NCERT Class 12 Maths: Applications of Derivative – Exercise 6.4 (Approximations)

Understanding Approximation using Differentials

The core idea behind approximation using differentials is to estimate the change in a function's value ($\Delta y$) when its input (x) changes by a small amount ($\Delta x$). Let's consider a function $y = f(x)$. If we want to find the value of $f(x + \Delta x)$, where $\Delta x$ is very small, we can approximate it using the derivative.

Recall that the derivative $f'(x)$ (or $dy/dx$) represents the instantaneous rate of change of $y$ with respect to $x$. For a very small change $\Delta x$, the average rate of change $\frac{\Delta y}{\Delta x}$ is approximately equal to the instantaneous rate of change $\frac{dy}{dx}$.

So, we can write:
$\frac{\Delta y}{\Delta x} \approx \frac{dy}{dx} = f'(x)$
This implies $\Delta y \approx f'(x) \Delta x$.

Now, we know that $f(x + \Delta x) = y + \Delta y$. Substituting the approximation for $\Delta y$, we get the fundamental formula for approximation:
$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$

Geometrically, $\Delta y$ is the actual change in the y-coordinate along the curve $y=f(x)$, while $f'(x) \Delta x$ represents the change in the y-coordinate along the tangent line to the curve at the point $(x, f(x))$. When $\Delta x$ is small, the tangent line lies very close to the curve, making the tangent's change in y a good approximation for the curve's actual change in y. This method is incredibly useful for estimating values like roots, powers, and trigonometric function values for numbers slightly different from easily calculable ones.

Key Definitions

Increment ($\Delta x$)

A small change or increase in the independent variable $x$. It can be positive or negative.

Differential of x ($dx$)

By convention, the differential of $x$, denoted as $dx$, is taken to be equal to the increment $\Delta x$. So, $dx = \Delta x$.

Differential of y ($dy$)

For a function $y = f(x)$, the differential of $y$, denoted as $dy$, is defined as $dy = f'(x) dx$. This represents the approximate change in $y$ corresponding to a change $dx$ in $x$.

Actual Change in y ($\Delta y$)

The actual change in the function's value when $x$ changes by $\Delta x$. It is given by $\Delta y = f(x + \Delta x) - f(x)$.

Approximation Formula

The formula $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$ is used to estimate the value of $f(x + \Delta x)$.

Step-by-Step Method for Approximation

1. Step 1: Define the Function $f(x)$ — Identify the base function $f(x)$ that resembles the given expression. For example, if you need to approximate $\sqrt{25.3}$, choose $f(x) = \sqrt{x}$.
2. Step 2: Identify $x$ and $\Delta x$ — Choose a value for $x$ close to the number you want to approximate, such that $f(x)$ is easy to calculate. $\Delta x$ will then be the small difference from $x$ to the target value. Ensure $x + \Delta x$ equals the number you're approximating.
3. Step 3: Calculate $f(x)$ — Substitute the chosen value of $x$ into your function $f(x)$ to find $f(x)$.
4. Step 4: Find the Derivative $f'(x)$ — Differentiate the function $f(x)$ with respect to $x$ to get $f'(x)$. Then, substitute the value of $x$ into $f'(x)$ to find $f'(x)$ at that point.
5. Step 5: Apply the Approximation Formula — Use the formula $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$. Substitute all the values you calculated in the previous steps to find the approximate value.

