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# Question 1

last edited by 16 years, 5 months 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

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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.

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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