I am new to calculus and cannot see the logic of the following question… Any feedback will be really appreciated!
The function $f(x,y,z)$ is differentiable at all points, and satisfies $f(x,y,2x^2+y^2)=4x+5y$.
Point P is defined as (1,2,6).
Unit vector $\mathbf{u}=(\frac{2}{3},\frac{1}{3},\frac{2}{3})$.
$D_\mathbf{u}f(P)=8.$
Find the gradient for f(x,y,z) at the given point. Give the sum of elements of the gradient you found.
Options:
- 4
- 4.5
- 5
- 5.5
I am lost at the very beginning: I understand that I need to calculate $f_x,f_y,f_z$ in order to calculate the gradient. $f_x=4$,$f_y=5$, but I am stuck at $f_z$. I have understood that z depends on x and y (but then I don't quite understand why it is included in the definition of the function $f(x,y,z)$, as $z$ is not independent of $x$ and $y$). It does not appear on the right side of $f(x,y,2x^2+y^2)=4x+5y$, so I cannot seem to find a way to find its derivative either.
Can anybody please help out?
Let's say $g(x,y) = 2x^2+y^2$. Then the problem says $f(x,y,g(x,y)) = 4x+5y$. Use the chain rule to differentiate both sides with respect to both $x$ and $y$ to get
$$ f_x + f_z g_x = f_x + 4xf_z = 4 $$
$$ f_y + f_z g_y = f_y + 2yf_z = 5 $$
Plugging in the point $P = (1,2,6)$, you get
$$ f_x + 4f_z = 4 $$
$$ f_y + 4f_z = 5 $$
The fact that $D_uf(P) = 8$ means that $\frac{1}{3} \left( 2f_x + f_y + 2f_z \right) = 8$. Or, multiplying by $3$, you get $2f_x + f_y + 2f_z = 24$. Now you have three equations in three unknowns. You can solve the linear system to get
$$f_x = 8.9 $$ $$f_y = 9.4 $$ $$f_z = -1.1 $$
The reason I think there may be a typo, and it might be $D_uf(P) = \frac{8}{3}$ is that in that case, you instead get $f_x = 2$, $f_y = 3$, $f_z = 0.5$.