Worked Examples

• Example 1: Approximate $\sqrt{25.3}$ Step 1: Define $f(x)$. Let $f(x) = \sqrt{x}$. Step 2: Identify $x$ and $\Delta x$. We choose $x = 25$ because $\sqrt{25}$ is easy to calculate. Then $x + \Delta x = 25.3$, so $25 + \Delta x = 25.3$, which gives $\Delta x = 0.3$. Step 3: Calculate $f(x)$. $f(25) = \sqrt{25} = 5$. Step 4: Find $f'(x)$. $f'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$. Now, calculate $f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{2 \times 5} = \frac{1}{10} = 0.1$. Step 5: Apply the approximation formula. $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$ $\sqrt{25.3} \approx f(25) + f'(25) \times (0.3)$ $\sqrt{25.3} \approx 5 + (0.1) \times (0.3)$ $\sqrt{25.3} \approx 5 + 0.03$ Final answer: $\sqrt{25.3} \approx 5.03$.
• Example 2: Approximate $(8.02)^{1/3}$ Step 1: Define $f(x)$. Let $f(x) = x^{1/3}$. Step 2: Identify $x$ and $\Delta x$. We choose $x = 8$ because $8^{1/3} = 2$ is easy to calculate. Then $x + \Delta x = 8.02$, so $8 + \Delta x = 8.02$, which gives $\Delta x = 0.02$. Step 3: Calculate $f(x)$. $f(8) = 8^{1/3} = 2$. Step 4: Find $f'(x)$. $f'(x) = \frac{d}{dx}(x^{1/3}) = \frac{1}{3}x^{(1/3) - 1} = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}$. Now, calculate $f'(8) = \frac{1}{3(8)^{2/3}} = \frac{1}{3(2^3)^{2/3}} = \frac{1}{3 \times 2^2} = \frac{1}{3 \times 4} = \frac{1}{12}$. Step 5: Apply the approximation formula. $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$ $(8.02)^{1/3} \approx f(8) + f'(8) \times (0.02)$ $(8.02)^{1/3} \approx 2 + (\frac{1}{12}) \times (0.02)$ $(8.02)^{1/3} \approx 2 + \frac{0.02}{12}$ $(8.02)^{1/3} \approx 2 + 0.001666...$ Final answer: $(8.02)^{1/3} \approx 2.00167$.
• Example 3: Approximate $\sin(29^{\circ})$ given $1^{\circ} = 0.01745$ radians. Step 1: Define $f(x)$. Let $f(x) = \sin(x)$. Remember, calculus functions work with radians. Step 2: Identify $x$ and $\Delta x$. We choose $x = 30^{\circ} = \frac{\pi}{6}$ radians, as $\sin(30^{\circ})$ is known. $\Delta x = 29^{\circ} - 30^{\circ} = -1^{\circ}$. Convert $\Delta x$ to radians: $\Delta x = -1 \times 0.01745 = -0.01745$ radians. Step 3: Calculate $f(x)$. $f(\frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2} = 0.5$. Step 4: Find $f'(x)$. $f'(x) = \frac{d}{dx}(\sin(x)) = \cos(x)$. Now, calculate $f'(\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \approx \frac{1.732}{2} = 0.866$. Step 5: Apply the approximation formula. $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$ $\sin(29^{\circ}) \approx f(\frac{\pi}{6}) + f'(\frac{\pi}{6}) \times (-0.01745)$ $\sin(29^{\circ}) \approx 0.5 + (0.866) \times (-0.01745)$ $\sin(29^{\circ}) \approx 0.5 - 0.0151127$ Final answer: $\sin(29^{\circ}) \approx 0.48489$.

Exam Tips for Approximations

1. Choosing $x$ and $\Delta x$: This is the most critical step. Always select x such that $f(x)$ is easily calculable and x is very close to the number you need to approximate. The value of x must make f(x) a 'perfect' value (e.g., perfect square for square roots, perfect cube for cube roots, standard angle for trigonometric functions). The sign of $\Delta x$ (positive or negative) must be correct.
2. Units for Trigonometric Functions: When dealing with trigonometric functions (like $\sin x$, $\cos x$, $\tan x$), remember that the derivative formulas ($d/dx (\sin x) = \cos x$) assume $x$ is in radians. If the problem gives angles in degrees, you must convert $\Delta x$ from degrees to radians before using it in the formula $f'(x) \Delta x$. The value of $x$ (e.g., $\pi/4$ for $45^{\circ}$) should also be in radians.
3. Careful Differentiation: Ensure your differentiation of $f(x)$ is accurate. A small error in $f'(x)$ will lead to an incorrect approximation.
4. Decimal Precision: Pay attention to decimal places, especially in intermediate calculations. Round only at the very final step, as specified in the question, to maintain accuracy.

