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91Ó°ÊÓ

Find the directions in which the functions increase and decrease most rapidly at \(P_{0} .\) Then find the derivatives of the functions in these directions. $$f(x, y, z)=(x / y)-y z, \quad P_{0}(4,1,1).$$

Short Answer

Expert verified
Increase direction: (1, -5, -1); Decrease direction: (-1, 5, 1); Maximum derivative: \(3\sqrt{3}\).

Step by step solution

01

Compute the Gradient of f

The first step is to find the gradient of the given function. The gradient of a function \(f(x, y, z)\) is a vector of partial derivatives, denoted as \(abla f\). So, calculate \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\). For \(f(x, y, z) = \frac{x}{y} - yz\):\[\frac{\partial f}{\partial x} = \frac{1}{y}\]\[\frac{\partial f}{\partial y} = -\frac{x}{y^2} - z\]\[\frac{\partial f}{\partial z} = -y\]So, the gradient is \(abla f = \left(\frac{1}{y}, -\frac{x}{y^2} - z, -y\right)\).
02

Evaluate the Gradient at P_0

Now, substitute the point \(P_0(4, 1, 1)\) into the gradient vector:\[x = 4, \quad y = 1, \quad z = 1\]\[abla f(P_0) = \left(\frac{1}{1}, -\frac{4}{1^2} - 1, -1\right) = (1, -5, -1)\]
03

Determine Directions of Maximum Increase and Decrease

The direction of maximum increase is in the direction of the gradient \(abla f\), which is \((1, -5, -1)\). The direction of maximum decrease is in the opposite direction of the gradient, which is \((-1, 5, 1)\).
04

Find the Derivative in the Direction of the Gradient

The derivative of the function in the direction of maximum increase is simply the magnitude of the gradient vector. Compute this magnitude:\[\| abla f \| = \sqrt{1^2 + (-5)^2 + (-1)^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3}\]
05

Conclusion

Therefore, the function increases most rapidly in the direction \((1, -5, -1)\) and decreases most rapidly in the direction \((-1, 5, 1)\). The derivative in the direction of the gradient, representing the rate of maximum increase, is \(3\sqrt{3}\).

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Key Concepts

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

Partial Derivatives
Understanding partial derivatives is essential for exploring functions with multiple variables like our example function, \( f(x, y, z) = \frac{x}{y} - yz \).
Partial derivatives represent how the function changes as you tweak one variable, keeping others constant.
For example:
  • \( \frac{\partial f}{\partial x} = \frac{1}{y} \) shows changes in \( f \) when only \( x \) changes. Here, since \( y \) is in the denominator, keep an eye on changes to it.
  • \( \frac{\partial f}{\partial y} = -\frac{x}{y^2} - z \) reflects changes when only \( y \) shifts.
  • \( \frac{\partial f}{\partial z} = -y \) affects \( f \) as only \( z \) varies.
By evaluating these at a specific point \( P_0 (4, 1, 1) \), you get an understanding of the behavior of the function at that particular spot in space.
It helps build towards finding the function's gradient, which is just a fancy extension of these partials.
Directional Derivative
The directional derivative extends the idea of a standard derivative by incorporating a specific direction into the calculation.
Imagine you are moving along a path, not just directly following the \( x, y, \) or \( z \) axes, but any direction. The directional derivative tells us how a function sails along this path.
To find it, you need:
  • The gradient, \( abla f \), which is the collection of all partial derivatives (think of it as a compass pointing to fastest slopes).
  • A direction, defined by a unit vector which provides the path's orientation.
In scenarios where the task asks to find derivatives in specific directions of increase (e.g., using the gradient \( abla f = (1, -5, -1) \)), directional derivatives are computed by dotting the gradient with the direction vector.
This tells us, numerically, how the height of \( f \) changes as we travel in that exact direction. There's no need to worry about complexities; the directional derivative simplifies it.
Function Optimization
Function optimization involves finding maximums or minimums, where directional derivatives play an important role.
In multivariable calculus, you often aim to discover at which points a function reaches its peak (maximum) or its lowest dip (minimum).
Here's how the gradient assists in this mission:
  • The gradient \( abla f \) is your guide, as it always points towards the direction of greatest increase.
  • If you seek local maximum, evaluate where \( abla f = \mathbf{0} \) (a zero vector), which potentially indicates a peak or trough, known commonly as critical points.
  • From these points, one can determine using test conditions if the point is a max, min, or a saddle point (not fully upward or downward, like a saddle on a horse).
Thus, knowing the gradient and where it vanishes is crucial in pinpointing these function characteristics. Remember, finding the rate of increase (directional derivative) provides insights into optimizing scenarios when simply "eyeballing" the visuals is not enough.

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Most popular questions from this chapter

Show that each function satisfies a Laplace equation. \(f(x, y)=e^{-2 y} \cos 2 x\)

Let \(f(x, y)=\left\\{\begin{array}{ll}x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}}, & \text { if }(x, y) \neq 0, \\ 0, & \text { if }(x, y)=0.\end{array}\right.\) a. Show that \(\frac{\partial f}{\partial y}(x, 0)=x\) for all \(x,\) and \(\frac{\partial f}{\partial x}(0, y)=-y\) for all \(y\) b. Show that \(\frac{\partial^{2} f}{\partial y \partial x}(0,0) \neq \frac{\partial^{2} f}{\partial x \partial y}(0,0)\) The graph of \(f\) is shown on page 788 . The three-dimensional Laplace equation $$\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}=0 $$ is satisfied by steady-state temperature distributions \(T=f(x, y, z)\) in space, by gravitational potentials, and by electrostatic potentials. The two- dimensional Laplace equation $$\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0$$ obtained by dropping the \(\partial^{2} f / \partial z^{2}\) term from the previous equation, describes potentials and steady-state temperature distributions in a plane (see the accompanying figure). The plane (a) may be treated as a thin slice of the solid (b) perpendicular to the \(z\) -axis.

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\). e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=x^{2}+y^{3}-3 x y, \quad-5 \leq x \leq 5, \quad-5 \leq y \leq 5$$

To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Functions: a. \(f(x, y)=2 x+3 y\) b. \(g(x, y)=x y\) c. \(h(x, y)=x^{2}+3 y^{2}\) Curves: i) The semiellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1, \quad y \geq 0\) ii) The quarter ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1, \quad x \geq 0, \quad y \geq 0\) Use the parametric equations \(x=3 \cos t, y=2 \sin t\).

Find the limits by rewriting the fractions first. $$\lim _{P \rightarrow(1,3,4)}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$$

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