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Thread: Related Rates, Ugh

  1. #1
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    Related Rates, Ugh

    Hey all, need help on the attached problems!
    Attached Thumbnails Attached Thumbnails Related Rates, Ugh-1.png   Related Rates, Ugh-2.png   Related Rates, Ugh-3.png   Related Rates, Ugh-4.png   Related Rates, Ugh-5.png  

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  2. #2
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    Re: Related Rates, Ugh

    For the first problem, #9, imagine a right triangle consisting of the line from the light perpendicular to the wall, the line from the light to the dot of light on the wall, and the line on the wall from that dot of light to the base of the perpendicular. The angle in the right triangle at the light is \theta. The length of the side perpendicular to the wall is 11 miles. The distance from the dot of light to the base of the perpendicular is x= 11 tan(\theta)

    So what is \frac{dx}{dt} in terms of \frac{d\theta}{dt}?

    2 "revolutions per minute" is 2\pi radians per minute so 120\pi radians per hour. \frac{d\theta}{dt}= 120\pi radians per hour.

    Now, try the others and post your work here.
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  3. #3
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    Re: Related Rates, Ugh

    For the first problem, #9, imagine a right triangle consisting of the line from the light perpendicular to the wall, the line from the light to the point on the wall, and the line on the wall from that dot of light to the base of the perpendicular. The angle in the right triangle at the light is \theta. The length of the side perpendicular to the wall is 11 miles. The distance from the dot of light to the base of the perpendicular is x= 11 tan(\theta)

    So what is \frac{dx}{dt} in terms of \frac{d\theta}{dt}?

    2 "revolutions per minute" is [tex]2(2\pi)= 4\pi[tex] radians per minute so 60(4\pi)= 240\pi radians per hour. \frac{d\theta}{dt}= 240\pi radians per hour.
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    Re: Related Rates, Ugh

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    Re: Related Rates, Ugh

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    Re: Related Rates, Ugh



    this problem is pretty much like the first one you posted ...
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  8. #8
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    Re: Related Rates, Ugh

    A helpful strategy I've found with related rates is to conjure a relationship between two variables, take the derivative with respect to time and you're done. The only thing left to do is to substitute the rate of change one of the quantitys is changing at.
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    Re: Related Rates, Ugh

    Quote Originally Posted by aardvark View Post
    Hey all, need help on the attached problems!
    Would be easier to help if you showed
    your work and where you're stuck...
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