (i.e. the angle between the ground and the sightline to the balloon). At what rate is the balloon rising when θ=45˚ and θ is increasing at 1˚/second?

    So, I'm thinking the job is to find:

    d
    (VP)/dt at the time when /dt=1.

    We already know that

    VP(
    θ)=90tanθ

    and

    θ(VP)=arctan(VP/90)

    So I take the derivative of both functions, accounting for the fact that both VP and θ are functions of time as well as of each other:


    Implicit differentiation problem-picture-23.png

    Knowing that /dt=1 and OP=90, I get that:

    Implicit differentiation problem-picture-24.png
    Which is 1/180. I notice that if I flip this over and solve for
    d(VP)/dt in the other equation, I get that

    d(VP)/dt=1

    Is this right?? If so, should I have been able to see the from the start? A lot of steps in this one, and I feel like I'm kinda guessing as to what I should do.
    Last edited by TwoPlusTwo; October 29th 2010 at 06:59 AM. Reason: typo
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  2. #2
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    Quote Originally Posted by TwoPlusTwo View Post
    A weather balloon V, rising vertically from a point P on the ground, is observed from a point O on the ground, 90 m from P. Let θ be the angle between OP and OV (i.e. the angle between the ground and the sightline to the balloon). At what rate is the balloon rising when θ=45˚ and θ is increasing at 1˚/second?

    So, I'm thinking the job is to find:

    d
    (VP)/dt at the time when /dt=1.

    We already know that

    VP(
    θ)=90tanθ

    and

    θ(VP)=arctan(VP/90)

    So I take the derivative of both functions, accounting for the fact that both VP and θ are functions of time as well as of each other:


    Click image for larger version. 

Name:	Picture 23.png 
Views:	17 
Size:	11.3 KB 
ID:	19519

    Knowing that /dt=1 and OP=90, I get that:

    Click image for larger version. 

Name:	Picture 24.png 
Views:	163 
Size:	7.0 KB 
ID:	19520
    Which is 1/180. I notice that if I flip this over and solve for
    d(VP)/dt in the other equation, I get that

    d(VP)/dt=1

    Is this right?? If so, should I have been able to see the from the start? A lot of steps in this one, and I feel like I'm kinda guessing as to what I should do.
    note that the rate of change in \theta has to be in radians ...

    \frac{d\theta}{dt} = \frac{\pi}{180} rad/sec

    let h = PV

    \theta = \arctan\left(\frac{h}{90}\right)

    \frac{d\theta}{dt} = \frac{\frac{1}{90}}{1 + \left(\frac{h}{90}\right)^2} \cdot \frac{dh}{dt}

    when \theta = 45^\circ , h = 90' ...

    \frac{\pi}{180} = \frac{1}{180} \cdot \frac{dh}{dt}

    \frac{dh}{dt} = \pi ft/sec
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  3. #3
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    Great! So how do we know the angle has to be in radians? That didn't even occur to me, since the problem was stated using degrees.
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  4. #4
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    Quote Originally Posted by TwoPlusTwo View Post
    Great! So how do we know the angle has to be in radians? That didn't even occur to me, since the problem was stated using degrees.
    Basically, you probably have learned that

    \frac{d}{dx} \tan x = \sec^2 x

    However, this statement is only true when x is expressed in radians. The derivatives/anti-derivatives of the trig functions are actually different if you express them in degrees.

    For example, if you REALLY wanted to take a derivative in degrees, you'd actually get this:

    \frac{d}{dy} \tan y = \frac{\pi}{180} \sec^2 y

    You could use this to solve this problem, but instead it's generally a lot easier to convert everything to radians first!
    Last edited by drumist; October 30th 2010 at 05:39 AM.
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