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Chapter 4: Problem 109

Find the line orthogonal to the graph of \(\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{xy}\) at the point \(\left(\mathrm{x}_{0}, \mathrm{y}_{0}, \mathrm{z}_{0}\right)=(-2,3,-6)\).

Short Answer

Expert verified
The line orthogonal to the graph of \(f(x, y) = xy\) at the point \((-2, 3, -6)\) is described by the parametric equations: \(x(t) = -2 + 3t\), \(y(t) = 3 - 2t\), and \(z(t) = -6 - t\).

Step by step solution

01

Find the gradient of the function f(x, y)

To find the gradient of the function \(f(x, y) = xy\), calculate the partial derivatives with respect to x and y: \[f_x = \frac{\partial}{\partial x}(xy) = y\] \[f_y = \frac{\partial}{\partial y}(xy) = x\]
02

Calculate the normal vector to the surface

Since we have the gradient of the function, we can now calculate the normal vector, denoted as \(\vec{n}\), to the surface at the point \((-2, 3, -6)\). The gradient vector at the point is given by the components \([\,f_x(x_0, y_0),\, f_y(x_0, y_0), \, -1\,]\): \[\vec{n} = \begin{bmatrix} y_0 \\ x_0 \\ -1 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \\ -1 \end{bmatrix}\]
03

Find the equation of the orthogonal line

To find the equation of the orthogonal line in parametric form, use the coordinates of the point \((-2, 3, -6)\) and normal vector \(\vec{n}\). The parametric equation of a line is given by: \[r(t) = \vec{p_0} + t \cdot \vec{n}\] where \(r(t)\) is the position vector of a point on the line, \(\vec{p_0}\) is the position vector of the given point (in our case, \((-2,3,-6)\)), and \(t\) is the parameter representing how far along the line the point is. Plugging in the values, we get: \[r(t) = \begin{bmatrix} -2 \\ 3 \\ -6 \end{bmatrix} + t \cdot \begin{bmatrix} 3 \\ -2 \\ -1 \end{bmatrix}\] Now, we can write the orthogonal line in parametric form: \[(x(t), y(t), z(t)) = (-2 + 3t, 3 - 2t, -6 - t)\] So, the line orthogonal to the graph of \(f(x, y) = xy\) at the point \((-2, 3, -6)\) is described by the parametric equations: \[x(t) = -2 + 3t\] \[y(t) = 3 - 2t\] \[z(t) = -6 - t\]

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Key Concepts

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

Gradient
The gradient of a function is an essential concept in multivariable calculus. It represents the direction of the steepest ascent for a scalar field. To find the gradient of a function, we compute its partial derivatives with respect to all variables involved.
For a given function, such as \( f(x, y) = xy \), the gradient vector \( abla f \) comprises the partial derivative of \( f \) with respect to each variable. Here, we calculate:
  • \( f_x = \frac{\partial}{\partial x}(xy) = y \)
  • \( f_y = \frac{\partial}{\partial y}(xy) = x \)
Thus, the gradient vector is \( abla f = [ y, x ] \), demonstrating how it points in the direction where the function increases most rapidly. This concept is crucial when determining normal vectors to surfaces defined by functions.
Partial Derivatives
Partial derivatives are like the building blocks of gradients. They measure how a function changes as we tweak one variable while keeping others fixed. In the context of the function \( f(x, y) = xy \), finding the partial derivatives involves taking the derivative first with respect to \( x \) and then with respect to \( y \).The partial derivative with respect to \( x \) is:
  • \( f_x = y \)
Here, \( y \) is treated as a constant. Similarly, the partial derivative with respect to \( y \) is:
  • \( f_y = x \)
In this case, \( x \) is the constant. These derivatives help in constructing the gradient vector, reflecting the function's rate of change along each axis. Recognizing these derivatives' significance allows for further exploration of how functions behave across multidimensional spaces.
Normal Vector
A normal vector is a vector that is perpendicular to a surface at a given point. For a function \( f(x, y) \), its gradient at any point provides a normal vector to the level curve through that point.
In our exercise, we derive the normal vector \( \vec{n} \) from the gradient of the function. Calculating it involves:
  • The gradient vector at point \( (-2, 3) \).
  • Here, \( f_x = y = 3 \) and \( f_y = x = -2 \).
The resulting normal vector formed is:
  • \( \vec{n} = \begin{bmatrix} 3 \ -2 \ -1 \end{bmatrix} \)
This vector is crucial, especially in calculating lines and planes orthogonal to the given surface. It simplifies understanding how the surface or function is oriented in space.
Orthogonal Line
An orthogonal line is one that intersects a surface at a right angle, closely related to the concept of normal vectors. To define an orthogonal line to a surface at a given point, we utilize the normal vector from the gradient of the function.
The equation for the orthogonal line is given in its parametric form:
  • \( r(t) = \vec{p_0} + t \cdot \vec{n} \)
Where \( \vec{p_0} \) is the position vector of the point \((-2, 3, -6)\), and \( \vec{n} = \begin{bmatrix} 3 \ -2 \ -1 \end{bmatrix} \) is the normal vector. With these, we construct the parametric equations:
  • \( x(t) = -2 + 3t \)
  • \( y(t) = 3 - 2t \)
  • \( z(t) = -6 - t \)
These equations help illustrate how the line moves through space, ensuring its perpendicular intersection with the surface, which is an invaluable tool for visualizing geometrical relationships in multivariable calculus.

