Another problem centered around implicit differentiation

Rumor

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Sep 8, 2009
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"Consider the equation x+xy+2y^2=6.

a) Find dy/dx in terms of x and y.

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

c) Find the coordinates of all other points on this curve where the tangent line is parallel to the tangent line at (2,1)."

I've already worked through some of this problem, but I'd like for my work to be checked as well.

a) So for dy/dx, I got this: 1+y+x(dy/dx)+4y(dy/dx)=0, so dy/dx(x+4y)=-1-y and therefore dy/dx=(-1-y)/(x+4y). Is this correct?

b) For b, I plugged the points back into the original equation and confirmed that they are on the curve. So for the equation, I evaluated the derivative at the points (2,1) and got y'=-1/3 and so the tangent line would be y=-1/3(x-2)+1. Of course, if I'm wrong in my differentiation, then I'm wrong in this as well.

And I don't know what to do for c. Any help would be appreciated!
 
Rumor said:
"Consider the equation x+xy+2y^2=6.

a) Find dy/dx in terms of x and y.

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

c) Find the coordinates of all other points on this curve where the tangent line is parallel to the tangent line at (2,1)."

I've already worked through some of this problem, but I'd like for my work to be checked as well.

a) So for dy/dx, I got this: 1+y+x(dy/dx)+4y(dy/dx)=0, so dy/dx(x+4y)=-1-y and therefore dy/dx=(-1-y)/(x+4y). Is this correct? <<< This is correct

b) For b, I plugged the points back into the original equation and confirmed that they are on the curve. So for the equation, I evaluated the derivative at the points (2,1) and got y'=-1/3 and so the tangent line would be y=-1/3(x-2)+1. Of course, if I'm wrong in my differentiation, then I'm wrong in this as well.

And I don't know what to do for c. Any help would be appreciated!

dy/dx=(-1-y)/(x+4y) = -1/3

x + 4y = 3+3y

x = 3 - y................................................(1)

Apply (1) into your original equation (Why?)

x+xy+2y^2=6

(3-y) + (3-y)y + 2y^2 - 6 = 0

y^2 + 2y - 3 = 0

(y+3)(y-1) = 0

Then the points (6, -3) & (2,1) will have tangents with slope (-1/3).
 
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