The temperature at any point in the plane is given by T(x, y) = 190/x^2 + y^2 +
ID: 2894771 • Letter: T
Question
The temperature at any point in the plane is given by T(x, y) = 190/x^2 + y^2 + 2. (a) What shape are the level curves of T? A. circles B. lines C. ellipses D. parabolas E. hyperbolas F. none of the above (b) At what point on the plane is it hottest? _______ What is the maximum temperature? ________ (c) Find the direction of the greatest increase in temperature at the point (3,-2). __________ What is the value of this maximum rate of change, that is, the maximum value of the directional derivative at (3, -2)? _________ (d) Find the direction of the greatest decrease in temperature at the point (3,-2). _________ What is the value of this most negative rate of change, that is, the minimum value of the directional derivative at (3,-2) _________Explanation / Answer
T(x,y)=190/(x2+y2+2)
a)
190/(x2+y2+2)=k
=>(x2+y2+2)=190/k
=>(x2+y2)=(190/k) -2
shape of level curves of T is circles
(b)
point in the plane is hottest where T(x,y)=190/(x2+y2+2) is maximum
T(x,y)=190/(x2+y2+2) is maximum when x2+y2+2 is minimum
minimum value of x2+y2+2 is 2 at (x,y)=(0,0)
so plane is hottest at (0,0)
maximum temperature =T(0,0)
maximum temperature =190/(02+02+2)
maximum temperature =95
(c)
T(x,y)=190/(x2+y2+2)
T=<-380x/(x2+y2+2),-380y/(x2+y2+2)>
at (3,-2)
T=<-380*3/(32+(-2)2+2),-380*-2/(32+(-2)2+2)>
T=<-380*3/15,-380*-2/15>
T=<-76,152/3>
direction of greatest increase of temperature =T
direction of greatest increase of temperature =<-76,152/3>
maximum rate of change =|T|
maximum rate of change =[(-76)2+(152/3)2]
maximum rate of change =(7613)/3
d)direction of greatest decrease of temperature =-T
direction of greatest decrease of temperature =<76,-152/3>
most negative rate of change =-|T|
most negative rate of change =-[(-76)2+(152/3)2]
most negative rate of change =-(7613)/3
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