12 Most VSAQ’s of Errors & Approximations Chapter in Inter 1st Year Maths-1B (TS/AP)

2 Marks

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

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

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

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

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

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

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