10.1 Errors and Approximations (VSAQs)

Maths-1B | 10.1 Errors and Approximations – VSAQs:
Welcome to VSAQs in Chapter 10: 10.1 Errors and Approximations. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

VSAQ-1 : Find ∆y and dy for the function y = x2 + x when x = 10, ∆x = 0.1

$$y = f(x) = x^2 + x \quad \text{and} \quad x = 10, \quad \Delta x = 0.1$$

$$\Delta y = f(x + \Delta x) – f(x)$$

$$= [(x + \Delta x)^2 + (x + \Delta x)] – x^2 – x$$

$$= [x^2 + (\Delta x)^2 + 2x\Delta x] + x + \Delta x – x^2 – x$$

$$= (\Delta x)^2 + 2x\Delta x + \Delta x$$

$$= \Delta x(\Delta x + 2x + 1)$$

$$= 0.1 (0.1 + 2(10) + 1) = (0.1)(21.1) = 2.11$$

$$dy = f'(x)\Delta x = (2x + 1)\Delta x$$

$$= 2(10) + 1 = 21(0.1) = 2.1$$

VSAQ-2 : If y = x2 + 3x + 6 then find ∆y and dy when x = 10, ∆x = 0.01

$$y = f(x) = x^2 + 3x + 6 \quad \text{and} \quad x = 10, \quad \Delta x = 0.01$$

$$\Delta y = f(x + \Delta x) – f(x)$$

$$= [(x + \Delta x)^2 + 3(x + \Delta x) + 6] – (x^2 + 3x + 6)$$

$$= [x^2 + (\Delta x)^2 + 2x\Delta x] + 3x + 3\Delta x + 6 – x^2 – 3x – 6$$

$$= (\Delta x)^2 + 2x\Delta x + 3\Delta x$$

$$= \Delta x(\Delta x + 2x + 3)$$

$$= (0.01)[0.01 + 2(10) + 3] = (0.01)[23.01] = 0.2301$$

$$dy = f'(x)\Delta x = (2x + 3)\Delta x$$

$$= 2(10) + 3 = (23)(0.01) = 0.23$$

VSAQ-3 : Find ∆y and dy for the function y = 1/(x + 2) when x = 8, ∆x = 0.02

$$y = f(x) = \frac{1}{x} + 2 \quad \text{and} \quad x = 8, \quad \Delta x = 0.02$$

$$\Delta y = f(x + \Delta x) – f(x) = \frac{1}{x + \Delta x} + 2 – \left(\frac{1}{x} + 2\right)$$

$$= \frac{1}{8 + 0.02} + 2 – \frac{1}{8} + 2 = \frac{1}{8.02} – \frac{1}{8}
= \frac{8 – 8.02}{(8.02)(8)} = \frac{-0.02}{64.16} = -0.0001996$$

$$dy = f'(x)\Delta x = \left(-\frac{1}{(x^2)}\right)\Delta x$$

$$= \left(-\frac{1}{(8 + 2)^2}\right)(0.02) = -\frac{0.02}{100} = -0.0002$$

VSAQ-4 : Find ∆y and dy for the function y = ex + x when x = 5, ∆x = 0.02

$$y = f(x) = e^x + x \quad \text{and} \quad x = 5, \quad \Delta x = 0.02$$

$$\Delta y = f(x + \Delta x) – f(x) = [e^{x+\Delta x} + (x + \Delta x)] – (e^x + x)$$

$$= [e^{5+0.02} + (5 + 0.02)] – (e^5 + 5)
= e^{5.02} + 5.02 – e^5 – 5
= e^{5}(e^{0.02} – 1) + 0.02$$

$$dy = f'(x)\Delta x = (e^x + 1)\Delta x$$

$$= (e^5 + 1)(0.02)$$

VSAQ-5 : If the increase in the side of a square is 4% then find the approximate percentage of increase in the area of the square

$$A = x^2$$

$$dA = 2x \, dx$$

$$dA = 2x \frac{dx}{x} x = 2 \frac{dx}{x} x^2$$

$$\frac{dA}{A} = \frac{2 \frac{dx}{x} x^2}{x^2} = 2 \frac{dx}{x}$$

$$\frac{dA}{A} \times 100 = 2 \left(\frac{dx}{x} \times 100\right) = 2(4) = 8$$

VSAQ-6 : If the increase in the side of a square is 2% then find the approximate percentage of increase in the area of the square

