Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 19

How fast is the base of the triangle changing when the altitude is 10cm and the area is 100cm$^2$

Illustration







Given:

$\qquad \displaystyle \frac{dh}{dt} = 1 cm/min$

$\qquad \frac{dA}{dt} = 2 cm^2 /min$

Required: $\displaystyle \frac{db}{dt}$ when $h = 10 cm$ and $A = 100cm^2$

Solution:


$\begin{equation} \begin{aligned} A =& \frac{1}{2} bh, &&\text{area of triangle} \\ \\ \frac{dA}{dt} =& \frac{1}{2} \left[ b \frac{dh}{dt} + h \frac{db}{dt} \right] && \text{(Derivative with respect to time)} \\ \\ \frac{db}{dt} =& \frac{\displaystyle 2 \frac{dA}{dt} - b \frac{dh}{dt}}{h} \end{aligned} \end{equation}$


To get the value of $b$, we will use the formula of area of triangle


$\begin{equation} \begin{aligned} A =& \frac{1}{2} bh \\ \\ b =& \frac{2A}{h} = \frac{2(100)}{10} = 20 cm \end{aligned} \end{equation}$


To solve for the unknown,


$\begin{equation} \begin{aligned} & \frac{db}{dt} = \frac{2(2) - 20(1)}{10} \\ \\ &\boxed{ \displaystyle \frac{db}{dt} = -1.6 cm /min} \qquad \text{(It means that the rate is decreasing)} \end{aligned} \end{equation}$


The length of the shadow is decreasing at a rate of $0.6 m/s$ when the man is $4m$ from the building.