At least one of the answers above is NOT correct. 2 of the questions remain unan
ID: 2877912 • Letter: A
Question
At least one of the answers above is NOT correct. 2 of the questions remain unanswered. The temperature at any point in the plane is given by 1 (x.y) = 190/x^2 + y^2 + 2. What shape are the level curves of T? Ellipses lines parabolas circles hyperbolas none of the above At what point on the plane is it hottest? What is the maximum temperature? Find the direction of the greatest increase in temperature at the point (3, -3). What is the value of this maximum rate of change, that is the maximum value of the directional derivative at (3, -3)? Find the direction of the greatest decrease in temperature at the point (3, -3). What is the value of this most negative rate of change, that is the minimum value of the directional derivative at (3, -3)?Explanation / Answer
(a) c = 190/ x ^2 + y ^2 + 2
x ^2 + y ^2 + 2 = 190/ c
so x ^2 + y ^2 = 190/ c 2
which is the formula for a circle in 2D LIKE x^ 2 + y^ 2 =r^2
(b) The maximum temperature occurs when x^2 + y^2 + 2 is minimal, namely when (x, y) = (0, 0).
This yields the maximal temperature T(0, 0) = 60.
(c) Use the gradient
T = <-190(2x)/(x^2 + y^2 + 2)^2, -190(2y)/(x^2 + y^2 + 2)^2>.
T = <-380x/(x^2 + y^2 + 2)^2, -380y/(x^2 + y^2 + 2)^2>.
u = T(3, -3)/||T(3, -3)||
(x^2+y^2+2)^2=> ((3)^2+(-3)^2+2) => (9+9+2)^2 =>400
T(3, -3= <-380(3)/400, -380(-3)/400>
T(3, -3 => <-27/20,27/20>/(18)
or
like the direction of the greatest increase in temperature is
=> -19 /20( 3i ) - 19 /20(-3 j)
=> (-19 /20)(3i-3j)
the magnitude is (19 /20)((3)^2+(-3)^2))=> (19 /20)(9+9)=> (19 /20)(18)
Du T(3, -3) = T(3, -3) · u = ||T(3, -3)|| => (19 /20)(18)
(d)The direction of greatest decrease in temperature is
This time,
use u = -T(3, -3)/||T(3, -3)||
= -(-19 /20)(3i-3j)
=> (19 /20)(3i-3j)
the magnitude is (19 /20)((3)^2+(-3)^2))=> (19 /20)(9+9)=> (19 /20)(18)
Du T(3, -3) = T(3, -3) · u = -||T(3, -3)|| => - (19 /20)(18)
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