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Chapter 12: Problem 3

\(\mathrm{T} / \mathrm{F}\) : Let \(z=f(x, y)\) be differentiable at \(P\). If \(\vec{n}\) is a normal vector to the tangent plane of \(f\) at \(P\), then \(\vec{n}\) is orthogonal to \(\ell_{x}\) and \(\ell_{y}\) at \(P\).

Short Answer

Expert verified
True: \( \vec{n} \) is orthogonal to \( \ell_{x} \) and \( \ell_{y} \) at \( P \).

Step by step solution

01

Define the problem

We are asked to determine whether the statement is true or false: If \( \vec{n} \) is a normal vector to the tangent plane of a differentiable function \( z = f(x, y) \) at point \( P \), then \( \vec{n} \) is orthogonal to the tangent vectors \( \ell_{x} \) and \( \ell_{y} \) at \( P \).
02

Understand the meaning of normal vector

A normal vector \( \vec{n} \) to the tangent plane at \( P \) indicates a vector perpendicular to every vector lying on the tangent plane.
03

Define tangent vectors

The tangent vectors \( \ell_{x} \) and \( \ell_{y} \) at \( P \) represent the directional derivatives of \( f \) with respect to \( x \) and \( y \) respectively, which lie on the tangent plane.
04

Orthogonality condition

A vector is orthogonal to another if their dot product is zero. Since \( \vec{n} \) is a normal vector, it is perpendicular to all vectors in the tangent plane, including \( \ell_{x} \) and \( \ell_{y} \).
05

Conclusion

Since \( \vec{n} \) is defined as normal to the tangent plane, it is orthogonal to both \( \ell_{x} \) and \( \ell_{y} \) at \( P \). Therefore, the statement is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Plane
A tangent plane to a surface, like the graph of a function \( z = f(x, y) \), at a point \( P \), is essentially a flat surface. This plane just 'touches' the surface without cutting through it at that point. Imagine a sheet of paper resting snugly on a balloon; the spot where it touches is what we mean by tangent. The plane's equation is typically derived using the functions' partial derivatives at the point \( P \), where the function is approximated linearly.
  • This means the tangent plane represents the same slope and direction as the surface itself does at that exact position.
  • The equation of the tangent plane to \( z = f(x, y) \) at a point \( (a, b) \) is usually given by: \( z = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \).
Normal Vector
The normal vector is a critical concept as it is perpendicular to the tangent plane. If the tangent plane is like a sheet of paper lying flat, the normal vector is an arrow pointing directly away from the paper, perpendicular to its surface.
  • This vector can often be found using the gradient of the function, that is the combination of all partial derivatives.
  • For the function \( z = f(x, y) \), the normal vector \( \vec{n} \) will typically be expressed as \( abla f = \left( -f_x, -f_y, 1 \right) \).
  • This vector provides a straightforward way to understand the orientation of the tangent plane relative to the surface.
Directional Derivatives
Directional derivatives generalize the concept of partial derivatives. They tell us how much a function's value changes as we move in a specific direction from a point. Instead of limiting movement to the axes as partial derivatives do, they explore any possible line of direction.
  • The directional derivative in the direction of a vector \( \mathbf{u} \) is given by the dot product \( abla f \cdot \mathbf{u} \), where \( abla f \) is the gradient and \( \mathbf{u} \) is a unit vector.
  • Directional derivatives show how quickly a function increases or decreases as you move from point \( P \) in a specified direction.
Orthogonality
Orthogonality refers to the concept of vectors being perpendicular. In the context of a tangent plane, orthogonality confirms that the normal vector is perpendicular to every vector lying in the plane.
  • This includes tangent vectors, like \( \ell_{x} \) and \( \ell_{y} \), which correspond to the partial derivatives of the function \( f \).
  • The dot product between any vector in the plane and the normal vector will be zero, which mathematically verifies their perpendicularity.
  • This property ensures our geometric concepts like the tangent plane are aligned with algebraic calculations.
Differentiability
Differentiability ensures that a function behaves 'smoothly' at a given point. A differentiable function suggests that near this point, the function can be represented (or approximated) very well by its tangent plane.
  • For a function \( z = f(x, y) \) to be differentiable at any point, its partial derivatives must exist around that point, and the function itself must closely follow the form laid out by its tangent plane.
  • This means differentiability depends on both continuity and the existence of partial derivatives.
  • In practical terms, this means there are no abrupt changes in direction at that point; it's smooth and continuous.

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Most popular questions from this chapter

A set \(S\) is given. (a) Give one boundary point and one interior point, when possible, of \(S\). (b) State whether \(S\) is open, closed, or neither. (c) State whether \(S\) is bounded or unbounded. $$ S=\left\\{(x, y) \mid x^{2}+y^{2}=1\right\\} $$

T/F: A point \(P\) is a critical point of \(f\) if \(f_{x}\) or \(f_{y}\) are undefined at \(P\).

Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point. $$ f(x, y)=\frac{1}{3} x^{3}-x+\frac{1}{3} y^{3}-4 y $$

Form a function \(z=f(x, y)\) such that \(f_{x}\) and \(f_{y}\) match those given. $$ f_{x}=6 x y-4 y^{2}, \quad f_{y}=3 x^{2}-8 x y+2 $$

Exercises \(13-16\) ask a variety of questions dealing with approximating error and sensitivity analysis. It is "common sense" that it is far better to measure a long distance with a long measuring tape rather than a short one. A measured distance \(D\) can be viewed as the product of the length \(\ell\) of a measuring tape times the number \(n\) of times it was used. For instance, using a \(3^{\prime}\) tape 10 times gives a length of \(30^{\prime}\). To measure the same distance with a \(12^{\prime}\) tape, we would use the tape 2.5 times. (l.e., \(30=12 \times 2.5 .\) Thus \(D=n \ell\). Suppose each time a measurement is taken with the tape, the recorded distance is within \(1 / 16 "\) of the actual distance. (l.e., \(d \ell=1 / 16^{\prime \prime} \approx 0.00 \mathrm{fft}\). Using differentials, show why common sense proves correct in that it is better to use a long tape to measure long distances.
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