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Find the directions of maximum and minimum change of \(f\) at the given point, and the values of the maximum and minimum rates of change. $$f(x, y)=x \cos 3 y,(2,0)$$

Short Answer

Expert verified
The direction of maximum change is [1, 0], and the direction of minimum change is [-1, 0]. The maximum rate of change is 1, and the minimum rate of change is 1.

Step by step solution

01

Find the Gradient of \(f\)

The gradient of a function is given by the vector of its partial derivatives. So, let's first calculate the partial derivatives of \(f\) with respect to \(x\) and \(y\).\n\n\(\frac{{\partial f}}{{\partial x}} = cos 3y\)\n\n\(\frac{{\partial f}}{{\partial y}} = -3x sin 3y\)\n\nThe gradient of \(f\) is then given by \(\nabla f = [cos 3y, -3x sin 3y]\).
02

Evaluate the Gradient at the Given Point

The given point is (2,0). We substitute these values into the gradient.\n\n \(\nabla f(2,0) = [cos 3*0, -3*2 sin 3*0] = [1, 0]\).\n\nThis is the direction of maximum change.
03

Find the Direction of Minimum Change

Recall that the direction of minimum change is the negative of the gradient. So, we simply take the negative of the result from Step 2 to find the direction of minimum change.\n\n -\(\nabla f = [-1, 0]\)
04

Calculate the Maximum and Minimum Rates of Change

In a similar vein, the rates of change are given by the magnitude of the gradients. \n\n maximum change = |\(\nabla f\)| = sqrt((1)^2 + (0)^2) = 1\n\n minimum change = |- \(\nabla f\)| = sqrt((-1)^2 + (0)^2) = 1

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Key Concepts

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

Gradient of a Function
In calculus, the gradient of a function represents the direction and rate of the steepest ascent of that function. Think of hiking up a hill: the gradient points you to the steepest path upwards. It is a vector that comprises all the partial derivatives with respect to each variable.

To calculate the gradient, we find the partial derivatives of the function for each variable. In our exercise, the function is given as f(x, y) = x \(\cos 3y\). The partial derivative with respect to x is \(\cos 3y\), and with respect to y, it is -3x \sin 3y. Hence, the gradient vector is \([\cos 3y, -3x \sin 3y]\).

At the point (2,0), this becomes \([1, 0]\), indicating that the steepest ascent at this point is in the direction of the x-axis and the rate of ascent (the magnitude of the gradient) is 1.
Partial Derivatives
Partial derivatives are used to measure how a function changes as only one of its input variables changes, holding the others constant. They are essentially the slopes of the function graph along the respective axes.

In our problem, we deal with the function f(x, y), which has two input variables: x and y. Thus, we calculate two partial derivatives: \(\frac{{\partial f}}{{\partial x}}\), which tells us how the function changes with x when y is held constant, and \(\frac{{\partial f}}{{\partial y}}\), indicating how it changes with y when x is held constant.

The partial derivatives we obtained are \(\cos 3y\) for x, and -3x \sin 3y for y. Importantly, these derivatives give us the components of the gradient vector, which is central to understanding the behavior of the function.
Maximum and Minimum Rates of Change
Determining the maximum and minimum rates of change of a function at a given point is akin to finding where you are on a hill. The maximum rate of change is akin to knowing the steepest way up, and the minimum rate is the steepest way down. These rates correspond to the magnitude of the gradient vector and its negative, respectively.

In the context of our example, the maximum rate of change at the point (2,0) occurs in the direction of the gradient [1, 0] which we evaluated earlier. The magnitude of this gradient, 1, gives us the maximum rate.

The minimum rate of change occurs in the opposite direction, [-1, 0], but interestingly, its magnitude is also 1, because we are dealing with a flat slope in the direction of y. This tells us that the function is not changing at all in the direction of the y-axis at the point (2,0).

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Most popular questions from this chapter

Find the absolute extrema of the function on the region. \(f(x, y)=x^{2}+y^{2}-4 x y,\) region bounded by \(y=x, y=-3\) and \(x=3\)

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Find the maximum of \(x^{2}+y^{2}\) on the square with \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1 .\) Use your result to explain why a computer graph of \(z=x^{2}+y^{2}\) with the graphing window \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1\) does not show a circular cross section at the top.

The accompanying data show the average number of points professional football teams score when starting different distances from the opponents' goal line. (For more information, see Hal Stern's "A Statistician Reads the Sports Pages" in Chance, Summer \(1998 .\) The number of points is determined by the next score, so that if the opponent scores next, the number of points is negative.) Use the linear model to predict the average number of points starting (a) 60 yards from the goal line and (b) 40 yards from the goal line. $$\begin{array}{|c|c|c|c|c|c|}\hline \text { Yards from goal } & 15 & 35 & 55 & 75 & 95 \\\\\hline \text { Average points } & 4.57 & 3.17 & 1.54 & 0.24 & -1.25 \\\\\hline\end{array}$$

In exercise \(3,\) there is a saddle point at \((0,0) .\) This means that there is (at least) one trace of \(z=x^{3}-3 x y+y^{3}\) with a local minimum at (0,0) and (at least) one trace with a local maximum at \((0,0) .\) To analyze traces in the planes \(y=k x\) (for some constant \(k\) ), substitute \(y=k x\) and show that \(z=\left(1+k^{3}\right) x^{3}-3 k x^{2} .\) Show that \(f(x)=\left(1+k^{3}\right) x^{3}-3 k x^{2}\) has a local minimum at \(x=0\) if \(k<0\) and a local maximum at \(x=0\) if \(k>0 .\) (Hint: Use the Second Derivative Test from section \(3.5 .)\)

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