Questions on partial derivatives used to find tangent line

[RM5152]

Hello… I have been stuck on these 2 questions. I understand I have to find the partial derivatives to solve them but I can’t figure out what the next step is past finding the partial derivatives. Can anyone help please? I’ve posted the questions and my workings so far … any help would be appreciated ?

 

[BigBeachBanana]

Please post one problem per thread to avoid cross-discussion and confusion. I'll use this thread for question 4. Repost number 5 in another thread.

Hint: Try implicit differentiation to find the tangent of the curve.

 

[Steven G]

BBB,
I hope that you don't mind my commenting on #5 as the OP has not made a new thread for it.

You have an equation, 2x^3 - 3y^2 =7, you need to compute d/dx for BOTH sides

6x^2 - 6ydy/dx = 0

Now compute d/dx for both sides again and then solve for d^2y/dx^2

 

[JeffM]

Why do you think these problems involve partial differentiation?

What are you currently studying?

Have you heard of implicit differentiation?

 

[topsquark]

You can do this with the gradient operation but it kind of messes up the concepts for 2D. You wrote [imath]\nabla f = (-11, 7)[/imath] but do you know what that means? [imath]\nabla f(x,y) = \dfrac{ \partial f}{ \partial x} \hat{i} + \dfrac{ \partial f}{ \partial y} \hat{j}[/imath]. So your result can be written as [imath]-11 \hat{i} + 7 \hat{j}[/imath]. What does that make the slope?

-Dan

 

[RM5152]

Please post one problem per thread to avoid cross-discussion and confusion. I'll use this thread for question 4. Repost number 5 in another thread.

Hint: Try implicit differentiation to find the tangent of the

Great . Thanks everyone for all the help… I think I’ve solved it . Posting my solution here.

 

[RM5152]

BBB,
I hope that you don't mind my commenting on #5 as the OP has not made a new thread for it.

You have an equation, 2x^3 - 3y^2 =7, you need to compute d/dx for BOTH sides

6x^2 - 6ydy/dx = 0

Now compute d/dx for both sides again and then solve for d^2y/dx^2

Great. Thanks so much. I think I’ve gotten it with your help. I posted my solution .. is this correct?

 

[JeffM]

There is a small technical flaw. You had an equation.

[math]f(x) = g(x) \implies f’(x) = g’(x).[/math]
In this case g(x) = 0 so g’(x) = 0.

I know you somehow realized that

[math]f(x) = y^3 + xy^2 - x^3 + 5 = 0 = g(x) \implies (3y^2 + 2xy) * \dfrac{dy}{dx} + y^2 - 3x^2 = \text {ZERO}.[/math]
Your mechanics were fine. But the absence of an equation made me wonder if you understand why the mechanics make sense.

 

[Steven G]

In taking the 2nd derivative you failed to use the product rule.

 

[RM5152]

In taking the 2nd derivative you failed to use the product rule.
So would the second line be 12x = 6d^2y/dx^2 -6dy/dx (dy/dx)’ ??

 

[Steven G]

No, that is not correct. You really need to show your work if you want help. If we see your work then we can tell you where you are making your mistake. Use the product rule carefully and you'll get the answer.