1. ## Differentiation Problem

Hello, I ran across this problem and want to make sure my answers are right.

Consider the curve defined by
x^2 + xy + y^2 = 27.

a. Write an expression for the slope of the curve at any point (x,y).
b. Determine whether the lines tangent to the curve at the x-intercepts of the curve are parallel. Show the analysis that leadds to your conclusion.
c. Find the points on the curve where the lines tangent to the curve are vertical.

2. Originally Posted by kittykat616
Hello, I ran across this problem and want to make sure my answers are right.

Consider the curve defined by
x^2 + xy + y^2 = 27.

a. Write an expression for the slope of the curve at any point (x,y).
b. Determine whether the lines tangent to the curve at the x-intercepts of the curve are parallel. Show the analysis that leadds to your conclusion.
c. Find the points on the curve where the lines tangent to the curve are vertical.

3. I've currently working on B but for A i got 1/xy.

4. Originally Posted by kittykat616
I've currently working on B but for A i got 1/xy.
that is incorrect ... what do you know about finding an implicit derivative?

5. I just redid it and got -3x - y / 2y.

my work:
2x + [(x)(1)+(1)(y)] + 2y (dy/dx) = 0

then solved for dy/dx

6. Originally Posted by kittykat616
I just redid it and got -3x - y / 2y.

my work:
2x + [(x)(1)+(1)(y)] + 2y (dy/dx) = 0
..........no
then solved for dy/dx
$\frac{d}{dx}[x^2 + xy + y^2 = 27]$

$2x + x \cdot \frac{dy}{dx} + y \cdot 1 + 2y \cdot \frac{dy}{dx} = 0
$

now solve for $\frac{dy}{dx}$

7. Originally Posted by skeeter
$\frac{d}{dx}[x^2 + xy + y^2 = 27]$

$2x + x \cdot \frac{dy}{dx} + y \cdot 1 + 2y \cdot \frac{dy}{dx} = 0
$

now solve for $\frac{dy}{dx}$
thank you for catching my mistake. for the final answer i got -4x-y/2y+x

8. Originally Posted by kittykat616
thank you for catching my mistake. for the final answer i got -4x-y/2y+x
...............no again
...

9. Originally Posted by skeeter
...
x(dy/dx) + 2y(dy/dx) = -2x - y
divide both sides by x
(dy/dx) + 2y(dy/dx) = -2x - y / x
divide both sides by 2y
1(dy/dx) + 1(dy/dx) = -2x -y/ 2xy
2(dy/dx) = -2xy -y/2xy
divide by 2 or times by 2/1.

= -4x -2y/2xy.

10. $2x + x \cdot \frac{dy}{dx} + y \cdot 1 + 2y \cdot \frac{dy}{dx} = 0
$

$x \cdot \frac{dy}{dx} + 2y \cdot \frac{dy}{dx} = -2x - y$

$\frac{dy}{dx}(x + 2y) = -2x - y$

$\frac{dy}{dx} = \frac{-2x-y}{x+2y}$

you need some more practice with the basics

11. it always makes sense when i see it done correctly.. you're right i do need more practice. Thank you for your help. How would i go about part b?

12. Originally Posted by kittykat616
it always makes sense when i see it done correctly.. you're right i do need more practice. Thank you for your help. How would i go about part b?
find the x-intercepts of the original curve ... parallel tangent lines should tell you that the respective slopes have to be equal.

13. So i would set -2x - y = 0, right?

14. Originally Posted by kittykat616
So i would set -2x - y = 0, right?
no ... x-intercepts are the points on the curve $x^2 + xy + y^2 = 27$ where $y = 0$

15. Gosh thanks for all your help.

So it would be x^2= 27, or x = 5.19 and -5.19

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