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    Five-year Analysis: Calculus Made Easy To Ace NEET

    Five-year Analysis: Calculus Made Easy To Ace NEET

    Ramraj SainiUpdated on 24 May 2022, 09:15 AM IST

    Calculus is an important subject for NEET aspirants. In 2021, a total of 12 questions asked in Physics and Chemistry were related to Calculus. Calculus is a branch of Mathematics that deals with mainly two operations, i.e, Differentiation and Integration. Calculus is widely used as a tool to determine quantities like the rate of change, gradient of curve, and area under curves. In Physics, most of the chapters in Class 11 and 12 have some topics relating to Calculus. Similarly, in Chemistry, a few chapters like Thermodynamics and Chemical Kinetics have some topics that demand a basic understanding of Calculus. But NEET aspirants don't need to master Calculus because it is very vast and the exam demands only a very basic knowledge of that. This article explains the physical significance of Calculus, the difference between differential and integral Calculus, and also the relation between them. This article also includes an analysis of the previous five year's papers to help you understand the concepts of Calculus easily.

    Five-year Analysis: Calculus Made Easy To Ace NEET
    Basic Calculus For NEET(Image: Shutterstock)

    Physical Significance Of Differential And Integral Calculus

    Whenever we differentiate any mathematical equation, we are basically finding the slope of the equation. Integral Calculus refers to the summation of uncountable or infinitely many elemental areas. Consider an example given in the graph below. Derivative of f(x) at point ‘a’ gives slope or rate of change of f(x) at point ‘a’. Another graph illustrates integration as integrating f(x) in limits b to c and gives the area under that curve.

    1653025894153


    Integration is of two types, indefinite and definite. If lower and upper limits are given then it is called a definite integral.

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    Differential V/S Integral Calculus

    Let there is a variable y that is changing with reference variable x. If there is a change in x that is ∆x and a corresponding change in y that is ∆y. The average rate of change of y with respect to x is given by ∆y/∆x. When the change in x is infinitely small then the rate can be represented as dy/dx or y’, known as differentiation of y with respect to x.

    Integration is the opposite process of differentiation represented as the symbol ”∫ “.

    ∫ (differentiation) = Initial function ⇒ ∫(dy/dx)dx = ∫dy = y.

    Let consider an example y = sinx

    Differentiation of y = dy/dx = cosx

    Integration = ∫cosx dx = sinx = Initial function.

    There are some fundamental formulas to compute differentiation and integration, in which a few of them are listed below.

    Differentiation And Integration Of Some Basic Functions

    Differential Calculus

    Indefinite Integral Calculus

    \frac{d}{dx}(x^n+c)=nx^{n-1}

    \int nx^{n-1}dx=x^n+c

    \frac{d}{dx}(lnx+c)=\frac{1}{x}

    \int \frac{1}{x}dx=ln\left | x \right |+c

    \frac{d}{dx}(e^x+c)=e^x

    \int (e^x)dx=e^x+c

    \frac{d}{dx}(sinx+c)=cos\space x

    \int (cosx)=sinx+c

    \frac{d}{dx}(cosx+c)=-sinx

    \int (-sinx)=cosx+c

    \frac{d}{dx}(tanx+c)=sec^2 x

    \int (sec^2x)dx=tan \space x+c

    Basic Differentiation And Integration Rules

    Rule

    Differentiation

    Integration

    Addition, f(x) + g(x)

    \frac{d}{dx}(f(x)+g(x)) =\frac{d}{dx}f(x) +\frac{d}{dx}g(x)

    \int (f(x)+g(x))dx=\int f(x)dx+\int g(x)dx

    Subtraction, f(x) - g(x)

    \frac{d}{dx}(f(x)-g(x)) =\frac{d}{dx}f(x) -\frac{d}{dx}g(x)

    \int (f(x)-g(x))dx=\int f(x)dx-\int g(x)dx

    Product Rule, f(x).g(x)

    \frac{d}{dx}(f(x).g(x)) = f'(x).g(x) + f(x).g'(x)


    Quotient Rule, \frac{f(x)}{g(x)}

    \frac{d}{dx}(\frac{f(x)}{g(x)}) = \frac{f'(x).g(x)-f(x)g'(x)}{(g(x))^2}


    Chain Rule, f(g(x))

    \frac{d}{dx}f(g(x)) =f'(g(x)).g'(x)


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    Analysis Of Previous Five Years’ NEET Paper

    The NEET paper includes a total of 200 questions and out of that 50 questions each are from Physics and Chemistry and 100 questions from Biology. Students need to attempt only 180 questions which include 45 from Physics, 45 from Chemistry, and 90 from Biology. Careers360 has analyzed the last 5 years' papers and extracted some important information, which is tabulated below.

    Year Wise Number Of Questions In which Calculus Is Used

    Year

    Physics

    Chemistry

    2021

    9

    3

    2020

    7

    2

    2019

    10

    2

    2018

    9

    0

    2017

    7

    3

    It is observed that in Physics out of 45 questions around 7-10 questions demand a basic understanding of Calculus whereas, in Chemistry out of 45 questions, 2-3 questions demand basic knowledge of Calculus. This means around 15-20 % of Physics sections required a basic understanding of Calculus. As Calculus accounts for around 20% of Physics questions, it becomes very important for NEET aspirants.

