Problem 165 - Calculus Volume 3

Question

As a useful review for techniques used in this section, find a normal vector and a tangent vector at point \(P\). \(x^{2}+x y+y^{2}=3, P(-1,-1)\)

Step-by-Step Solution

Answer

Normal vector: \((-3, -3)\); Tangent vector: \((3, -3)\) or \((-3, 3)\).

Step 1: Understanding the Problem

We need to find the normal and tangent vectors to the curve defined by the equation \(x^{2}+xy+y^{2}=3\) at the point \(P(-1,-1)\). A normal vector is perpendicular to the tangent vector at the given point on the curve.

Step 2: Finding the Gradient (Normal Vector)

The gradient vector of a function \(f(x, y)\) is normal to the level curves. Compute the gradient of \(f(x, y) = x^2 + xy + y^2 - 3\). The partial derivatives are:\[f_x = \frac{\partial}{\partial x}(x^{2}+xy+y^{2}) = 2x + y\] \[f_y = \frac{\partial}{\partial y}(x^{2}+xy+y^{2}) = x + 2y\] Thus, the gradient vector \(abla f = (2x + y, x + 2y)\). At point \((-1, -1)\), \(abla f = (2(-1) + (-1), (-1) + 2(-1)) = (-3, -3)\). Thus, a normal vector at \(P\) is \((-3, -3)\).

Step 3: Computation of the Tangent Vector

To find a tangent vector at \(P\), we can take a vector that is orthogonal to the normal vector \((-3, -3)\). For simplicity, we can pick \((3, -3)\), since \((-3)(3) + (-3)(-3) = 0\). Alternatively, \((-3, 3)\) is also a correct choice.

Key Concepts

Gradient Vector

The gradient vector is an essential concept in understanding functions of several variables. It provides a vector that points in the direction of the steepest increase of a function. For a function defined as \(f(x, y)\), the gradient vector \(abla f\) is composed of the partial derivatives with respect to each variable, which are fundamental tools that measure how the function changes as each variable slightly changes.

• The partial derivative of \(x\), denoted as \(f_x\), shows how the function \(f\) changes as \(x\) changes, keeping other variables constant. In the problem, \(f_x = 2x + y\).
• The partial derivative of \(y\), \(f_y\), indicates how \(f\) changes with respect to \(y\), while keeping \(x\) constant. Here, \(f_y = x + 2y\).

Combining these gives the gradient vector \(abla f = (f_x, f_y) = (2x + y, x + 2y)\). At any given point, such as \((-1, -1)\), substituting the coordinates into the gradient equations provides a vector \((-3, -3)\). This vector is often used to find the normal vector to a surface, as it points perpendicular to the level curve of the function.

Tangent Vector

Understanding the tangent vector at a given point provides insight into the direction along the curve at that point. A tangent vector is parallel to the curve and is orthogonal to the normal vector. In simpler terms, if you imagine standing on a hill, the tangent vector would point along the steepest path you could walk without changing elevation.In the given exercise, the normal vector at point \((-1, -1)\) was calculated as \((-3, -3)\). To find a tangent vector, we need a vector that is perpendicular to this normal vector. If two vectors have a dot product of zero, they are orthogonal, meaning they meet at a right angle.

• If the normal vector is \((-3, -3)\), a simple tangent vector can be \((3, -3)\). This is because the dot product \((-3)(3) + (-3)(-3) = 0\), confirming their orthogonality. Alternatively, \((-3, 3)\) also works for the same reason.

These tangent vectors lay flat along the curve, illustrating the possible directions one could move while remaining on the curve without ascending or descending.

Partial Derivatives

Partial derivatives are a crucial tool in multivariable calculus. They help us to understand how a function of multiple variables changes concerning only one of those variables while keeping others constant. It's like focusing on one direction at a time to see how a landscape changes.For a function \(f(x, y) = x^2 + xy + y^2\), we can dissect it into its individual effects:

• The partial derivative with respect to \(x\) is found by differentiating \(f(x, y)\) with \(y\) held constant. This gives \(f_x = 2x + y\).
• Similarly, the partial derivative with respect to \(y\) treats \(x\) as constant, resulting in \(f_y = x + 2y\).

These derivatives help us form the gradient vector, which can be evaluated at specific points to obtain the directional rate of change of the function.The versatility of partial derivatives allows them to be used in calculating things like the gradient and understanding the orientation of curves, as well as solving problems involving rates of change in applied mathematics.