Practice Questions with Solutions

• Q: Using differentials, find the approximate value of $\sqrt{49.5}$. A: Step 1: Let $f(x) = \sqrt{x}$. Step 2: Choose $x = 49$ and $\Delta x = 0.5$. Step 3: Calculate $f(x) = f(49) = \sqrt{49} = 7$. Step 4: Find $f'(x) = \frac{1}{2\sqrt{x}}$. So, $f'(49) = \frac{1}{2\sqrt{49}} = \frac{1}{2 \times 7} = \frac{1}{14}$. Step 5: Apply $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$. $\sqrt{49.5} \approx 7 + (\frac{1}{14}) \times (0.5) = 7 + \frac{0.5}{14} = 7 + \frac{1}{28} \approx 7 + 0.0357$. Final answer: $\sqrt{49.5} \approx 7.0357$.
• Q: Using differentials, find the approximate value of $(25)^{1/3}$. A: Step 1: Let $f(x) = x^{1/3}$. Step 2: Choose $x = 27$ and $\Delta x = -2$. (Since $27^{1/3} = 3$ is easily calculable and 27 is closest to 25 with a perfect cube root). Step 3: Calculate $f(x) = f(27) = (27)^{1/3} = 3$. Step 4: Find $f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}$. So, $f'(27) = \frac{1}{3(27)^{2/3}} = \frac{1}{3(3^3)^{2/3}} = \frac{1}{3 \times 3^2} = \frac{1}{3 \times 9} = \frac{1}{27}$. Step 5: Apply $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$. $(25)^{1/3} \approx 3 + (\frac{1}{27}) \times (-2) = 3 - \frac{2}{27} \approx 3 - 0.07407$. Final answer: $(25)^{1/3} \approx 2.92593$.
• Q: Approximate $\tan(44^{\circ})$, given that $1^{\circ} = 0.01745$ radians. A: Step 1: Let $f(x) = \tan(x)$. Step 2: Choose $x = 45^{\circ} = \frac{\pi}{4}$ radians. $\Delta x = 44^{\circ} - 45^{\circ} = -1^{\circ}$. Convert to radians: $\Delta x = -1 \times 0.01745 = -0.01745$ radians. Step 3: Calculate $f(x) = f(\frac{\pi}{4}) = \tan(\frac{\pi}{4}) = 1$. Step 4: Find $f'(x) = \sec^2(x)$. So, $f'(\frac{\pi}{4}) = \sec^2(\frac{\pi}{4}) = (\sqrt{2})^2 = 2$. Step 5: Apply $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$. $\tan(44^{\circ}) \approx 1 + (2) \times (-0.01745) = 1 - 0.0349$. Final answer: $\tan(44^{\circ}) \approx 0.9651$.
• Q: If $y = x^3 - 7x^2 + 15$, find the approximate change in $y$ when $x$ changes from 5 to 5.01. A: Step 1: The function is $y = f(x) = x^3 - 7x^2 + 15$. Step 2: Identify $x = 5$ and $\Delta x = 5.01 - 5 = 0.01$. Step 3: Calculate $f'(x) = \frac{dy}{dx} = 3x^2 - 14x$. Step 4: Calculate $f'(5) = 3(5)^2 - 14(5) = 3(25) - 70 = 75 - 70 = 5$. Step 5: The approximate change in $y$ is $\Delta y \approx f'(x) \Delta x$. $\Delta y \approx 5 \times (0.01) = 0.05$. Final answer: The approximate change in $y$ is $0.05$.

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