**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}$

$\qquad=0.48e^{1.2}$

$\qquad=0.48 \times 3.3201$

$\quad=1.594 ^{\circ}c/s$