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Most popular questions from this chapter

Show that if a function \(\mathrm{F}: \mathrm{R}^{2} \rightarrow \mathrm{R}\) satisfies \(\partial \mathrm{F} / \partial \mathrm{x}=\partial \mathrm{F} / \partial \mathrm{y}\), then \(F(x, y)=f(x+y)\) where \(f\) is some differentiable function of one variable, \(\mathrm{s}=\mathrm{x}+\mathrm{y}\)

If \(\mathrm{G}(\mathrm{s}, \mathrm{t})=\mathrm{F}\left(\mathrm{e}^{\mathrm{s}} \cos \mathrm{t}, \mathrm{e}^{\mathrm{s}} \sin \mathrm{t}\right)\), show that \(\mathrm{G}_{11}+\mathrm{G}_{22}=\mathrm{e}^{2 \mathrm{~s}}\left(\mathrm{~F}_{11}+\mathrm{F}_{22}\right)\), where \(\mathrm{G}_{11}\) and \(\mathrm{G}_{22}\) are evaluated at \((\mathrm{s}, \mathrm{t})\) and \(\mathrm{F}_{11}\) and \(\mathrm{F}_{22}\) are evaluated at \(\left(\mathrm{e}^{3} \cos \mathrm{t}, \mathrm{e}^{\mathrm{s}} \sin \mathrm{t}\right)\). Note: \(\mathrm{G}_{11}, \mathrm{G}_{22}\) are abbreviations for \(\left(\partial^{2} \mathrm{G} / \partial \mathrm{s}^{2}\right),\left(\partial^{2} \mathrm{G} / \partial \mathrm{t}^{2}\right)\) respectively and \(F_{11}, F_{22}\) are abbreviations for \(\left(\partial^{2} F / \partial s^{2}\right)\) and \(\left(\partial^{2} \mathrm{~F} / \partial \mathrm{t}^{2}\right)\), respectively.

Let \(u=r^{3}\) be a scalar field where \(r\) is the distance \(O P\) from the origin 0 to a variable point \(\mathrm{P}\), in \(\mathrm{R}^{3}\). Find the gradient of \(u\) at \(P\) without resorting to rectangular coordinates.

Determine whether the following vector fields have potential functions and if so find them: a) \(F(x, y)=\left(3 x y, \sin x y^{3}\right)\) b) \(F(x, y)=\left(2 x y, x^{2}+3 y^{2}\right)\)

Determine whether or not the vector field \(F(x, y)=\left[\left\\{\left(e^{T} x\right) / r\right\\},\left\\{\left(e^{\mathrm{r}} y\right) / r\right\\}\right], r=\sqrt{\left(x^{2}+y^{2}\right)}\) has a potential function (i.e., if it is conservative). If a potential function exists, find it.
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