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Question 1

Page history last edited by PBworks 17 years, 1 month ago

Question 1

 

You MAY use a calculator to solve this problem.

 

Consider the equation x2 - 2xy + 4y2 = 64

 

(a) Write an expression for the slope of the curve at the point (x, y).

 

(b) Find the equation of the tangent lines to the curve at the point x = 2.

 

(c) Find  at (0, 4).

 

Solution

 

 

a) differentiate implicitly with respect to x and remembering that y is a function of x

 

2x-[2(xy' - y(1))] +8yy'= 0

2x - 2xy' - 2y +8yy' = 0

 

2xy' +8yy'= 2y-2x

y'(2x-8y) = 2y-2x

 

dy= 2y-2x

dx      2x-8y

 

 

________________________________________________________________________________________

b) First find the y-coordinate of the equation x2 - 2xy + 4y2 = 64 at x=2

 

 22 - 2(2)y + 4y2 = 64 

4-4y+4y2 = 64

 

4(y2 +2y-20) = 0

 

4(y+6)(y-4)=0

 

y=-6, y=4 (original answer: y= 6, 4)

 

we have two points (2,-6) and (2,4)

 

y' (2,-6) = 2(-6)-2(2) 

                 2(2)-8(-6)

 

y' (2,6) = -4/13 <--- the slope at point (2,6)  (original answer: y(2,6) = -1/6)

 

y'(2,4) = 2(4)-2(2)

              2(2)-8(4)

 

y'(2,4) = -1/7 <--- slope of the line at (2,4)

 

I THINK I MIXED SOMETHING UP HERE.

 

 

 

________________________________________________________________________________________

c) differentiate the slope equation first.

 

y' = 2y-2x

       2x-8y

 

 

Think Lodihi-hidilo...

 

   =  (2x-8y)D(2y-2x) - (2y-2x)D(2x-8y)

                                 (2x-8y)2

 

          = (2x-8y)(2y'-2) - (2y-2x)(2-8y')

                              (2x-8y)2

 

          = 4xy'-4x-16yy'+16y - (4y-16yy'-4x+16xy')

                                    (2x-8y)2

 

 

         = 4xy'-4x-16yy'+16y +4y+16yy'+4x-16xy'

                                    (2x-8y)2

 

         = 12xy'+20y

              (2x-8y)2

 

Substitute y' into this equation.

 

 

 

 

        = 12x(2y-2x)+20y

 

                 (2x-8y)        

                 (2x-8y)2 

 

  Simplify by finding a common denominator

 

        = 12x(2y-2x)+20y (2x-8y)

                 (2x-8y)         (2x+8y)

                 (2x-8y)2 

                     1

 

 

        = 12x(2y-2x) +20y(2x-8y)

                       2x+8y_______

                      (2x-8y)2 

                          1

 

        =24xy-24x2 + 40xy-160y   *  (   1   )

                       2x-8y                      (2x-8y)2 

 

        = 16xy-24x2 -160y 

                (2x+8y)^3

 

 

    = 8(2xy- 3x2 - 20y)

                   (2x+8y)^3

 

Now substitute the point (0,4) into the equation

 

    = 8(2(0)(4)- 3(0)2 - 20(4))

                   (2(0)+8(4))^3

 

 

           =     -640

                32768

 

           =  -5_

              256

 

 

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