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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:

  1. 4
  2. 4.5
  3. 5
  4. 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?

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  • $\begingroup$ I don't think you can say $f_x = 4$ and $f_y = 5$. It says $f$ is equal to $4x+5y$ only when $z = 2x^2+y^2$. $\endgroup$ – Nick Apr 21 at 16:18
  • $\begingroup$ Also, are you sure the problem says $D_uf(P) = 8$? Is it possible this is a typo, and it should be $\frac{8}{3}$? If it is indeed 8/3, I get the answer 5.5. If it is 8, I get the answer is not one of the 4 choices. $\endgroup$ – Nick Apr 21 at 16:30
  • $\begingroup$ It definitely says 8, but I will check if it's a typo…! $\endgroup$ – dalta Apr 21 at 17:12
  • $\begingroup$ @nick But in the meanwhile, could you tell me how you computed the gradient? $\endgroup$ – dalta Apr 21 at 17:13
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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 $f_x = 2$, $f_y = 3$, $f_z = 0.5$.

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