- #1

Poetria

- 267

- 42

- Homework Statement
- You are swimming along the surface of a large natural hot water spring. The temperature is hottest near the geothermal heat sources, and cools off inversely proportional to the distance from the heat source as you move away. The hot spring you have found has two heat sources. One is located below x=0, y =0, the other is located x=-10, y=20

Approximation of the temperature:

##T(x,y) = \frac {450} {\sqrt{x^2+y^2+1}}+\frac {420} {\sqrt{(x+10)^2+(y-20)^2+1}}##

You enter the pool at the edge x=20, y=20, where the temperature is 30 degrees Celsius.

What is the gradient of the temperature?

- Relevant Equations
- $$\nabla (\frac {450} {\sqrt{x^2+y^2+1}}+\frac {420} {\sqrt{x+10)^2+(y-20)^2+1}})$$

I have computed

##T_x## and ##T_y## and evaluated it at the point (20, 20).

## \frac {-450*x}{x^2 + y^2 + 1)^(\frac {3} {2}} - \frac {420*(x + 10)} {(x + 10)^2 + (y - 20)^2 + 1)^(\frac {3} {2}}, \frac {-450*y}{x^2 + y^2 + 1)^(\frac {3} {2}} - \frac {420*(y - 20)} {(x + 10)^2 + (y - 20)^2 + 1)^(\frac {3} {2}}##

I got [-0.8628, -0.3970]. But it is not correct.

So the answer to the next question: At what rate does the temperature rise per unit distance in that direction?

is not correct:

-2.1732 degrees per meter

##T_x## and ##T_y## and evaluated it at the point (20, 20).

## \frac {-450*x}{x^2 + y^2 + 1)^(\frac {3} {2}} - \frac {420*(x + 10)} {(x + 10)^2 + (y - 20)^2 + 1)^(\frac {3} {2}}, \frac {-450*y}{x^2 + y^2 + 1)^(\frac {3} {2}} - \frac {420*(y - 20)} {(x + 10)^2 + (y - 20)^2 + 1)^(\frac {3} {2}}##

I got [-0.8628, -0.3970]. But it is not correct.

So the answer to the next question: At what rate does the temperature rise per unit distance in that direction?

is not correct:

-2.1732 degrees per meter

Last edited: