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Math Help - How is this done? 2 questions - Curves + Derivatives and Related Rates

  1. #1
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    How is this done? 2 questions - Curves + Derivatives and Related Rates

    I'm doing some Calculus homework, but these two problems are a bit confusing. Thanks for taking the time to look:

    Consider the curve defined by 2y^3 + 6x^(2)y - 12x^2 + 6y = 1

    1. Find dy/dx (Found using implicit differentiation; (4x-2xy)/(x^2+y^2+1))
    2. Write an equation of each horizontal tangent line to the curve (Don't understand)
    3. The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates at point P. (Also don't understand)
    --
    Here is the second question:
    Ship A is traveling due west toward Blunt Rock at a speed of 15 km/hr. Ship B is traveling due north away from Blunt Rock at a speed of 10 km/hr. Let x be the distance between Ship A and Blunt Rock at time t, and let y be the distance between Ship B and Blunt Rock at time t.

    1. Find the distance, in km, between Ship A and Ship B when x=4 km and y=3 km (seems simple, use pythagorean theorem to get hypotenuse = 5 km)
    2. Find the rate of change, in km/hr, of the distance between the two ships when x=4km and y=3km. (Think I got this, using implicit on pytahgorean theorem I get -6km/hr)
    3. Let theta be the angle [at ship A, between the x axis and hypotenuse]. Find the rate of change of theta, in radians per hour, when x=4km and y=3km. (Not sure how this is done)

    Thank you in advance
    Last edited by mr fantastic; September 23rd 2010 at 03:49 PM.
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  2. #2
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    Quote Originally Posted by Harryhit4 View Post
    Sorry if my title sucked! I'm doing some Calculus homework, but these two problems are a bit confusing. Thanks for taking the time to look:

    Consider the curve defined by 2y^3 + 6x^(2)y - 12x^2 + 6y = 1

    1. Find dy/dx (Found using implicit differentiation); (4x-2xy)/(x^2+y^2+1))
    2. Write an equation of each horizontal tangent line to the curve (Don't understand)
    3. The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates at point P. (Also don't understand)
    --

    Thank you in advance
    2. The equation of a horizontal line is of the form y=k.

    If it is a tangent to the curve, the derivative must be zero,
    since the derivative formulates the slope of the tangent,
    hence for what values of x or y is the numerator of the derivative zero?


    3. This is the line y=-x.

    The slope of a line is it's derivative.

    Hence the derivative is -1, and you can substitute y=-x into the derivative to solve for x, and thus y.
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  3. #3
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    Quote Originally Posted by Harryhit4 View Post
    Here is the second question:
    Ship A is traveling due west toward Blunt Rock at a speed of 15 km/hr. Ship B is traveling due north away from Blunt Rock at a speed of 10 km/hr. Let x be the distance between Ship A and Blunt Rock at time t, and let y be the distance between Ship B and Blunt Rock at time t.

    1. Find the distance, in km, between Ship A and Ship B when x=4 km and y=3 km (seems simple, use pythagorean theorem to get hypotenuse = 5 km)
    2. Find the rate of change, in km/hr, of the distance between the two ships when x=4km and y=3km. (Think I got this, using implicit on pytahgorean theorem I get -6km/hr)
    3. Let theta be the angle [at ship A, between the x axis and hypotenuse]. Find the rate of change of theta, in radians per hour, when x=4km and y=3km. (Not sure how this is done)

    Thank you in advance
    2. On your drawing, ship A is moving to left, towards the 90 degree corner at 15km/hr.

    Ship B is moving up, away from the corner at 10km/hr.

    \displaystyle\frac{dx}{dt}=-15km/hr

    \displaystyle\frac{dy}{dt}=10km/hr

    \frac{dx}{dt} is negative as it is moving towards the triangle's right-angle,
    instead of moving away from it.

    hence, we ought to expect a negative value for the rate of change of the hypotenuse,
    since ship A is going a faster rate of knots than ship B.

    L^2=x^2+y^2

    \displaystyle\frac{d}{dt}L^2=\frac{d}{dt}x^2+\frac  {d}{dt}y^2

    2L\frac{dL}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}

    You have the values for L, x, y, dx/dt and dy/dt, hence dL/dt can be solved for.
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  4. #4
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    Hello, Harryhit4!

    \text{Ship A is traveling due west toward Blunt Rock at a speed of 15 kph.}
    \text{Ship B is traveling due north away from Blunt Rock at a speed of 10 kph.}
    \text{Let }x\text{ be the distance between Ship A and Blunt Rock at time }t.
    \text{Let }y\text{ be the distance between Ship B and Blunt Rock at time }t.

    \text{1. Find the distance, in km, between Ship A and Ship B}
    . . . \text{when }x=4\text{ and }y=3.
    \text{(seems simple, use Pythagorus to get hypotenuse = 5 km)} .Good!

    \text{2. Find the rate of change, in kph, of the distance between the ships}
    . . . \text{when }x=4 \text{ and }y=3.
    \text{(Think I got this, using implicit on Pythagorus I get: -6 kph)} .Yes!

    \text{3. Let }\theta\text{ be the angle at ship A, between the }x\text{-axis and hypotenuse.}
    \text{Find the rate of change of }\theta\text{, in radians/hr, when }x =4\text{ and }y=3.
    Code:
        B *
        : |  *
        : |     *
        y |        *
        : |           *
        : |            @ *
        R *-----------------*
          : - - -  x  - - - A

    Blunt Rock is at \,R.

    \text{Ship A is at }\,A\!:\;x = AR,\;\dfrac{dx}{dt} = \text{-}6\text{ kph}

    \text{Ship B is at }\,B\!:\;y = BR,\;\dfrac{dy}{dt} = 10\text{ kph}

    \,\theta \,=\,\angle A

    Note: .when x = 4,\;y = 3\!:\;\cos\theta \:=\:\frac{4}{5}



    \text{We have: }\;\tan\theta \:=\:\dfrac{y}{x}

    \text{Then: }\;\sec^2\!\theta\,\dfrac{d\theta}{dt} \;=\;\dfrac{x\frac{dy}{dt} - y \frac{dx}{dt}}{x^2}

    . . . . . . . . . . \dfrac{d\theta}{dt} \;=\;\dfrac{x\frac{dy}{dt} - y\frac{x}{dt}}{x^2}\cos^2\!\theta

    . . . . . . . . . . \dfrac{d\theta}{dt} \;=\;\dfrac{4(10) - 3(\text{-}15)}{4^2}\cdot \left(\frac{4}{5}\right)^2

    . . . . . . . . . . \displaystyle \frac{d\theta}{dt} \;=\;\frac{85}{16}\cdot\frac{16}{25} \;=\;\frac{17}{5} \;=\;3\tfrac{2}{5}\text{ radians/hour}
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