The function T = ln(x^2 + y^2) gives the temperature at each...

The function $T = \ln(x^2 + y^2)$ gives the temperature at each point in the plane (except (0,0)). a) At the point P: (1,2), in which direction should you go to get the most rapid increase in T? b) At P, about how far should you go in the direction found in part (a) to get an increase of .20 in T? c) At P, approximately how far should you go in the direction of i+j to get an increase of about .12? d) At P, in which direction(s) will the temperature change rate be 0?

Transcript

00:01 Hello students, the gradient is given by gradient of t x, y, z equal to del t by del x, del t by del y, del t by del z.

00:21 Given the temperature function is t of x, y, z which is equal to 200 into e to the power exponential of minus x square minus y square by 4 minus z square by 9.

00:40 Now we compute the partial derivatives with respect to x, y and z.

00:45 So del t by del x equal to minus 400 x into exponential of minus x square minus y square by 4 minus z square by 9 minus z square by 9 del t by del y equal to minus 100 y into exponential of minus x square minus y square by 4 minus z square by 9 and del t by del z equal to minus 400 z by 9 to exponential of minus x square minus y square by 4 minus z square by 9.

01:30 Now at the point at the point 1 comma 1 comma 1 we have del t by del x del t by del x equal to minus 101 .64 del t by del y equal to minus 25 .41 and del t by del z is approximately equal to minus 11 .29.

02:04 So the gradient at 1 comma 1 comma 1 point is equal to minus 101 .64 comma minus 25 .41 comma minus 11 .29 to find the unit vector in the direction toward the point 5 comma 5 comma minus 1 5 comma 5 comma minus 1 we need to calculate the direction vector first.

02:35 So the direction vector is equal to 5 minus 1 comma 5 minus 1 comma minus 1 minus 1.

02:49 This is equal to 4 comma 4 comma minus 2.

02:54 Then normalize this direction vector to get the unit vector.

02:58 So the magnitude of the direction vector is equal to 6.

03:13 So the unit vector is equal to 4 by 6 comma 4 by 6 comma minus 2 by 6 that is equal to 2 by 3 comma 2 by 3 comma minus 1 by 3.

03:29 Now we can find the rate of change of the temperature at 1 comma 1 comma 1 in the direction toward the given point by taking the dot product of the gradient and unit vector.

03:43 So rate of change is equal to gradient of t into 1 comma 1 comma 1 dot unit vector.

03:57 So this is equal to minus 101 .64 comma minus 25 .41 comma minus 11 .29 dot 2 by 3 comma 2 by 3 comma minus 1 by 3.

04:15 So this is equal to minus 67 .67 minus 67 .76 comma minus 16 .94 is plus plus 3 .77.

04:38 So this is equal to minus 80 .93...

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