# Implicite Differentiation

• Dec 15th 2009, 09:29 AM
thekrown
Implicite Differentiation
6. a) The equation of a curve is defined implicitly is x^3y^2 = -3xy. Verify that the point (-1,-3) belongs to the curve. Find an equation of the tangent line to the curve at this point.

My work:

Step 1. Verify point. I did and got -9 = -9.

Step 2. Differentiate.

3x^2y^2 + 2x^3yy' = -3x -3xy'

Step 3. Sub pt into above equation.

This is where I am stuck... you see I am able to differentiate it properly but then when it comes time to sub the points in, I get a totally different answer.

I got received tutoring about this question and was able to get it, but a week later here I am reviewing it and cannot. I keep getting dy/dx = 18/3 = 6. It's supposed to be -12 and I got that once, but now I keep getting 6 for some reason. What could I be doing wrong?

My work:
.
.
.
a) 3x^2y^2dx + 2x^3ydy = -3ydx + -3xdy

b) Put dx's and dy's together. I get (2x^3y+3x)dy = (-3y-3x^2y^2)dx

c) dy/dx = (-3y-3x^2y^2)dx / (2x^3y+3x)dy

d) 9-27 / 6-3 = -18/3 = -6.

e) Teacher solution is -12. How come I get -6?
• Dec 15th 2009, 10:35 AM
Jester
Quote:

Originally Posted by thekrown
6. a) The equation of a curve is defined implicitly is x^3y^2 = -3xy. Verify that the point (-1,-3) belongs to the curve. Find an equation of the tangent line to the curve at this point.

My work:

Step 1. Verify point. I did and got -9 = -9.

Step 2. Differentiate.

3x^2y^2 + 2x^3yy' = -3x -3xy' (1)

Step 3. Sub pt into above equation.

This is where I am stuck... you see I am able to differentiate it properly but then when it comes time to sub the points in, I get a totally different answer.

I got received tutoring about this question and was able to get it, but a week later here I am reviewing it and cannot. I keep getting dy/dx = 18/3 = 6. It's supposed to be -12 and I got that once, but now I keep getting 6 for some reason. What could I be doing wrong?

My work:
.
.
.
a) 3x^2y^2dx + 2x^3ydy = -3ydx + -3xdy

b) Put dx's and dy's together. I get (2x^3y+3x)dy = (-3y-3x^2y^2)dx

c) dy/dx = (-3y-3x^2y^2)dx / (2x^3y+3x)dy

d) 9-27 / 6-3 = -18/3 = -6.

e) Teacher solution is -12. How come I get -6?

Might want to check the term in red above. It's alos easier to sub in your x and y values into (1) above in red and then solve for y'.
• Dec 15th 2009, 10:46 AM
thekrown
I think the x in red was added to correct a typo on my part but I do not understand why there is a (1) added in red.
• Dec 15th 2009, 10:55 AM
Jester
Quote:

Originally Posted by thekrown
I think the x in red was added to correct a typo on my part but I do not understand why there is a (1) added in red.

It just to reference the equaton - that's all. As in eqn. (1).
• Dec 15th 2009, 10:56 AM
thekrown
Okay. Have you had the chance to verify my work below in c and d?
• Dec 15th 2009, 11:27 AM
Jester
Quote:

Originally Posted by thekrown
6. a) The equation of a curve is defined implicitly is x^3y^2 = -3xy. Verify that the point (-1,-3) belongs to the curve. Find an equation of the tangent line to the curve at this point.

My work:

Step 1. Verify point. I did and got -9 = -9.

Step 2. Differentiate.

3x^2y^2 + 2x^3yy' = -3x -3xy'

Step 3. Sub pt into above equation.

This is where I am stuck... you see I am able to differentiate it properly but then when it comes time to sub the points in, I get a totally different answer.

I got received tutoring about this question and was able to get it, but a week later here I am reviewing it and cannot. I keep getting dy/dx = 18/3 = 6. It's supposed to be -12 and I got that once, but now I keep getting 6 for some reason. What could I be doing wrong?

My work:
.
.
.
a) 3x^2y^2dx + 2x^3ydy = -3ydx + -3xdy

b) Put dx's and dy's together. I get (2x^3y+3x)dy = (-3y-3x^2y^2)dx

c) dy/dx = (-3y-3x^2y^2)dx / (2x^3y+3x)dy

d) 9-27 / 6-3 = -18/3 = -6.

e) Teacher solution is -12. How come I get -6?

Outside of the red dx and dy which shouldn't be there, it's correct.
• Dec 15th 2009, 11:40 AM
thekrown
Okay so I am correct up until the point where I substitute the values (-1,-3) because once I do, I get -6 as you can see. The answer however is -12.

It's possible I am mixing a sign somewhere because 9+27 is 36 and 36/3 is 12.

I on the other hand get 9-27 and so that is -18, -18/3 is -6.

From what I can see though, I'm applying everything properly, how could I be getting -9 and not +9 in order to get 36/3?
• Dec 15th 2009, 12:58 PM
Jester
Quote:

Originally Posted by thekrown
Okay so I am correct up until the point where I substitute the values (-1,-3) because once I do, I get -6 as you can see. The answer however is -12.

It's possible I am mixing a sign somewhere because 9+27 is 36 and 36/3 is 12.

I on the other hand get 9-27 and so that is -18, -18/3 is -6.

From what I can see though, I'm applying everything properly, how could I be getting -9 and not +9 in order to get 36/3?

From what I see, you are correct. I can only think that you've written the problem down wrong or the answer that you're comparing yours with (i.e. -12) is wrong.