Help with basic Calculus (Equation of a Tangent Line) Please?

[Spencero94]

This is my first post, and I'm excited to be able to receive quality help from what seems to be a good place. I'll attempt to start using the proper format and make it easy for everyone to read, thank you for the help! This should be a simple problem, but I seem to have forgotten how to do it.

Homework Statement

Write the equation of the line tangent to the curve 3x³+y³=10xy at the point (1,3)

Homework Equations

3x³+y³=10xy

The Attempt at a Solution

my attempt revolved around getting x and y on separate sides of the equation, then taking the derivative of each side. I ended up getting:

y'(x)=(10y-9x²)/(-10x+3y²)

For some reason I thought to plug set that equal to 0 and then plug in my x and y coordinates and then plug that value into x for my original equation, but I know something went completely wrong somewhere, so I'm wondering at what point I went wrong and was wondering if there were a few starting steps I could get to do a problem like this. Thanks!

 

[Dick]

This is my first post, and I'm excited to be able to receive quality help from what seems to be a good place. I'll attempt to start using the proper format and make it easy for everyone to read, thank you for the help! This should be a simple problem, but I seem to have forgotten how to do it.

Homework Statement

Write the equation of the line tangent to the curve 3x³+y³=10xy at the point (1,3)

Homework Equations

3x³+y³=10xy

The Attempt at a Solution

my attempt revolved around getting x and y on separate sides of the equation, then taking the derivative of each side. I ended up getting:

y'(x)=(10y-9x²)/(-10x+3y²)

For some reason I thought to plug set that equal to 0 and then plug in my x and y coordinates and then plug that value into x for my original equation, but I know something went completely wrong somewhere, so I'm wondering at what point I went wrong and was wondering if there were a few starting steps I could get to do a problem like this. Thanks!

The tangent to the curve at (1,3) is given by y'(1). That doesn't have to be zero. Find what it is by putting x=1 and y=3 into the derivative formula you so correctly derived.

 

[scurty]

This is my first post, and I'm excited to be able to receive quality help from what seems to be a good place. I'll attempt to start using the proper format and make it easy for everyone to read, thank you for the help! This should be a simple problem, but I seem to have forgotten how to do it.

Homework Statement

Write the equation of the line tangent to the curve 3x³+y³=10xy at the point (1,3)

Homework Equations

3x³+y³=10xy

The Attempt at a Solution

my attempt revolved around getting x and y on separate sides of the equation, then taking the derivative of each side. I ended up getting:

y'(x)=(10y-9x²)/(-10x+3y²)

For some reason I thought to plug set that equal to 0 and then plug in my x and y coordinates and then plug that value into x for my original equation, but I know something went completely wrong somewhere, so I'm wondering at what point I went wrong and was wondering if there were a few starting steps I could get to do a problem like this. Thanks!

Hi Spencer! Welcome to PF! This is a great resource, and not only for HW help.

Your equation for the derivative is correct. I'm not sure what you're setting equal to 0; at this point plug in your values for x and y to get the value for the derivative, or equivalently the slope. What would you do next, now that you have a point and a slope?

 

[Spencero94]

Thank you so much for your replies!

I believe that the number I get after plugging in would then be slope (m) and then I could easily plug that into y=mx+b and get my full answer. Such a simple problem that I over-thought!

 

[scurty]

Thank you so much for your replies!

I believe that the number I get after plugging in would then be slope (m) and then I could easily plug that into y=mx+b and get my full answer. Such a simple problem that I over-thought!

Almost. That should be the final form of the equation after simplifying. The equation you plug the numbers into is slightly different: $y - y_1 = m (x - x_1)$

 

[Spencero94]

Ohh, thank you. Is it also possible to find b first using y=mx+b and then to plug b back into the slope intercept form, or is that considered incorrect procedure?

 

[Dick]

Ohh, thank you. Is it also possible to find b first using y=mx+b and then to plug b back into the slope intercept form, or is that considered incorrect procedure?

That is a fine procedure.