Problem 10
Question
A function \(z=f(x, y)\) and a point \(P\) are given. Find the equation of the normal line to \(f\) at \(P\). Note: these are the same functions as in Exercises \(5-8\). \(f(x, y)=3 \cos x \sin y, P=(\pi / 3, \pi / 6)\)
Step-by-Step Solution
Verified Answer
The normal line has parametric equations: \(x = \frac{\pi}{3} - \frac{3\sqrt{3}}{4}t\), \(y = \frac{\pi}{6} + \frac{3\sqrt{3}}{4}t\), \(z = \frac{3}{4} - t\).
1Step 1: Calculate the Partial Derivatives
To find the normal line, we first need to compute the partial derivatives of the function. The function given is \(f(x, y) = 3 \cos x \sin y\). We will find \(f_x\) and \(f_y\). - The partial derivative with respect to \(x\) (\(f_x\)) is \(-3 \sin x \sin y\).- The partial derivative with respect to \(y\) (\(f_y\)) is \(3 \cos x \cos y\).
2Step 2: Evaluate the Partial Derivatives at P
Substitute the coordinates of point \(P\), \(\left(\frac{\pi}{3}, \frac{\pi}{6}\right)\), into the partial derivatives calculated.- \(f_x\left(\frac{\pi}{3}, \frac{\pi}{6}\right) = -3 \sin\left(\frac{\pi}{3}\right) \sin\left(\frac{\pi}{6}\right) = -3 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = -\frac{3\sqrt{3}}{4}\).- \(f_y\left(\frac{\pi}{3}, \frac{\pi}{6}\right) = 3 \cos\left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{6}\right) = 3 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4}\).
3Step 3: Determine the Normal Vector
The normal vector to the surface at a point is given by \(\langle f_x, f_y, -1 \rangle\). Using the derivatives evaluated at point \(P\), the normal vector is:\(\langle -\frac{3\sqrt{3}}{4}, \frac{3\sqrt{3}}{4}, -1 \rangle\).
4Step 4: Write the Equation of the Normal Line
The equation of the normal line can be expressed in parametric form using the normal vector and the point \(P\):- \(x = \frac{\pi}{3} - \frac{3\sqrt{3}}{4}t\)- \(y = \frac{\pi}{6} + \frac{3\sqrt{3}}{4}t\)- \(z = 3 \cos\left(\frac{\pi}{3}\right) \sin\left(\frac{\pi}{6}\right) - t = \frac{3}{4} - t\)These equations describe the normal line to the surface \(f(x, y)\) at point \(P\).
Key Concepts
Partial DerivativesParametric EquationsNormal Vector
Partial Derivatives
Partial derivatives are a foundational concept in multivariable calculus. They describe how a function of several variables changes as one of those variables changes, while the others are held constant.
Imagine a function of two variables, like our example function, which depends on both \(x\) and \(y\): \(f(x, y) = 3 \cos x \sin y\). To find out how the function \(f(x, y)\) changes when you alter \(x\), you'd compute the partial derivative with respect to \(x\), represented as \(f_x\). Similarly, to understand how changes in \(y\) affect the function, you'd calculate \(f_y\).
Imagine a function of two variables, like our example function, which depends on both \(x\) and \(y\): \(f(x, y) = 3 \cos x \sin y\). To find out how the function \(f(x, y)\) changes when you alter \(x\), you'd compute the partial derivative with respect to \(x\), represented as \(f_x\). Similarly, to understand how changes in \(y\) affect the function, you'd calculate \(f_y\).
- \(f_x\): Represents the rate of change of \(f\) with respect to \(x\), treating \(y\) as a constant.
- \(f_y\): Represents the rate of change of \(f\) with respect to \(y\), treating \(x\) as a constant.
Parametric Equations
Parametric equations help us describe a set of related quantities as continuous functions of a parameter. They're particularly useful in defining curves and lines where each coordinate depends on a single parameter.
In the context of our exercise, we used parametric equations to write the equation of the normal line at point \(P\). The idea is to express \(x\), \(y\), and \(z\) as functions of a parameter \(t\), utilizing the normal vector, which dictates the direction of the line.
In the context of our exercise, we used parametric equations to write the equation of the normal line at point \(P\). The idea is to express \(x\), \(y\), and \(z\) as functions of a parameter \(t\), utilizing the normal vector, which dictates the direction of the line.
- Each equation represents how a coordinate changes as the parameter \(t\) changes.
- This approach simplifies describing complex shapes or directions using simple equations.
Normal Vector
A normal vector is perpendicular to the surface of a function at a given point. It plays a key role in defining the orientation of the surface in 3D space.
For surfaces defined by functions like \(z = f(x,y)\), the normal vector can be found using the partial derivatives. In our exercise, the normal vector at point \(P\) was derived from the tuple \(\langle f_x, f_y, -1 \rangle\). This vector gives the direction in which the normal line extends away from the surface at that point.
For surfaces defined by functions like \(z = f(x,y)\), the normal vector can be found using the partial derivatives. In our exercise, the normal vector at point \(P\) was derived from the tuple \(\langle f_x, f_y, -1 \rangle\). This vector gives the direction in which the normal line extends away from the surface at that point.
- Normal vectors: Crucial for computing angles, reflections, and even for rendering objects in computer graphics.
- They help in understanding surface geometry and in calculating normals to curves and surfaces.
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