Problem 30 - Calculus with Concepts in Calculus

Question

Find a vector that is normal to the graph of \(f\) at the given point. $$ f(x, y)=y^{2} e^{x} ;(0,-3,9) $$

Step-by-Step Solution

Answer

The normal vector is \((9, -6)\).

Step 1: Understand the concept of normal vector

A normal vector to the graph of a function at a given point is a vector that is perpendicular to the tangent plane of the surface at that point. In other words, this vector can be found by calculating the gradient of the function.

Step 2: Find the gradient of the function

The gradient of a function, denoted as \(abla f\), is a vector comprised of its partial derivatives with respect to each variable. For the function \(f(x, y) = y^2 e^x\), the gradient is given by:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]

Step 3: Calculate the partial derivative with respect to \(x\)

The partial derivative of \(f(x, y)\) with respect to \(x\) is found by differentiating \(f(x, y) = y^2 e^x\) while treating \(y\) as a constant:\[ \frac{\partial f}{\partial x} = y^2 e^x \]

Step 4: Calculate the partial derivative with respect to \(y\)

The partial derivative of \(f(x, y)\) with respect to \(y\) is found by differentiating \(f(x, y) = y^2 e^x\) while treating \(x\) as a constant:\[ \frac{\partial f}{\partial y} = 2y e^x \]

Step 5: Evaluate the gradient at the given point

The given point is \((x, y) = (0, -3)\). Substitute these values into the gradient:\[ abla f(0, -3) = (y^2 e^x, 2y e^x) \] \[ abla f(0, -3) = ((-3)^2 e^0, 2(-3) e^0) \] \[ abla f(0, -3) = (9, -6) \]

Step 6: Interpret the result

The vector \((9, -6)\) represents the normal vector to the surface \(f(x, y) = y^2 e^x\) at the point \((0, -3)\). This vector is perpendicular to the tangent plane at the given point.

Key Concepts

Gradient of a function

The gradient of a function is a key concept in multivariable calculus. It is like a compass that points in the direction of the steepest increase of a function. The gradient, denoted by \( abla f \), is a vector that consists of all the first partial derivatives of a function. This means that for a function \( f(x, y) \), the gradient is \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). When you compute the gradient, you are actually finding a vector field that shows how the function changes at every point.
In the context of the given problem, the function under consideration is \( f(x, y) = y^2 e^x \). The gradient tells us how \( f \) changes as you move in the \( x \) and \( y \) directions. For this function, the gradient \( abla f \) at any point can be understood as telling how steep the slope is in the \( x \)-direction and \( y \)-direction.
The gradient not only helps in calculating the normal vector but also in finding the tangent plane to a surface, since the gradient is perpendicular to the tangent plane.

Partial derivatives

Partial derivatives are the building blocks of the gradient of a function. They show how a function changes as one variable changes, while the other variables are held constant. For a function \( f(x, y) \), the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), shows the rate of change of the function when \( x \) changes while \( y \) is kept constant.
In the exercise, we have \( f(x, y) = y^2 e^x \). The partial derivative \( \frac{\partial f}{\partial x} = y^2 e^x \), meaning it depends on both \( y^2 \) and \( e^x \). This expression shows that the effect of \( x \)-change on \( f \) is influenced by the value of \( y \).
The partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} = 2ye^x \), tells us how \( f \) changes as \( y \) changes, keeping \( x \) constant. It consists of two terms: \( 2y \) which accounts for the linear change with respect to \( y \), and \( e^x \), which remains constant if \( x \) does not change.

• Understanding partial derivatives is crucial for calculating the gradient and, in turn, important for finding normal vectors and tangent planes.
• They provide specific information about the behavior of multivariable functions at a point.

Tangent plane

In calculus, the tangent plane to a surface at a given point provides the best linear approximation of the surface near that point. If you imagine the surface \( z = f(x, y) \), the tangent plane at a particular point is simply a flat plane that just "touches" the surface at that point without cutting through it. Think of it like trying to lay a flat sheet of paper on a curved surface so it fits perfectly without folding.
The equation of a tangent plane can be derived using the gradient of the function. At a point \((x_0, y_0, z_0)\), the equation is given by \( z - z_0 = abla f(x_0, y_0) \cdot ((x - x_0), (y - y_0)) \). Here, \( abla f \) is the gradient of the function at that particular point, acting like the plane's normal vector.
For the function \( f(x, y) = y^2 e^x \), and at the point \((0,-3)\), this plane explains how the surface can be linearized around the point \(-3\) on the \( y \)-axis. The calculated normal vector \((9, -6)\) is perpendicular to this tangent, thus validating the accuracy of the surface's local approximation.

• Understanding the tangent plane helps in visualizing how a small area of a surface behaves like a plane.
• The concept is instrumental in fields requiring optimization and approximation methods.