$$A = x^2$$

$$dA = 2x \, dx$$

$$\frac{dA}{A} = \frac{2 \frac{dx}{x} x^2}{x^2} = 2 \frac{dx}{x}$$

$$\frac{dA}{A} \times 100 = 2 \left(\frac{dx}{x} \times 100\right) = 2(2) = 4$$

VSAQ-7 : The side of a square is increased from 3cm to 3.01 cm. Find the approximate increase in the area of the square

$$A = x^2$$

$$dA = \frac{dA}{dx} \Delta x = (2x)\Delta x$$

$$= 2(3)(0.01) = 0.06 \text{ sq.cm}$$

VSAQ-8 : If the radius of a sphere is increased from 7 cm to 7.02 cm then find the approximate increase in the volume of the sphere

$$V = \frac{4\pi}{3} r^3$$

$$dV = \frac{dV}{dr} \Delta r = 4\pi r^2 \Delta r$$

$$= \frac{4\pi}{3} (3r^2) \Delta r = 4\pi r^2 \Delta r$$

$$= 4 \pi (7^2)(0.02) = 4 \times \frac{22}{7} \times 49 \times 0.02 = 12.32 \, \text{cm}^3$$

VSAQ-9 : Find the approximate value of √82

$$x = 81, \quad \Delta x = 1, \quad f(x) = \sqrt{x}$$

$$f'(x) = \frac{1}{2\sqrt{x}}$$

$$f(x + \Delta x) = f(x) + f'(x) \Delta x \quad \text{at} \quad x = 81$$

$$\sqrt{82} \approx \sqrt{81} + \frac{1}{2\sqrt{81}} (1)$$

$$= 9 + \frac{1}{2 \times 9} = 9 + \frac{1}{18}$$

$$= 9 + 0.0555 = 9.0555$$

VSAQ-10 : Find the approximate value of 3√65

$$x = 64, \quad \Delta x = 1, \quad f(x) = x^{1/3}$$

$$f'(x) = \frac{1}{3} x^{1/3 – 1} = \frac{1}{3} x^{-2/3}$$

$$f(x + \Delta x) = f(x) + f'(x) \Delta x \quad \text{at} \quad x = 64$$

$$\sqrt[3]{65} \approx \sqrt[3]{64} + \frac{1}{3} \cdot 64^{-2/3} \cdot 1$$

$$= 4 + \frac{1}{3} \cdot (64^{-2/3}) \cdot 1 = 4 + \frac{1}{3} \cdot (4^{-2})$$

$$= 4 + \frac{1}{3} \cdot \frac{1}{16} = 4 + \frac{1}{48}$$

$$= \frac{192}{48} + \frac{1}{48} = \frac{193}{48} \approx 4.0208$$

VSAQ-11 : Find the approximate value of 3√999

$$x = 1000, \quad \Delta x = -1, \quad f(x) = x^{1/3}$$

$$f'(x) = \frac{1}{3} x^{1/3-1} = \frac{1}{3} x^{-2/3} = \frac{1}{3} x^{2/3}$$

$$f(x + \Delta x) = f(x) + f'(x) \Delta x$$

$$\sqrt[3]{999} \approx \sqrt[3]{1000} + \frac{1}{3} \cdot 1000^{2/3} \cdot (-1)$$

$$= 10 + \frac{1}{3} \cdot (10^2) \cdot (-1)$$

$$= 10 – \frac{1}{3} \cdot 100 = 10 – \frac{100}{3}$$

$$= 10 – 33.3333 = 9.6667$$

VSAQ-12 : Find the approximate value of 4√17

$$x = 16, \quad \Delta x = 1, \quad f(x) = x^{1/4}$$

$$f'(x) = \frac{1}{4} x^{1/4-1} = \frac{1}{4} x^{-3/4} = \frac{1}{4} x^{3/4}$$

$$f(x + \Delta x) = f(x) + f'(x) \Delta x$$

$$\sqrt[4]{17} \approx \sqrt[4]{16} + \frac{1}{4} \cdot 16^{3/4} \cdot 1$$

$$= 2 + \frac{1}{4} \cdot (2^3) = 2 + \frac{1}{4} \cdot 8$$

$$= 2 + \frac{8}{4} = 2 + 2 = 2.0312$$