    Previous Year's Questions And Solutions

    Here are some previous year's questions with their solution in which concepts of Calculus were used. After going through these questions and analyzing the previous year's papers you will realize that the NEET doesn't go deep to Calculus. Most of the questions are formulas-based or very normal calculations with basic Differentiation and Integration.

    Differential Calculus

    Q.1 (NEET - 2021)

    A capacitor of capacitance C is connected across an ac source of voltage V.given by 1652953325954. The displacement current between the plates of the capacitor would then be given by:

    Solution:

    We know that charge on the capacitor’s plate is given by Q = CV. Current is the rate of flow of charge which means the differentiation of charge function with respect to time gives current.

    1652953324713

    Concept Used From Differentiation-

    We know the chain rule is

    \frac{d}{dx}f(g(x)) =f'(g(x)).g'(x)

    \frac{d}{dt}sin\omega t=(coswt)(w)=wcoswt

    That is chain rule and differentiation of sin function

    Q.2 (NEET - 2020)

    The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:

    Solution:

    We know that rate of change of displacement is velocity and rate of change of velocity is acceleration. This means the differentiation of the displacement function with time gives velocity and also that acceleration can be calculated by differentiating the velocity function with time.

    1652953324273

    (Since -sin⍵t=sin(⍵t+π))

    Concept Used From Calculus-

    \\\frac{d}{dt}sin\omega t=wcoswt\ and\\\frac{d}{dt}wcoswt=w(-sinwt)(w)=-w^{2}sinwt

    That is chain rule and differentiation of sin and cos function.

    Q.3 (NEET - 2017)

    The x and y coordinates of the particle at any time are x = 5t - 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is.

    Solution:

    We know that velocity is the rate of change of displacement and acceleration is the rate of change of velocity. When we differentiate the displacement function we get the velocity function and when we differentiate the velocity function we get the acceleration function.

    x = 5t-2t2
    y = 10t
    dx/dt=5-4t
    dy/dt=10
    vx=5 - 4t
    vy = 10
    ax= -4
    ay= 0
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    Acceleration of particle at (t=2)= -4m/s2

    Concept Used From Differentiation-

    \frac{d}{dt} ( t^{n})={n} t^{n-1}

    and the concept that differentiation of a constant function is zero is also used.

    Integral Calculus

    Q.1 (NEET - 2019)

    A force F = 20 + 10 y acts on a particle in y-direction where F is in newton and y in meter Work done by this force to move the particle from y = 0 to y = 1 is:

    Solution:

    Definition of work done by variable force -

    W=int vec{F}cdot vec{ds}

    wherein vec{F} is variable force and vec{ds} is small displacement

    \\ F = 20+ 10 y \\\\ W = F\cdot ds \\\\ here \ W = \int_{0}^{1}F\cdot dy =\int_{0}^{1} ( 20 + 10 y ) dy \\\\ W = [ 20 y ]^1 _0 + \left [ \frac{10 y^2}{2} \right ]^1_0 \\\\ W = 20+5 = 25 J

    Concept Used From Integral Calculus-

    Here we used the integration formula

    \int x^n dx=\frac{x^{n+1}}{n+1}+c

    Putting upper and lower limits and then subtracting them, we can calculate the value of the definite integral.

    Q.2 (NEET - 2017)

    A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy \Delta U of the gas in joules will be:

    Solution:

    Work Done for Irreversible Isothermal Expansion of an ideal gas -

    \\ W = - \int_{V_{i}}^{V_{f}} P_{ext} \ dv \\ \\ \ = - P_{ext} \left ( V_{f} -V_{i}\right )

    wherein P_{ext} may be equal to P_{f} or may not be equal to P_{f}, but work done is always calculated by P_{ext} and in Adiabatic Process, Heat exchange between system and surrounding is zero i.e. q= 0

    wherein

    \Delta E= q+w

    q= 0

    \Delta E=w

    Work done in this process will be

    \\ W = - \int_{V_{i}}^{V_{f}} P_{ext} \ dv \\ \\ \ = - P_{ext} \left ( V_{f} -V_{i}\right )

    = -2.5[4.5 - 2.5] = -5 L atm

    = - 5 \times 101 J = -505J

    since the system is well-insulated

    q = 0

    \Rightarrow \Delta U-W=0

    or ΔU = +W = -505 J

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    How A Student Can Get Hold On Calculus Concepts

    Two words “Why” and “How” are widely mentioned. If the reason or “why” is clear it means a student is aware of why s/he must have command on Calculus then s/he can find a way or “How” to get it. First of all stop fearing the Calculus part in calculations. Just face it, set a goal, work smart and practise well. Then Calculus becomes easy.

    As you have already gone through the previous years’ questions you will understand that only tricky questions are asked in NEET. Having a good command of Physics, Chemistry, and Biology concepts as well as a basic understanding of Calculus is essential for premiere exams like NEET. A basic understanding of Calculus cannot be ignored because without it would be difficult to obtain good marks in Physics. Students can easily get hold of Calculus concepts through disciplined study and practise.

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