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Chapter 12: Q. 32 (page 989)

Find the gradient of the given function, and find the direction in which the function increases most rapidly at the specified point P.

f(x,y)=xy鈥�P=(4,3)

Short Answer

Expert verified

13,鈭�49

Step by step solution

01

Step 1. Given Information

The gradient of a functionf(x,y)is the vector function defined by鈭�f(x,y)=鈭�f鈭�xi+鈭�f鈭�yj,Heref(x,y)=xy.

02

Step 2. Solution

So its gradient is given by,

鈭�f(x,y)=鈭�鈭�xxyi+鈭�鈭�yxyj=1yi鈭�xy2jTherefore鈭�f(x,y)=1y,鈭�xy2

As the gradient of a functionfat a pointp, points in the direction in whichfincreases mostrapidly.Atp=(4,3),鈭�f(4,3)=13i鈭�49jHence the direction in whichfincreases most rapidly atp=(4,3)is鈭�f(4,3)=<13,鈭�49>

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Most popular questions from this chapter

Describe the meanings of each of the following mathematical expressions:

鈭�f(x,y,z)

Use Theorem 12.33 to find the indicated derivatives in Exercises 27鈥�30. Express your answers as functions of two variables.

鈭�z鈭�swhenz=x2y3,x=tsins,andy=scost

In Exercises 37-42, use the partial derivatives of fx,y=xsinyand the point 2,蟺3specified to

afind the equation of the line tangent to the surface defined by the function in the xdirection,

bfind the equation of the line tangent to the surface defined by the function in the ydirection, and

cfind the equation of the plane containing the lines you found in parts aand b.

Gradients: Find the gradient of the given function, and find the direction in which the function increases most rapidly at the specified point P.

f(x, y ,z) = ln(x + y + z), P = (e, 0, 鈭�1) .

In Example 4 we found that the function f(x,y)=x3+9x2+6y2has stationary points at (0,0)and (-6,0).

(a) Use the second-derivative test to show that \(f\) has a saddle point at(-6,0).

(b) Use the second-derivative test to show that \(f\) has a relative minimum at(0,0).

(c) Use the value of \(f(-10,0)\) to argue that \(f\) has a relative minimum at (0,0),and not an absolute minimum, without using the second-derivative test.
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