# Prob. 4, Exercises 8.14, in Apostol's CALCULUS Vol II: Computing the gradient vector and a directional derivative for a scalar field

Here is Prob. 4, Exercises 8.14, in the book Calculus Vol II by Tom M. Apostol, 2nd edition:

A differentiable scalar field $f$ has, at the point $(1, 2)$, directional derivative $+2$ in the direction toward $(2, 2)$ and $-2$ in the direction toward $(1, 1)$. Determine the gradient vector at $(1, 2)$ and compute the directional derivative in the direction toward $(4, 6)$.

My Attempt:

Let $(a, b)$ be the gradient vector $\nabla f(1, 2)$ of the scalar field $f$ at the point $(1, 2)$.

The vector $\mathbf{u}$ from $(1, 2)$ to $(2, 2)$ is given by $$\mathbf{u} = (2, 2) - (1, 2) = (1, 0).$$ Now as $f$ is differentiable at $(1, 2)$, so the directional derivative of $f$ in the direction of $\mathbf{u}$ is $$\nabla f(1, 2) \cdot \mathbf{u} = (a, b) \cdot (1, 0) = a.$$ Therefore we obtain $a = 2$.

The vector $\mathbf{v}$ from $(1, 2)$ to $(1, 1)$ is given by $$\mathbf{v} = (1, 1) - (1, 2) = (0, -1).$$ Once again as $f$ is differentiable at $(1, 2)$, so the directional derivative of $f$ in the direction of $\mathbf{v}$ is $$\nabla f(1, 2) \cdot \mathbf{v} = (a, b) \cdot (0, -1) = -b.$$ Therefore we obtain $b = 2$.

Thus the gradient vector of $f$ at $(1, 2)$ is given by $$\nabla f(1, 2) = (2, 2).$$

Now the vector $\mathbf{w}$ from $(1, 2)$ toward $(4, 6)$ is given by $$\mathbf{w} = (4, 6)- (1, 2) = (3, 4).$$ So the directional derivative of $f$ at $(1, 2)$ in the direction toward $\mathbf{w}$ is $$\nabla f(1, 2) \cdot \mathbf{w} = (2, 2) \cdot (3, 4) = 14.$$

Is there any error --- either in logic or calculation --- in this solution?

$$\nabla f(1, 2) \cdot \dfrac{\mathbf{w}}{\left|\mathbf{w}\right|} = (2, 2) \cdot \dfrac{1}{5}(3, 4) = \dfrac{14}{5}$$