# Newton's law of cooling is given by $\theta=\theta_0^{\circ}e^{-kt},$ where the excess of temperature at zero time is $0^{\circ}_{\circ}C$ and at time $t$ seconds is $\theta ^{\circ}C$. Determine the rate of change of temperature after $40 s,$ given that $\theta_{\cdot}=16^{\circ}C$ and $k=-0.03.\qquad (e^{1.2}=3.3201)$

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}=f'(x)$ is the rate of change of $y$. w.r.t $x$
• $\large\frac{dy}{dx_{(x_1,y_1}}$ is the slope of the tangent to the curve at the point $(x_1,y_1)$ on the curve. It is the slope of the curve at that point.
• The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
$\theta=\theta_0e^{-kt}$($\theta$ in $^{\circ}c$)
Step 1:
Rate of change of temperature is
$\large\frac{d\theta}{dt}$$=\theta_0.e^{-kt}.(-k) \qquad=-k\theta_0e^{-kt} Step 2: When t=40\;secs, \; \theta_0=16 ^{\circ}c and k=-0.03 \large\frac{d \theta}{dt}$$=+(0.03)(16)e^{(0.03)40}$
$\qquad=0.48e^{1.2}$
$\qquad=0.48 \times 3.3201$
$\quad=1.594 ^{\circ}c/s$