To solve this problem, we need to do the following steps:

Step 1: Equation of the level curve

The level curve of a function $f(x, y)$ is a curve where the function has a constant value. For a given point $P(x_0, y_0)$, the level curve is the set of points $(x, y)$ where $f(x, y) = f(x_0, y_0)$.

Given the function $f(x, y) = 4x - 2y + 3$ and the point $P(1, 2)$, we can substitute $x = 1$ and $y = 2$ into the function to find the constant value of the level curve at point $P$:

$f(1, 2) = 4(1) - 2(2) + 3 = 4 - 4 + 3 = 3$

Thus, the level curve passing through $P(1, 2)$ corresponds to the equation: $f(x, y) = 3$

Substituting the function into this equation: $4x - 2y + 3 = 3$ Simplifying: $4x - 2y = 0 \quad \Rightarrow \quad 2x = y$ So, the level curve is the line: $y = 2x$

Step 2: Gradient Vector at $P$

The gradient vector $\nabla f(x, y)$ points in the direction of the greatest rate of change of the function. The gradient is calculated by taking the partial derivatives of $f(x, y)$ with respect to $x$ and $y$.

Given $f(x, y) = 4x - 2y + 3$, the partial derivatives are:

$\frac{\partial f}{\partial x} = 4$ $\frac{\partial f}{\partial y} = -2$

Thus, the gradient vector is: $\nabla f(x, y) = (4, -2)$

At point $P(1, 2)$, the gradient vector is: $\nabla f(1, 2) = (4, -2)$

Step 3: Sketching the Level Curve and Gradient

• The level curve is the line $y = 2x$, which passes through the point $P(1, 2)$.
• The gradient vector $\nabla f(1, 2) = (4, -2)$ is a vector that points in the direction of the greatest increase in $f(x, y)$ at point $P$, and it is perpendicular to the level curve.

Would you like a detailed visualization or more clarification on any specific part?

Related Questions:

1. How would the level curve change if $f(x, y)$ had a different constant term?
2. What happens to the gradient vector if the function changes to $f(x, y) = 2x^2 - y^2 + 5$?
3. How can we use the gradient to find the direction of steepest descent?
4. What is the significance of the gradient being perpendicular to the level curve?
5. How do we sketch level curves for multivariable functions that don't represent straight lines?

Tip: Always remember that the gradient vector at any point on a level curve will be perpendicular to the curve itself.