Problem 8 - Advanced Engineering Mathematics

Question

Sketch the level curve or surface passing through the indicated point. Sketch the gradient at the point. $$ f(x, y)=\frac{y-1}{\sin x} ;\left(\pi / 6, \frac{3}{2}\right) $$

Step-by-Step Solution

Answer

The level curve is \(y = \sin x + 1\), passing through \(\left(\frac{\pi}{6}, \frac{3}{2}\right)\). The gradient vector at this point is \((-\sqrt{3}, 2)\).

Step 1: Understand the Problem

We want to find and sketch the level curve for the function \(f(x, y)\). A level curve for a function of two variables, \(f(x, y)\), with a specific value \(k\) is the set of points that satisfy the equation \(f(x, y) = k\). Here we need to find the level curve that passes through the given point \(\left(\frac{\pi}{6}, \frac{3}{2}\right)\).

Step 2: Calculate the Level Value

We substitute \(x = \frac{\pi}{6}\) and \(y = \frac{3}{2}\) into the function \(f(x, y)\) to determine the value \(k\): \[ k = \frac{\frac{3}{2} - 1}{\sin\left(\frac{\pi}{6}\right)} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \]. Thus, the level curve will be for \(f(x, y) = 1\).

Step 3: Find the Equation of the Level Curve

The level curve equation is \(f(x, y) = 1\), which is \(\frac{y-1}{\sin x} = 1\). Solving for \(y\) gives \(y - 1 = \sin x\) or \(y = \sin x + 1\). This is the equation of the level curve.

Step 4: Sketch the Level Curve

Plot the function \(y = \sin x + 1\) across a suitable range of \(x\). This curve is simply the standard sine wave shifted up by 1 unit. Include the given point \(\left(\frac{\pi}{6}, \frac{3}{2}\right)\) on the sketch.

Step 5: Calculate the Gradient

The gradient vector \(abla f\) is \(\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). Calculating these, we get \(\frac{\partial f}{\partial x} = -\frac{(y-1)\cos x}{\sin^2 x}\) and \(\frac{\partial f}{\partial y} = \frac{1}{\sin x}\). Substitute \(x = \frac{\pi}{6}\) and \(y = \frac{3}{2}\): \(\frac{\partial f}{\partial x} = -\frac{\frac{1}{2} \cdot \frac{\sqrt{3}}{2}}{\left(\frac{1}{2}\right)^2} = -\sqrt{3}\) and \(\frac{\partial f}{\partial y} = 2\). Thus, \(abla f\left(\frac{\pi}{6}, \frac{3}{2}\right) = (-\sqrt{3}, 2)\).

Step 6: Sketch the Gradient Vector

Draw the gradient vector \((-\sqrt{3}, 2)\) starting from the point \(\left(\frac{\pi}{6}, \frac{3}{2}\right)\) on the level curve sketch. The gradient vector indicates the direction of the steepest ascent from the point and is perpendicular to the level curve at that point.

Key Concepts

Gradient Vector

The gradient vector is fundamental in multivariable calculus, representing the direction and rate of the fastest increase of a function. For a function of two variables, say \(f(x, y)\), the gradient is denoted as \(abla f\) and is a vector composed of partial derivatives:

• \(\frac{\partial f}{\partial x}\) - represents how \(f\) changes with respect to \(x\) while keeping \(y\) constant.
• \(\frac{\partial f}{\partial y}\) - shows how \(f\) changes with respect to \(y\) while \(x\) is constant.

These combined derivatives form the gradient vector \((abla f)\) as \((\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})\). The gradient points in the direction of steepest ascent on the level curve, perpendicular to the contour line. Hence, it’s crucial for understanding changes in a surface or curve graphically.
In our exercise, the gradient vector \(abla f(\frac{\pi}{6}, \frac{3}{2}) = (-\sqrt{3}, 2)\) was calculated from the function \(f(x, y) = \frac{y-1}{\sin x}\), showcasing how each variable influences the function's behavior at that specific point.

Partial Derivatives

Partial derivatives are essential tools when dealing with functions of more than one variable. They show how a function changes as one variable changes, while others are held constant, enabling us to understand the function's behavior more deeply. For a function \(f(x, y)\), the partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x}\), and with respect to \(y\) is \(\frac{\partial f}{\partial y}\).
For the function given in the exercise, \(f(x, y) = \frac{y-1}{\sin x}\), these derivatives are determined as:

• \(\frac{\partial f}{\partial x} = -\frac{(y-1)\cos x}{\sin^2 x}\) - This derives from considering \(y\) constant and analyzing change in \(f\) when \(x\) varies.
• \(\frac{\partial f}{\partial y} = \frac{1}{\sin x}\) - This explains how \(f\) shifts with variations in \(y\) when \(x\) is constant.

Partial derivatives play a vital role in sketching the gradient and understanding how the level curves are shaped. They help reveal the nuances of how each variable is interacting within the function.

Function of Two Variables

Functions of two variables, such as \(f(x, y)\), output a single value from two inputs. This type of function can be used to depict surfaces and curves, where each point \((x, y)\) results in a height or depth value \(f(x, y)\). The output can represent anything from temperature over a region to topographical elevation.
Level curves are contour lines on a surface formed by a function of two variables. They help visualize how the function behaves without needing a 3D plot.
A level curve \(f(x, y) = k\) is created by holding the function's output constant, producing curves on the \(xy\)-plane. For our function \(f(x, y) = \frac{y-1}{\sin x}\), and given point \((\frac{\pi}{6}, \frac{3}{2})\), the respective level curve was found where \(f(x, y) = 1\), leading to the equation \(y = \sin x + 1\). This highlights simple shifts in the function’s surface. These concepts are crucial for exploring changes across two variables, which are often interconnected in real-world problems, each reflecting a different dimension of the scenario under study.