Question

E temperature t of a metal sphere is inversely proportional to the distance from the centre of the sphere (the origin (0, 0, 0)). E temperature at the point (1, 0, 0) is 120○c

Answer 1

(-x, -y, -z) the vector form (x, y, z) to the origin, in the direction of the greatest increase.

How to calculate temperature?

The quantity or amount of radiation contained in material or item, as measured by a temperature or sensed by touch and stated on a numerical scale.

The temperature T of a ball bearing is inverse to the length first from the origin, which we consider to be the center of the ball. The temperature at the exact location

(x, y, z) is a point on the sphere, the temperature at this point is given by

$T(x,y,z)= \dfrac{k}{(x^2 +y^2+z^2)^{1/2}}$

Where k is constant, then we have

T(1, 2, 2) = k/3 = 170

k = 510

So

$T(x,y,z)= \dfrac{510}{(x^2 +y^2+z^2)^{1/2}}$

Then we have

$\rm \triangledown T = \left ( -\dfrac{510x}{(x^2 +y^2+z^2)^{1/2}}, -\dfrac{510y}{(x^2 +y^2+z^2)^{1/2}}, \dfrac{510z}{(x^2 +y^2+z^2)^{1/2}} \right )\\\\\\\triangledown T (1, 2, 2) = -\dfrac{510}{27} (1, 2, 2) =-\dfrac{170}{9} (1, 2, 2)$

Then the direction will be from (1, 2, 2) to (4, 3, 5) will be (3, 1, 3)

So u = (3/√19, 1/√19, 3/√19)

Then we get

$\rm \triangledown T \cdot u = \dfrac{-170}{9}(1, 2, 2) \cdot \left (\dfrac{3}{\sqrt{19}}, \dfrac{1}{\sqrt{19}}, \dfrac{3}{\sqrt{19}} \right )\\\\\\\triangledown T \cdot u = - \dfrac{170}{9\sqrt{19}} (3 + 2 + 6)\\\\\\\triangledown T \cdot u = -\dfrac{1870}{9\sqrt{19}}$

The direction of the greatest inverse in the temperature is given by any vector parallel to and having the same direction $\triangledown$T. (-x, -y, -z) the vector form (x, y, z) to the origin, in the direction of the greatest increase.

More about the temperature link is given below.

brainly.com/question/11464844

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