In this page rate of change question4 we are going to see solution of some practice question of the worksheet.

**Question 4:**

Newton's law of cooling is given by θ = θ₀° e^(⁻kt), where the excess of temperature at zero time is θ₀° C and at time t seconds is θ° C. Determine the rate of change of temperature after 40 s ,given that θ₀ = 16° C and k = -0.03.(e^1.2 = 3.3201)

**Solution:**

Newton's law of cooling θ = θ₀° e^(⁻kt)

θ₀° = 16° C

k = -0.03

rate of change of temperature with respect to time

dθ/dt = - k θ₀° e^(⁻kt)

t = 40

dθ/dt = - (-0.03) (16)e^(⁻0.03) (40)

= 0.48 e^(⁻1.2)

= 0.48 (3.3201)

= 1.5936° C/s

- Back to worksheet
- First Principles
- Implicit Function
- Parametric Function
- Substitution Method
- logarithmic function
- Product Rule
- Chain Rule
- Quotient Rule
- Rolle's theorem
- Lagrange's theorem
- Finding increasing or decreasing interval
- Increasing function
- Decreasing function
- Monotonic function
- Maximum and minimum
- Examples of maximum and minimum