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A function \(z=f(x, y),\) a vector \(\vec{v}\) and a point \(P\) are given. Give the parametric equations of the following directional tangent lines to \(f\) at \(P\) : (a) \(\ell_{x}(t)\) (b) \(\ell_{y}(t)\) (c) \(\ell_{\vec{u}}(t),\) where \(\vec{u}\) is the unit vector in the direction of \(\vec{v}\). $$ f(x, y)=3 x-5 y, \vec{v}=\langle 1,1\rangle, P=(4,2) $$

Short Answer

Expert verified
(a) \(\ell_x(t) = (4 + t, 2, 2 + 3t)\); (b) \(\ell_y(t) = (4, 2 + t, 2 - 5t)\); (c) \(\ell_{\vec{u}}(t) = (4 + \frac{t}{\sqrt{2}}, 2 + \frac{t}{\sqrt{2}}, 2 - \frac{2t}{\sqrt{2}})\).

Step by step solution

01

Find Partial Derivatives

First, find the partial derivatives of the function \(f(x, y) = 3x - 5y\). Calculate: \(f_x\) and \(f_y\).\[f_x = \frac{\partial}{\partial x}(3x - 5y) = 3\] \[f_y = \frac{\partial}{\partial y}(3x - 5y) = -5\].
02

Evaluate Partial Derivatives at Point P

Evaluate the partial derivatives at the given point \(P(4,2)\). \[f_x(4,2) = 3\] \[f_y(4,2) = -5\].
03

Parametric Equation for \(\ell_x(t)\)

Construct the parametric equation for the directional tangent line along the x-axis. \(\ell_x(t) = (4 + t, 2, f(4,2) + t f_x(4,2))\)Since \(f(4,2) = 3 \times 4 - 5 \times 2 = 2\), then \(\ell_x(t) = (4 + t, 2, 2 + 3t)\).
04

Parametric Equation for \(\ell_y(t)\)

Construct the parametric equation for the directional tangent line along the y-axis. \(\ell_y(t) = (4, 2 + t, f(4,2) + t f_y(4,2))\)Thus, \(\ell_y(t) = (4, 2 + t, 2 - 5t)\).
05

Normalize the Direction Vector \(\vec{v}\)

Find the unit vector \(\vec{u}\) in the direction of \(\vec{v} = \langle 1,1 \rangle \).\[\|\vec{v}\| = \sqrt{1^2 + 1^2} = \sqrt{2}\].Thus, \(\vec{u} = \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle\).
06

Parametric Equation for \(\ell_{\vec{u}}(t)\)

Construct the parametric equation for the directional tangent line in the direction of \(\vec{u}\). \(\ell_{\vec{u}}(t) = (4 + \frac{t}{\sqrt{2}}, 2 + \frac{t}{\sqrt{2}}, 2 + t(\frac{3}{\sqrt{2}}) - t(\frac{5}{\sqrt{2}}))\)Simplify to get \(\ell_{\vec{u}}(t) = (4 + \frac{t}{\sqrt{2}}, 2 + \frac{t}{\sqrt{2}}, 2 - \frac{2t}{\sqrt{2}})\).

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental concept in calculus, essential for understanding how functions behave when more than one variable is involved. They represent how the function changes in relation to one variable while keeping other variables constant.
In the given function, \(f(x, y) = 3x - 5y\), we can find partial derivatives with respect to \(x\) and \(y\), denoted as \(f_x\) and \(f_y\). For \(f_x\), differentiate \(f\) treating \(y\) as a constant:
  • \(f_x = \frac{\partial}{\partial x}(3x - 5y) = 3\)
Similarly, for \(f_y\), treat \(x\) as a constant:
  • \(f_y = \frac{\partial}{\partial y}(3x - 5y) = -5\)
By evaluating these derivatives at the point \(P(4,2)\), you obtain the rates of change of the function in both the \(x\) and \(y\) directions at that point:
  • \(f_x(4,2) = 3\)
  • \(f_y(4,2) = -5\)
Partial derivatives are crucial for constructing parametric equations for tangent lines to surfaces, which help in understanding the geometry of the function.
Parametric Equations
Parametric equations are used to describe a path of points in a coordinate system. Instead of relying on a single equation in terms of x and y, parametric equations express each coordinate as a separate function of an independent variable, usually represented by \(t\).
For directional tangent lines to the function \(f(x, y)\) at point \(P(4,2)\), we construct the parametric equations \(\ell_x(t)\) and \(\ell_y(t)\), which move horizontally and vertically from \(P\), respectively.
  • For \(\ell_x(t)\): Move along the x-axis while keeping the y-coordinate constant. \[\ell_x(t) = (4 + t, 2, f(4,2) + 3t) = (4 + t, 2, 2 + 3t)\]
  • For \(\ell_y(t)\): Move along the y-axis while keeping the x-coordinate constant. \[\ell_y(t) = (4, 2 + t, f(4,2) - 5t) = (4, 2 + t, 2 - 5t)\]
These parametric forms trace directions of change, showing how the function behaves right from the point \(P\) along specified axes.
Unit Vector
A unit vector is a vector with a magnitude (or length) of one. It indicates direction and is often used in directional derivatives, where the change in a function is measured in a specific direction.
To find the unit vector \(\vec{u}\) in the direction of a given vector \(\vec{v}\), we first calculate the magnitude of \(\vec{v}\). For \(\vec{v} = \langle 1, 1 \rangle\), the magnitude is calculated as:
  • \[\|\vec{v}\| = \sqrt{1^2 + 1^2} = \sqrt{2}\]
The unit vector \(\vec{u}\) is obtained by dividing each component of \(\vec{v}\) by its magnitude:
  • \[\vec{u} = \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle\]
Using this unit vector, we can build the parametric equation \(\ell_{\vec{u}}(t)\), which describes the line moving through point \(P\) in the direction of \(\vec{v}\). This helps understand how the function changes in that specific vector direction.

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Most popular questions from this chapter

In Exercises \(19-22,\) functions \(z=f(x, y), x=g(s, t)\) and \(y=h(s, t)\) are given. (a) Use the Multivariable Chain Rule to compute \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\) (b) Evaluate \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\) at the indicated \(s\) and \(t\) values.$ 12. $$ z=x^{2} y, \quad x=s-t, \quad y=2 s+4 t ; \quad s=1, t=0 $$

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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(a) Find the domain \(D\) of the given function. (b) State whether \(D\) is an open or closed set. (c) State whether \(D\) is bounded or unbounded. $$ f(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}} $$

Evaluate the limit along the paths given, then state why these results show the given limit does not exist. \(\lim _{(x, y) \rightarrow(1,2)} \frac{x+y-3}{x^{2}-1}\) (a) Along the path \(y=2\). (b) Along the path \(y=x+1\).

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