Did I do these exercises correctly? (Orthogonal vectors and gradients)

[itsrayex]

The question is the same for both exercises: "for which values of the parameter t (belonging to R) is the gradient of the [function] calculated in a certain [point] orthogonal to a certain [vector]?" (sorry if translation is awkward but i am translating directly from another language. and sorry for my hideous handwriting, I hope you can understand).

Here are the exercises and my solving:



Is it correct? What does it mean that I get a zero at the end? does it mean the gradient is orthogonal to the vector in the value t=0, or does it mean that the vectors are never orthogonal?

Thank you for your help

 

[blamocur]

I've checked your first problem, and you have an error where you write [imath]f^\prime(3,-1) = 0[/imath] -- where did you get this from?

 

[itsrayex]

I've checked your first problem, and you have an error where you write [imath]f^\prime(3,-1) = 0[/imath] -- where did you get this from?

From the fact that the derivative of any number is zero

 

[blamocur]

View attachment 31056
From the fact that the derivative of any number is zero

This would make all derivatives of all the functions 0, would not it?

In one line you say [imath]f^\prime(x, -1) = -2x+7[/imath], but in the next line you claim that [imath]f(3,-1) = -5[/imath] -- do you see an error here?

 

[topsquark]

View attachment 31056
From the fact that the derivative of any number is zero

[imath]f(x, y) = x^2 y + x y^2[/imath]

What do you mean by f'(x, y)? Do you mean [imath]\nabla f(x, y)[/imath]? Otherwise we need to know how x and y are changing over some other variable t.

Either way, f'(3, -7) = f'(x, y) for x = 3 and y = -7. So you have to find f'(x, y) then plug in 3 and -7.

-Dan

 

[blamocur]

[imath]f(x, y) = x^2 y + x y^2[/imath]

What do you mean by f'(x, y)? Do you mean [imath]\nabla f(x, y)[/imath]? Otherwise we need to know how x and y are changing over some other variable t.

Either way, f'(3, -7) = f'(x, y) for x = 3 and y = -7. So you have to find f'(x, y) then plug in 3 and -7.

-Dan

I believe it is "-1" , not "-7" in the "hideous handwriting".

I agree: [imath]f^\prime[/imath] would not be my choice for [imath]\frac{\partial f}{\partial x}[/imath] or [imath]\frac{\partial f}{\partial y}[/imath], but I don't know how @itsrayex is taught about partial derivatives. Maybe they define partial derivatives by fixing the value of one of the variables and defining the parital as a regular derivative when there is only one variable left?

 

[itsrayex]

I believe it is "-1" , not "-7" in the "hideous handwriting".

I agree: [imath]f^\prime[/imath] would not be my choice for [imath]\frac{\partial f}{\partial x}[/imath] or [imath]\frac{\partial f}{\partial y}[/imath], but I don't know how @itsrayex is taught about partial derivatives. Maybe they define partial derivatives by fixing the value of one of the variables and defining the parital as a regular derivative when there is only one variable left?

I have gathered wrong information about the way to do this and my teacher's notes were confusing, but yesterday I learnt how to do it the proper way after some more research. As you were suggesting, I did partial derivatives in the wrong way (the way I went about this exercise is kinda wrong overall). Anyway, I solved the exercise and the result was correct. Do you know by any chance how one can close a thread?