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watchmath

  • 5 years ago

Find the height of the lamp post (see the attachment).

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  1. watchmath
    • 5 years ago
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    welcome to answer everybody :).

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  2. myininaya
    • 5 years ago
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    did you make this problem up or did you find it somewhere

  3. watchmath
    • 5 years ago
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    It is from a book :). We give this as an assignment for our student :).

  4. myininaya
    • 5 years ago
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    i think i'm fixing to figure out lol

  5. myininaya
    • 5 years ago
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    if i can figure out P i'm done

  6. myininaya
    • 5 years ago
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    so far i have y1=2x0/y0 where P=(-x0,y0) this is the way i'm suppose to go about it right?

  7. myininaya
    • 5 years ago
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    and where L representing the point of the bulb of the lamp post at (3,y1)

  8. myininaya
    • 5 years ago
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    so found the slope of the line containing P and the x intercept (-5,0) and got m=(y1-0)/(3+5)=y1/8 and i found the derivative of the ellipse which is y'=-x/4y i evaluated y' at (-x0,y0) and got y'=x0/4y0 so this has to be equal to y1/8 so we have y1=2x0/y0

  9. myininaya
    • 5 years ago
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    omg omg P=(-1,1) so y1=2

  10. myininaya
    • 5 years ago
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  11. myininaya
    • 5 years ago
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    the height is 2 :)

  12. myininaya
    • 5 years ago
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    ty so much for the problem :) i don't know if i went about the long way or not but i feel good about it

  13. watchmath
    • 5 years ago
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    Good job Miyaniya :) So you like it?

  14. myininaya
    • 5 years ago
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    yes im going to use it wow it took while for me to figure how to get P lol that was my problem for the longest time

  15. myininaya
    • 5 years ago
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    is this way that you were thinking about finding the height? or is there a shorter way?

  16. watchmath
    • 5 years ago
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    Here what I will do to find \(P\). Let \(P=(a,b)\). Using \((a,b)\) and \((-5,0)\) the slope is \(m=\frac{b}{a+5}\). Using implicit differentiation the slope is \(m=-\frac{a}{4b}\). Setting the two to be equal, we have \(4b^2=-a^2-5a\). But since \((a,b)\) is on the ellipse, \(4b^2=5-a^2\). So \(-a^2-5a=-a^2+5\) which implies \(a=-1\). It follows that \(4b^2=5-a^2=4\). So \(b=\pm 1\). But \(b\) is above the \(x\)-axis. So \(b=1\). Hence \(P=(-1,1)\). Let \(h\) be the height that we are looking for. Using similar triangle we have \(\frac{h}{5+3}=m=\frac{1}{4}\). Then \(h=2\).

  17. myininaya
    • 5 years ago
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    ok nice. you have less text than i have writing lol

  18. myininaya
    • 5 years ago
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    thank you so much watchmath. i'm totally going to give this to my calculus students and let them work on it as a group project

  19. watchmath
    • 5 years ago
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    My pleasure :)

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