Question:

If the area of a circle increases at a uniform rate, then its perimeter varies

Updated On: May 12, 2024
  • directly as its radius
  • inversely as its radius
  • directly as the square of its radius
  • inversely as the square of the radius
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The Correct Option is A

Solution and Explanation

Area of circle = $\pi r^2$ and perimeter = $2\pi r$
Let $f(r) = \pi r^2 \Rightarrow \: f'(r) = 2 \pi r$
$f(r)$ is increases $\Rightarrow \:\: f'(r) \geq 0$
Now $f"(r) = 2\pi > 0$
$\Rightarrow \: f' (r)$ is also an increasing function
$\Rightarrow \:\: f' (r)$ varies directly as its radius
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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives