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Chapter 15: Problem 20

Computing gradients Compute the gradient of the following functions and evaluate it at the given point \(P\). $$h(x, y)=\ln \left(1+x^{2}+2 y^{2}\right) ; P(2,-3)$$

Short Answer

Expert verified
Answer: The gradient of the function h at point P is \(\left(\frac{4}{23}, -\frac{12}{23}\right)\).

Step by step solution

01

Find the partial derivatives

Differentiate the given function \(h(x, y)=\ln \left(1+x^{2}+2 y^{2}\right)\) with respect to the variables x and y, i.e., find \(\frac{\partial h}{\partial x}\) and \(\frac{\partial h}{\partial y}\). Using chain rule, we have: $$\frac{\partial h}{\partial x}=\frac{1}{1+x^{2}+2 y^{2}}\left(\frac{\partial}{\partial x}(1+x^{2}+2 y^{2})\right)$$ $$\frac{\partial h}{\partial y}=\frac{1}{1+x^{2}+2 y^{2}}\left(\frac{\partial}{\partial y}(1+x^{2}+2 y^{2})\right)$$ Now, take the derivatives inside the parentheses: $$\frac{\partial}{\partial x}(1+x^{2}+2 y^{2}) = 0 + 2x + 0 = 2x$$ $$\frac{\partial}{\partial y}(1+x^{2}+2 y^{2}) = 0 + 0 + 4y = 4y$$ So, the partial derivatives are: $$\frac{\partial h}{\partial x}=\frac{2x}{1+x^{2}+2 y^{2}}$$ $$\frac{\partial h}{\partial y}=\frac{4y}{1+x^{2}+2 y^{2}}$$
02

Evaluate the partial derivatives at the point P

Now, substitute the given point \(P(2, -3)\) into the partial derivatives we found in Step 1: $$\frac{\partial h}{\partial x}(2, -3)=\frac{2(2)}{1+(2)^{2}+2 (-3)^{2}}=\frac{4}{1+4+18}=\frac{4}{23}$$ $$\frac{\partial h}{\partial y}(2, -3)=\frac{4(-3)}{1+(2)^{2}+2 (-3)^{2}}=-\frac{12}{23}$$
03

Compute the gradient

Finally, we can write the gradient of the function \(h\) as a vector with the evaluated partial derivatives: $$\nabla h(P)=\left(\frac{\partial h}{\partial x}(P), \frac{\partial h}{\partial y}(P)\right)=\left(\frac{4}{23}, -\frac{12}{23}\right)$$ So, the gradient of the function \(h(x, y)=\ln \left(1+x^{2}+2 y^{2}\right)\) evaluated at point \(P(2, -3)\) is \(\left(\frac{4}{23}, -\frac{12}{23}\right)\).

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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 a fundamental skill in multivariate calculus. They represent the rate of change of a multivariable function with respect to one variable while holding the other variables constant. For a function like \( h(x, y) = \ln(1 + x^2 + 2y^2) \), we compute partial derivatives separately for \( x \) and \( y \).

When we differentiate \( h \) with respect to \( x \), the \( y \)-related components are treated as constants. Similarly, when differentiating with respect to \( y \), all \( x \)-components are constants. In the case of \( h \), the partial derivative with respect to \( x \) is \( \frac{2x}{1+x^2+2y^2} \) and with respect to \( y \) is \( \frac{4y}{1+x^2+2y^2} \).

These computations are crucial for analyzing the function's behavior in the vicinity of a specific point, like \( P(2, -3) \) in the exercise.
Chain Rule
The chain rule is a powerful derivative tool that allows us to differentiate compositions of functions. When a function, such as the natural logarithm in our example, is applied to another function, such as \( 1 + x^2 + 2y^2 \), the chain rule guides us to first differentiate the outer function, leaving the inner function alone, and then multiply that by the derivative of the inner function.

Utilizing the chain rule, we acknowledge that the derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \), and then we find \( u \)'s derivative with respect to \( x \) or \( y \), depending on which partial derivative we are calculating. This step-by-step process ensures we account for how each variable's change affects the overall function.
Gradient of a Function
The gradient of a function paints a complete picture of the function's rate of change in all directions. It is a vector comprised of all the partial derivatives of the function with respect to its variables. For the given exercise, the gradient is not just a set of numbers; it's an arrow pointing in the direction of the steepest ascent from the point \( P(2, -3) \), and its magnitude tells us how steep the hill is at that point.

To compute the gradient \( abla h \), we found the partial derivatives and evaluated them at \( P \). This gives us the gradient vector \( \left(\frac{4}{23}, -\frac{12}{23}\right) \), a crucial concept in fields like machine learning, physics, and optimization as it facilitates understanding of how to navigate the terrain of a function most efficiently.

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Most popular questions from this chapter

Batting averages Batting averages in bascball are defined by \(A=x / y,\) where \(x \geq 0\) is the total number of hits and \(y>0\) is the total number of at- bats. Treat \(x\) and \(y\) as positive real numbers and note that \(0 \leq A \leq 1\) a. Use differentials to estimate the change in the batting average if the number of hits increases from 60 to 62 and the number of at-bats increases from 175 to 180 . b. If a batter currently has a batting average of \(A=0.350,\) does the average decrease more if the batter fails to get a hit than it increases if the batter gets a hit? c. Does the answer to part (b) depend on the current batting average? Explain.

Use Lagrange multipliers in the following problems. When the constraint curve is unbounded, explain why you have found an absolute maximum or minimum value. Maximum area rectangle in an ellipse Find the dimensions of the rectangle of maximum area with sides parallel to the coordinate axes that can be inscribed in the ellipse \(4 x^{2}+16 y^{2}=16\)

Find the absolute maximum and minimum values of the following functions over the given regions \(R .\) \(f(x, y)=2 x^{2}-4 x+3 y^{2}+2\) \(R=\left\\{(x, y):(x-1)^{2}+y^{2} \leq 1\right\\}\) (This is Exercise 51 Section \(15.7 .)\)

A snapshot (frozen in time) of a set of water waves is described by the function \(z=1+\sin (x-y),\) where \(z\) gives the height of the waves and \((x, y)\) are coordinates in the horizontal plane \(z=0\) a. Use a graphing utility to graph \(z=1+\sin (x-y)\) b. The crests and the troughs of the waves are aligned in the direction in which the height function has zero change. Find the direction in which the crests and troughs are aligned. c. If you were surfing on one of these waves and wanted the steepest descent from the crest to the trough, in which direction would you point your surfboard (given in terms of a unit vector in the \(x y\) -plane)? d. Check that your answers to parts (b) and (c) are consistent with the graph of part (a).

The output \(Q\) of an economic system subject to two inputs, such as labor \(L\) and capital \(K,\) is often modeled by the Cobb-Douglas production function \(Q(L, K)=c L^{a} K^{b},\) where \(a, b,\) and \(c\) are positive real numbers. When \(a+b=1,\) the case is called constant returns to scale. Suppose \(a=1 / 3, b=2 / 3,\) and \(c=40\). a. Graph the output function using the window \([0,20] \times[0,20] \times[0,500]\). b. If \(L\) is held constant at \(L=10,\) write the function that gives the dependence of \(Q\) on \(K\). c. If \(K\) is held constant at \(K=15,\) write the function that gives the dependence of \(Q\) on \(L\).
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