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JEE Main Chemistry High Weightage Chapters and Topics 2025

Derivative as Rate Measure - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • Derivative as Rate Measure is considered one of the most asked concept.

  • 24 Questions around this concept.

Solve by difficulty

The real number x when added to its inverse gives the minimum value of the sum at x equal to

If x is real, the maximum value of  \frac{3x^{2}+9x+17}{3x^{2}+9x+7} is :

The function g\left ( x \right )= \frac{x}{2}+\frac{2}{x} has a local minimum at 

Concepts Covered - 1

Derivative as Rate Measure

Derivative as Rate Measure

If two related quantities are changing over time, the rates at which the quantities change are related. For example, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon are increasing. 

If a variable quantity y depends on and varies with a quantity x, then the rate of change of y with respect to x is  \frac{dy}{dx}.

A rate of change with respect to time is simply called the rate of change.

For example, the rate of change of displacement (s) of an object w.r.t. time is velocity (v). 

\mathit{v=\frac{ds}{dt}}

 

Illustration:

Consider balloon example again, a spherical balloon is being filled with air at a constant rate of 2 cm3 /sec. How fast is the radius increasing when the radius is 3 cm?

The volume of a sphere of radius r centimeters is,

V=\frac{4}{3} \pi r^{3} \mathrm{cm}^{3}

Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, t seconds after beginning to fill the balloon with air, the volume of air in the balloon is

V(t)=\frac{4}{3} \pi[r(t)]^{3} \mathrm{cm}^{3}

Differentiating both sides of this equation with respect to time and applying the chain rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equation

\\\frac{d}{dt}\left (V(t) \right )=\frac{4}{3} \pi\;\frac{d}{dt}\left ([r(t)]^{3} \right ) \mathrm{cm}^{3}\\\\\text{ }\;\;\;\;\;V^{\prime}(t)=4 \pi[r(t)]^{2} r^{\prime}(t)

The balloon is filled with air at a constant rate of 2 cm3 /sec. So, V’(t) = 2 cm3 /sec

\\\therefore \text{}\;\;\;\;\;\;\;2\;\text{cm}^3/\text{sec}=\left (4 \pi[r(t)]^{2}\text{cm}^2 \right )\cdot r^{\prime}(t)\\\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;r^{\prime}(t)=\frac{1}{2 \pi[r(t)]^{2}} \mathrm{cm} / \mathrm{sec}\\\\\text{When the radius r = 3 cm}\\\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;r^{\prime}(t)=\frac{1}{18 \pi} \mathrm{cm} / \mathrm{sec}

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Derivative as Rate Measure

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Derivative as Rate Measure

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 5.12

Line : 38

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