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Chapter 1: Problem 59

Find a normal vector \(\mathbf{N}\) and the tangent plane \(H\) to each surface at the given point: (a) surface \(x^{2} y+3 y z=20\) and point \(P(1,3,2)\); (b) surface \(x^{2}+3 y^{2}-5 z^{2}=160\) and point \(P(3,-2,1)\).

Short Answer

Expert verified
(a) For the surface \(x^2y+3yz=20\) at point P(1,3,2), the normal vector \(\mathbf{N}\) is \(\langle 6, 7, 9 \rangle\) and the tangent plane H has the equation \(6(x-1)+7(y-3)+9(z-2)=0\). (b) For the surface \(x^2+3y^2-5z^2=160\) at point P(3,-2,1), the normal vector \(\mathbf{N}\) is \(\langle 6, -12, -10 \rangle\) and the tangent plane H has the equation \(6(x-3)-12(y+2)-10(z-1)=0\).

Step by step solution

01

Find the Gradient Vector of the Surface

To find the gradient vector of the surface \(x^{2} y+3 y z=20\), we need to calculate the partial derivatives of the function with respect to each variable: \(f_{x},f_{y},\) and \(f_{z}\). \[f_{x}=\frac{\partial}{\partial x}(x^{2} y+3 y z) = 2xy\] \[f_{y}=\frac{\partial}{\partial y}(x^{2} y+3 y z) = x^{2} + 3z\] \[f_{z}=\frac{\partial}{\partial z}(x^{2} y+3 y z) = 3y\] The gradient vector \(\nabla f\) is: \[\nabla f = \langle 2xy, x^{2} + 3z, 3y \rangle\]
02

Evaluate the Gradient Vector at the Given Point

Now, we evaluate the gradient vector at the given point P(1,3,2): \[\nabla f(1,3,2) = \langle 2(1)(3), (1)^{2} + 3(2), 3(3) \rangle = \langle 6, 7, 9 \rangle\] The normal vector \(\mathbf{N}\) at point P(1,3,2) is \(\mathbf{N}=\langle 6, 7, 9 \rangle\).
03

Find the Tangent Plane Equation at the Given Point

To find the equation of the tangent plane to the surface at P(1,3,2), we will use the point-normal form of a plane: \[(\mathbf{r} - \mathbf{P})\cdot \mathbf{N} = 0\] Here, \(\mathbf{r} = \langle x, y, z \rangle\), \(\mathbf{P} = \langle 1, 3, 2 \rangle\), and \(\mathbf{N} = \langle 6, 7, 9 \rangle\). Substituting the values, we get: \[(\langle x, y, z \rangle - \langle 1, 3, 2 \rangle)\cdot \langle 6, 7, 9 \rangle = 0\] Expanding the dot product, we have: \[6(x-1)+7(y-3)+9(z-2)=0\] Thus, the tangent plane H to the surface at P(1,3,2) has the equation: \[6(x-1)+7(y-3)+9(z-2)=0\] #B. Normal Vector and Tangent Plane for Surface x^2+3y^2-5z^2=160 at Point (3,-2,1):#
04

Find the Gradient Vector of the Surface

To find the gradient vector of the surface \(x^2+3y^2-5z^2=160\), we need to calculate the partial derivatives of the function with respect to each variable: \(f_{x},f_{y},\) and \(f_{z}\). \[f_{x}=\frac{\partial}{\partial x}(x^2+3y^2-5z^2) = 2x\] \[f_{y}=\frac{\partial}{\partial y}(x^2+3y^2-5z^2) = 6y\] \[f_{z}=\frac{\partial}{\partial z}(x^2+3y^2-5z^2) = -10z\] The gradient vector \(\nabla f\) is: \[\nabla f = \langle 2x, 6y, -10z \rangle\]
05

Evaluate the Gradient Vector at the Given Point

Now, we evaluate the gradient vector at the given point P(3,-2,1): \[\nabla f(3,-2,1) = \langle 2(3), 6(-2), -10(1) \rangle = \langle 6, -12, -10 \rangle\] The normal vector \(\mathbf{N}\) at point P(3,-2,1) is \(\mathbf{N}=\langle 6, -12, -10 \rangle\).
06

Find the Tangent Plane Equation at the Given Point

To find the equation of the tangent plane to the surface at P(3,-2,1), we will use the point-normal form of a plane: \[(\mathbf{r} - \mathbf{P})\cdot \mathbf{N} = 0\] Here, \(\mathbf{r} = \langle x, y, z \rangle\), \(\mathbf{P} = \langle 3, -2, 1 \rangle\), and \(\mathbf{N} = \langle 6, -12, -10 \rangle\). Substituting the values, we get: \[(\langle x, y, z \rangle - \langle 3, -2, 1 \rangle)\cdot \langle 6, -12, -10 \rangle = 0\] Expanding the dot product, we have: \[6(x-3)-12(y+2)-10(z-1)=0\] Thus, the tangent plane H to the surface at P(3,-2,1) has the equation: \[6(x-3)-12(y+2)-10(z-1)=0\]

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Key Concepts

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

Normal Vector
In calculus and differential geometry, a **normal vector** to a surface at a particular point is a vector that is perpendicular to the tangent plane of the surface at that point. It provides crucial information about the directional properties of the surface at that point.

  • The normal vector is often found using the gradient vector of a surface, as the gradient is perpendicular to the level curves or surfaces.
  • A normal vector is essential in defining planes and lines in 3D space.
  • In the context of the exercise provided, the normal vector \(\mathbf{N}\) at a given point is computed as \(\langle 6, 7, 9 \rangle\) for the function \(x^2 y+3yz=20\) at \(P(1,3,2)\).
The process typically involves evaluating the gradient vector at the particular point. The calculated normal vector can then be used to find the equation of the tangent plane, which is fundamental for understanding the geometry and orientation of surfaces in space.
Gradient Vector
The **gradient vector**, often denoted as \(abla f\), of a function at a point is a vector composed of partial derivatives of each variable present in the function. It represents both the direction and rate of fastest increase.

  • The gradient vector always points in the direction of steepest ascent from a point on the surface.
  • It is a crucial concept in multivariable calculus, especially for finding tangent planes and normal lines.
  • In the problem context, the gradient of \(x^2 y + 3yz = 20\) is \(\langle 2xy, x^2 + 3z, 3y \rangle\).
To find it, compute the partial derivatives with respect to all the variables. This process yields a vector field that assists in visualizing how a function changes across its domain. At the point of interest, evaluate these derivatives to find the specific gradient vector. This vector plays a pivotal role in finding the normal vector and subsequently, the tangent plane.
Partial Derivatives
**Partial derivatives** are derivatives of functions with multiple variables, with respect to one variable at a time, treating other variables as constants. They are fundamental for understanding how multi-variable functions change.

  • These derivatives help in analyzing the rate of change of the function along each axis of the domain.
  • They are the building blocks of the gradient vector and thus play a crucial role in applications like finding tangent planes.
  • In the given problem, for the function \(x^2 y + 3yz = 20\), partial derivatives were calculated as follows: for \(x\), \(f_x = 2xy\); for \(y\), \(f_y = x^2 + 3z\); and for \(z\), \(f_z = 3y\).
These derivatives provide the necessary components for constructing the gradient vector. By evaluating these derivatives at a given point, one gains insights into the surface's slope along various directions and thus can deduce the surface's tangent properties. Understanding partial derivatives is key to mastering multi-variable calculus and its applications.

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Most popular questions from this chapter

Find \(x, y, z\) where \((x, y+1, y+z)=(2 x+y, 4,3 z)\).

Prove the following properties of the cross product: (a) \(u \times v=-(v \times u)\) (d) \(u \times(v+w)=(u \times v)+(u \times w)\) (b) \(u \times u=0\) for any vector \(u\) (e) \((v+w) \times u=(v \times u)+(w \times u)\) (c) \(\quad(k u) \times v=k(u \times v)=u \times(k v)\) (f) \((u \times v) \times w=(u \cdot w) v-(v \cdot w) u\)

Prove Theorem \(1.4\) (Minkowski): \(\|u+v\| \leq\|u\|+\|v\|\). By the Schwarz inequality and other properties of the dot product, $$ \|u+v\|^{2}=(u+v) \cdot(u+v)=(u \cdot u)+2(u \cdot v)+(v \cdot v) \leq\|u\|^{2}+2\|u\|\|v\|+\|v\|^{2}=(\|u\|+\|v\|)^{2} $$ Taking the square root of both sides yields the desired inequality.

Simplify: (a) \(i^{0}, i^{3}, i^{4},(\mathrm{b}) \quad i^{5}, i^{6}, i^{7}, i^{8},(\mathrm{c}) \quad i^{39}, i^{174}, i^{252}, i^{317}\)

Prove: For any vectors \(u, v \in \mathbf{C}^{n}\) and any scalar \(z \in \mathbf{C}\), (i) \(u \cdot v=\overline{v \cdot u}\), (ii) \((z u) \cdot v=z(u \cdot v)\), (iii) \(u \cdot(z v)=\bar{z}(u \cdot v)\) Suppose \(u=\left(z_{1}, z_{2}, \ldots, z_{n}\right)\) and \(v=\left(w_{1}, w_{2}, \ldots, w_{n}\right)\). (i) Using the properties of the conjugate, $$ \begin{aligned} \overline{v \cdot u} &=\overline{w_{1} \bar{z}_{1}+w_{2} \bar{z}_{2}+\cdots+w_{n} \bar{z}_{n}}=\overline{w_{1} \bar{z}_{1}}+\overline{w_{2} \bar{z}_{2}}+\cdots+\overline{w_{n} \bar{z}_{n}} \\\ &=\bar{w}_{1} z_{1}+\bar{w}_{2} z_{2}+\cdots+\bar{w}_{n} z_{n}=z_{1} \bar{w}_{1}+z_{2} \bar{w}_{2}+\cdots+z_{n} \bar{w}_{n}=u \cdot v \end{aligned} $$ (ii) Because \(z u=\left(z z_{1}, z z_{2}, \ldots, z z_{n}\right)\) $$ (z u) \cdot v=z z_{1} \bar{w}_{1}+z z_{2} \bar{w}_{2}+\cdots+z z_{n} \bar{w}_{n}=z\left(z_{1} \bar{w}_{1}+z_{2} \bar{w}_{2}+\cdots+z_{n} \bar{w}_{n}\right)=z(u \cdot v) $$ (Compare with Theorem \(1.2\) on vectors in \(\mathbf{R}^{n}\).) (iii) Using (i) and (ii), $$ u \cdot(z v)=\overline{(z v) \cdot u}=z(\overline{v \cdot u})=\bar{z}(\overline{v \cdot u})=\bar{z}(u \cdot v) $$
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