Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. If we measure temperature in degrees Celsius and time in minutes, the constant of proportionality k equals 0.3. Suppose the ambient temperature TA(t) is equal to a constant 70 degrees Celsius. Write the differential equation that describes the time evolution of the temperature T of the object.

a. dT/dt =

b. Suppose the ambient temp TA(t) = 66cos((?/30)t) degrees Celsius (time measured in minutes). Write the DE that describes the time evolution of temperature T of the object.

dT/dt =

c. If we measure time in hours the differential equation in part (b) changes. What is the new differential equation?

dT/dt =

d. If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more. What is the new differential equation?

dT/dt =

a) (dT/dt) = -0.3 [T - 70]

b) (dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}

c) (dT/dt) = -18 {T - [66 cos (2πt)]}

with t in hours

d) (dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]

with T in Fahrenheit and t in hours

Explanation:

The Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings.

If the temperature of the object = T

Temperature of the surroundings = Ambient temperature = TA(t)

(dT/dt) ∝[T - TA(t)]

Introducing the constant of proportionality, k

(dT/dt) = k [T - TA(t)]

Temperature is in degree Celsius and time is in minutes.

Because the temperature of the body is decreasing, we introduce a minus sign

(dT/dt) = -k [T - TA(t)]

a) If TA(t) = 70°C, k = 0.3

(dT/dt) = -0.3 [T - 70]

b) The ambient temp TA(t) = 66 cos ((π/30)t) degrees Celsius (time measured in minutes).

(dT/dt) = -k [T - TA(t)]

(dT/dt) = -k {T - [66 cos ((π/30)t)]}

(dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}

c) If we measure time in hours the differential equation in part (b) changes.

1 hour = 60 mins

If t is now expressed in hours,

t hours = (60t) mins

(dT/dt) = -k {T - [66 cos ((π/30)t)]}

dT = -k {T - [66 cos ((π/30)t)]} dt

dT = -k {T - [66 cos ((π/30)60t)]} d(60t)

(dT) = -60k {T - [66 cos ((π/30)60t)]} dt

(dT/dt) = -60k {T - [66 cos (2πt)]}

with t in hours, k = 0.3, 60k = 18

(dT/dt) = -18 {T - [66 cos (2πt)]}

d) If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more.

If T is in degree Fahrenheit

T°F = (5/9)(T°F - 32) degrees Celsius

T°F = [(5T/9) - 17.78] degrees Celsius

(dT/dt) = -60k {T - [66 cos (2πt)]}

time already converted to hours.

dT = -60k {T - [66 cos (2πt)]} dt

66 cos (2πt) degrees Celsius = {(9/5) [66 cos (2πt)] + 32} degrees Fahrenheit = {[118.8 cos (2πt)] + 57.6} degrees Fahrenheit

d[(5T/9) - 17.78] = -60k {T - [118.8 cos (2πt) + 57.6]} dt

(5/9) dT = -60k [T - 57.6 - 118.8 cos (2πt)] dt

(5/9) (dT/dt) = -60k [T - 57.6 - 118.8 cos (2πt)]

(dT/dt) = -108k [T - 57.6 - 118.8 cos (2πt)]

k = 0.3, 108k = 32.4

(dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]

with T in Fahrenheit and t in hours

Hope this Helps!!!

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)