We are given that the curve is $y = x^3$, and the tangent touches the curve at the point $(1, 1)$ with a gradient (slope) of 3.

Steps to solve:

1. Find the derivative of the curve equation: The slope of the tangent to the curve is given by the derivative $\frac{dy}{dx}$. $y = x^3 \implies \frac{dy}{dx} = 3x^2$

2. Evaluate the slope at the point $(1, 1)$: We substitute $x = 1$ into $\frac{dy}{dx} = 3x^2$ to find the slope of the curve at that point. $\frac{dy}{dx} \Bigg|_{x=1} = 3(1)^2 = 3$ So the slope of the tangent at $(1, 1)$ is indeed 3, which matches the given gradient of the tangent.

3. Use the point-slope form of the tangent line equation: The point-slope form of the equation of a line is given by: $y - y_1 = m(x - x_1)$ where $m$ is the slope and $(x_1, y_1)$ is the point of tangency. Here, $m = 3$ and $(x_1, y_1) = (1, 1)$. Substituting these values into the point-slope form: $y - 1 = 3(x - 1)$

4. Simplify the equation: $y - 1 = 3x - 3$ $y = 3x - 2$

Thus, the equation of the tangent is: $y = 3x - 2$

Would you like further details or have any questions?

Here are five related questions:

1. How do you compute the derivative of more complex functions?
2. What happens if the point of tangency changes, say to $(2, 8)$?
3. Can you find the normal to the curve at the point $(1, 1)$?
4. How would you compute the tangent to an implicit curve like $x^2 + y^2 = 1$?
5. What does it mean geometrically if the derivative at a point is zero?

Tip: The point-slope form is very useful for finding the equation of a line when you know the slope and a specific point on the line.