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anonymous

  • 5 years ago

The hypotenuse of a right triangle is growing at a constant rate of a cm per second and one leg is decreasing at a constant rate of b cm per second. How fast is the acute angle between the hypotenuse and the other leg changing at the instant when both legs are 1 cm?

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  1. amistre64
    • 5 years ago
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    we know that dr/dt = a; dy/dt = b; and at the moment both legs are equal we have x=1; y=1; and r=sqrt(2). not all that is pertinent tho :)

  2. amistre64
    • 5 years ago
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    make that: dy/dt= -b the angle involved can be the tan(a) = y/x: tan(a) = y/x Dt(tan(a)) = Dt(y/x)

  3. amistre64
    • 5 years ago
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    da/dt (sec^2(a)) = [ x(dy/dt) -(dx/dt)y ] / x^2

  4. amistre64
    • 5 years ago
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    x(dy/dt) -(dx/dt)y da/dt = --------------- (xsec(a))^2

  5. amistre64
    • 5 years ago
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    or should we do that with sin(a)? sin(a) = y/r... lets try that out... r(dy/dt) - y(dr/dt) da/dt = ---------------- x^2 cos(a)

  6. amistre64
    • 5 years ago
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    x^2 should be r^2 .... my brain locked :)

  7. anonymous
    • 5 years ago
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    yes thanks i think it should be sin! since we are given the opposite and the hyp

  8. amistre64
    • 5 years ago
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    sqrt(2)(-b) - (1)(a) da/dt = --------------- sqrt(2)

  9. amistre64
    • 5 years ago
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    i got 2 "a"s in that that are supposed to be different variables.... :)

  10. amistre64
    • 5 years ago
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    d(theta)/dt = ......

  11. amistre64
    • 5 years ago
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    x^2 + y^2 = r^2 (dx/dt)2x + (dy/dt)2y = (dr/dt)2r ; divide everything by 2 (dx/dt)x + (dy/dy)y = (dr/dt)r (dr/dt)r - (dy/dt)y (dx/dt) = --------------- x

  12. amistre64
    • 5 years ago
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    dx/dt = (a)(sqrt(2)) - (-b)(1) ---------------- 1 Hows that look?

  13. amistre64
    • 5 years ago
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    if we had the rates for "a" and "b" we'd be sittin' pretty :)

  14. anonymous
    • 5 years ago
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    thanks! thats how i approached it too i just had some trouble with the simplification

  15. amistre64
    • 5 years ago
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    dy/dy is sposed ta be dy/dt.... fatfingers...tiny little keyboard :)

  16. amistre64
    • 5 years ago
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    I need to double check this... for x=1' y=1; r=sqrt(2) r(dy/dt) - y(dr/dt) da/dt = ---------------- x^2 cos(a) sqrt(2)(-b) - (a) ------------- (sqrt(2)/2) that should be the answer there....

  17. amistre64
    • 5 years ago
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    1^2 does not equal 2 :)

  18. amistre64
    • 5 years ago
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    2(sqrt(2))(-b)/(sqrt(2)) - 2a/sqrt(2) -2b - 2a/sqrt(2) -2b - 2a(sqrt(2))/2 -2b - a(sqrt(2)) = d(theta)/dt Check my math on that.... but that should be the simplified version of d(theta)

  19. amistre64
    • 5 years ago
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    i beg of you ...PLEASE check my math :) if I get any more "idiot" braincells marching around in myhead, im gona scream :)

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