We are tasked with determining whether there exist points $(a, b)$ such that the direction of maximum increase of the function $f(x, y) = xy^3$ is aligned with the vector $\langle 1, 1 \rangle$, and the slope in this direction is $\sqrt{18}$.

Step 1: Gradient of the Function

The direction of maximum increase of a function is given by its gradient. We start by computing the gradient of $f(x, y) = xy^3$.

The gradient $\nabla f(x, y)$ is the vector of partial derivatives:

$\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

• $\frac{\partial f}{\partial x} = y^3$
• $\frac{\partial f}{\partial y} = 3xy^2$

So, the gradient is: $\nabla f(x, y) = \langle y^3, 3xy^2 \rangle$

Step 2: Aligning the Gradient with $\langle 1, 1 \rangle$

To find the points where the direction of maximum increase aligns with $\langle 1, 1 \rangle$, we need the gradient to be proportional to $\langle 1, 1 \rangle$, meaning:

$\nabla f(a, b) = k \langle 1, 1 \rangle$

where $k$ is some constant. This gives two equations:

$y^3 = k \quad \text{and} \quad 3xy^2 = k$

From the first equation, we have $k = y^3$. Substituting this into the second equation:

$3xy^2 = y^3$

Dividing both sides by $y^2$ (assuming $y \neq 0$):

$3x = y$

Thus, we have the relation $y = 3x$.

Step 3: Slope of the Direction of Maximum Increase

Next, we are told that the slope in the direction of maximum increase is $\sqrt{18}$. The magnitude of the gradient gives the slope in the direction of maximum increase, so:

$|\nabla f(a, b)| = \sqrt{18}$

The magnitude of the gradient is:

$|\nabla f(x, y)| = \sqrt{(y^3)^2 + (3xy^2)^2}$

Substituting $y = 3x$ into this expression:

$|\nabla f(x, y)| = \sqrt{(27x^3)^2 + (9x^3)^2} = \sqrt{729x^6 + 81x^6} = \sqrt{810x^6} = \sqrt{810} |x^3|$

For the slope to be $\sqrt{18}$, we need:

$\sqrt{810} |x^3| = \sqrt{18}$

Squaring both sides:

$810 x^6 = 18$

$x^6 = \frac{18}{810} = \frac{1}{45}$

Taking the cube root of both sides:

$x^2 = \frac{1}{\sqrt[3]{45}}$

Finally, solving for $x$, we get two possible values for $x$. Substituting these into $y = 3x$, we will get corresponding $y$-values.

Step 4: Number of Solutions

Since $x^2 = \frac{1}{\sqrt[3]{45}}$ yields two distinct values for $x$, and for each $x$, there is a corresponding $y = 3x$, there are two distinct points where the conditions are satisfied.

Thus, the correct answer is:

$\boxed{d. \text{There exist two points } (a, b).}$

---

Would you like more details on the steps? Here are some related questions you might find interesting:

1. How does the gradient determine the direction of maximum increase?
2. What does it mean for two vectors to be proportional?
3. Why does the magnitude of the gradient give the slope of maximum increase?
4. How would the problem change if the slope in the direction of maximum increase were different?
5. Can the gradient be aligned with any arbitrary vector?

Tip: When solving optimization problems, always check the gradient and critical points carefully to ensure all conditions are satisfied.