Can you please point out my mistake here? (partial derivatives and equations)

I

itsrayex

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Jan 18, 2022
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The equality should yeld two values but it turns out to have zero solutions to me. Can you look at what I did and tell me what I did wrong? thanks
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It would help if you would state what you are trying to find!

Here's what I got from your work:
[imath]f_x(x, y) = 2xy + 4x \implies f_x(1, 1) = 6[/imath]
(Note: [imath]f'_x(x, y)[/imath] has no meaning. Leave the "prime" off here.)

[imath]f_y(x, y) = x^2 - 2y \implies f_y(1, 1) = -1[/imath]

We have the vector [imath]\textbf{v} = ( t^2 - t + 1, 6t + 6)[/imath]

It looks like you are asking what values of t makes [imath]\textbf{v}[/imath] perpendicular to [imath]\textbf{\nabla}f(x, y)[/imath] at x = 1, y = 1. Note that 6 - 6 = 0...

-Dan
 
topsquark said:
It would help if you would state what you are trying to find!

Here's what I got from your work:
[imath]f_x(x, y) = 2xy + 4x \implies f_x(1, 1) = 6[/imath]
(Note: [imath]f'_x(x, y)[/imath] has no meaning. Leave the "prime" off here.)

[imath]f_y(x, y) = x^2 - 2y \implies f_y(1, 1) = -1[/imath]

We have the vector [imath]\textbf{v} = ( t^2 - t + 1, 6t + 6)[/imath]

It looks like you are asking what values of t makes [imath]\textbf{v}[/imath] perpendicular to [imath]\textbf{\nabla}f(x, y)[/imath] at x = 1, y = 1. Note that 6 - 6 = 0...

-Dan
Click to expand...
Thank you for pointing it out, I solved it! How do you close/archive threads btw?
 
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