Question

6. The temperature at (x,y) is T(x,y) = 20 + 100%+++") degrees (in Celsius). A bug carries a tiny thermometer along the path c(t) = (cos(t - 2), sin(t -- 2)) where t is in seconds. What is the initial rate of change of the temperature t = 0.6s?

Answer 1

Given: Tx,y) = 20 + 10e -0.3(+2+y2)

The path moved by the bug or displacement of bug is given by ;

c(t) = (cos(t – 2), sin(t – 2)

  
= cos(t - 2)i + sin(t - 2)

The displacement is usually denoted by  
ci+yi

• Therefore the x component is ,
  2 = cos(t - 2
  
• Therefore the y component is ,
  y = sin(t – 2
  

We have to find rate of change of temperature at t=0.6

dT dt t=0.6

To find  rate of change of temperature T , we use the chain rule

dT dt dT do * dr dt + dT dy dy dt

Now we have to find each part

• 
  dT do 20 + 10e -0.3(rº+y) dir
  

C (20) + de [10 -0.3672 +3 ***] 10e

  
d 0+10 dic [e -0.3(x2 +ya e

-0.3(r+y?) = 10 dr e

= 10e -0.3(x2 +y*) * (-0.3(x2 + y?) *

  
= 10e-0.3(zº+y®)* -0.3 (22 + y²)

  
10e-0.3(r2 +y) * -0.3(2.c + 0)

  
= 10e-0.3(zº+y)* -0.6.2

  
-6.ce-0.3(x2+y)

• 
  dT dy 20 + 10e-0.3(r+y) dy
  

  
0 + 10---0.3(x2+y2) dy

$=\left [10e^{-0.3(x^{2}+y^{2})}\frac{\mathrm{d} }{\mathrm{d} y}{-0.3(x^{2}+y^{2})} \right ]$

$=\left [10e^{-0.3(x^{2}+y^{2})}*{-0.3(0+2y)} \right ]$

= 10e -0.3(+2+y?) -0.6y

= –буе -0.3(x2+y2),

• 
  2 = cos(t - 2
  

dir dt (cos(t dt 2)]

$=sin(t-2)\frac{\mathrm{d} }{\mathrm{d} t}\left [ t-2\right ]$

$=-sin(t-2)(1-0)$

  
= -sin(t - 2)

• 
  y = sin(t – 2
  

hip sin(t dt 2

= cos(t – 2) (t - 2) dt

$=cos(t-2)(1-0)$

$=cos(t-2)$

dT dt dT do * dr dt + dT dy dy dt

$=-6xe^{-0.3(x^{2}+y^{2})}*-sin(t-2)+-6ye^{-0.3(x^{2}+y^{2})}*cos(t-2)$

dT dt t=0.6
  

-6*cos(0.6-2)* -0.3(rº+y)*-sin(0.6–2)+-6*sin(0.6-2)* -0.3(r2+ya)

  
= 6 * sin(-1.4)cos(-1.4) *e-0.3(r2+y) +-6 * sin(-1.4)cos(-1.4)*
​​​​​​​
-0.3(32 +y) e

$=0$