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October 20, 2014

October 20, 2014

Posted by **Fareha** on Saturday, December 3, 2011 at 10:22pm.

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

b. Write an equation for the line tangent to the curve at the point (2, 1)

c. Find the coordinates of all other points on this curve with slope equal to the slope at (2, 1)

- calculus -
**Fareha**, Saturday, December 3, 2011 at 10:23pmsorry here's the equation:

x+xy+2y^2=6

- calculus -
**MathMate**, Sunday, December 4, 2011 at 8:30amx+xy+2y^2=6

(a)

x(1+y)=6-2y²

x=(6-2y²)/(1+y)

differentiate with respect to y to get

dx/dy=-4x/(x+1)-(6-2x^2)/(x+1)^2

So

dy/dx

= 1/(dx/dy)

= -(1/2)(x^2+2x+1)/(x^2+2x+3)

(b) For the point (2,1), first check that it lies on the curve.

f'(2)=-(1/2)(2^2+2*2+1)/(2^2+2*2+3)

=-9/22

(c)

Now solve for

f'(x)=-9/22

to get x=-4 or x=2

Verify all the numerical values.

- calculus - correction -
**MathMate**, Sunday, December 4, 2011 at 8:20pmThere was a mistake in the calculation of dx/dy. It should have been expressed in terms of y and not x.

Given : x+xy+2y^2=6

(a)

x(1+y)=6-2y²

x=(6-2y²)/(1+y)

differentiate with respect to y to get

dx/dy=-4y/(y+1)-(6-2y^2)/(y+1)^2

So

f'(y)=dy/dx

= 1/(dx/dy)

= -(1/2)(y^2+2y+1)/(y^2+2y+3)

(b) For the point (2,1), first check that it lies on the curve.

f'(y)=

f'(1)=-(1/2)(1^2+2*1+1)/(1^2+2*1+3)

=-1/3

Tangent line with slope -1/3 passing through (2,1) is:

(y-1) = (-1/3)(x-2)

(c)

Now solve for

f'(y)=-1/3

to get x=-3 or x=1

Please verify all the numerical values.

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