Derivative as a Rate Measure, Tangents and Normals
1.0 Derivative as a rate of change
1.0 Derivative as a rate of change
Whenever we heard this term "rate of change", one thing comes to our mind is "time". In this topic we are not only restricted to rate of change with respect to time but we will learn how one variable is dependent on the other variable and the rate at which one variable is changing with respect to other.
Generally we represent any function by $f(x)$. So, the rate of change of function with respect to $x$ is represented by $f'(x)$. We categorised rate of change in two types:
1. Average rate of change: The average rate of change of function $f(x)$ with respect to $x$ over an interval $[a,a+h]$ is given by $$f'(x) = \frac{{f(a + h) - f(a)}}{h}$$
2. Instantaneous rate of change: The insyantaneous rate of change of function $f(x)$ with respect to $x$ over an interval $[a,a+h]$ is given by $$f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(a + h) - f(a)}}{h}$$
Example 1. If area of circle increases at a rate of $1{\text{ }}c{m^2}/\sec $, then find the rate at which area of the inscribed square increases.
Solution: As we know that Area of circle $({A_1})$ = $\pi {r^2}$ and Area of Square $({A_2})$= $2{r^2}$ as shown in figure. Differentiate the equations with respect to time, we get $$\frac{{d{A_1}}}{{dt}} = 2\pi r\frac{{dr}}{{dt}},\quad \frac{{d{A_2}}}{{dt}} = 4r.\frac{{dr}}{{dt}}$$ It is given in the question $\frac{{d{A_1}}}{{dt}} = 1$, therefore, using this we get $$\begin{equation} \begin{aligned} 1 = 2\pi r\frac{{dr}}{{dt}} \\ r\frac{{dr}}{{dt}} = \frac{1}{{2\pi }} \\\end{aligned} \end{equation} $$ Put the value in other equation, we get $$\frac{{d{A_2}}}{{dt}} = 4 \times \frac{1}{{2\pi }} = \frac{2}{\pi }$$ Area of square increases at the rate of $\frac{2}{\pi }{\text{ }}c{m^2}/\sec $.
