TS Inter Maths1B | Page 2 of 2

Errors and Approximations || V.S.A.Q’S||

Errors and Approximations || V.S.A.Q’S|| Errors and approximations: These solutions were designed by the ‘Basics in Maths‘ team. These notes to do help intermediate First-year Maths students. Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations. Errors and Approximations Question 1 Find dy and ∆y for the following functions for the values of x and ∆x which are shown against each of the functions (i) y = f(x) = x2 + x at x = 10 when ∆x = 0.1. Sol: Given y = f(x) = x2 + x at x = 10, ∆x = 0.1 ∆y = f (x + ∆x) – f (x) = f (10 + 0.1) – f (10) = f (10.1) – f (10) = (10. 1)2 + 10.1 – (102 + 10) = 102.01 + 10.1 – 100 – 10 = 112.11 – 110 = 2.11 dy = f’ (x) ∆x = (2x + 1) (0.1) = [2(10) + 1] (0.1) = 21 × 0.1 = 2.1 (ii) y = cos x at x = 600 with ∆x = 10 (10 = 0.0174 radians) Sol: Given y = cos x, x = 600 and ∆x = 10 ∆y = f (x + ∆x) – f (x) = cos (600 + 10) – cos 600 = cos (610) – 0.5 = 0.4848 – 0.5 = – 0.0152 dy = f’ (x) ∆x = – sin x (10) = – sin 600 × 0.0174 =– 0.8660 × 0.0174 = – 0.0150 (iii) y = x2 + 3x + 6, x = 10 with ∆x = 0.01 Sol: y = x2 + 3x + 6 ∆y = f (x + ∆x) – f (x) = f (10 + 0.01) – f (10) = f (10.01) – f (10) = (10.01)2 + 3 (10.01) + 6 – (102 + 3 (10) + 6) = 100. 2001 + 30.03 + 6 – 100 – 30 – 6 =130. 2301 – 130 = 0.2301 dy = f’ (x) ∆x = (2x + 3 + 0) (0.01) = (2× 10 + 3) (0.01) = 23 × 0.01 = 0.23 (iv) y = , x = 8 and ∆x =0.02 Sol: Question 2 The side of a square is increased from 3cm to 3.01cm find the approximate increase in the area of the square. Sol: Let x be the side of the square and the area be A Area of the square A = x2 x = 3 and ∆x = 0.01 ∆A = 2x × ∆x = 2(3) (0.01) = 6 × 0.01 = 0.06 Question 3 If an increase in the side of a square is 2% then find the approximate percentage of increase in its area. Sol: Let x be the side of the square and A be its area = 2 A = x2 ∆A = 2x × ∆x The approximate percentage error in area A = 2 × 2 =4 https://www.basicsinmaths.com/inter-mathematics-1a-and-1b-pdf-files/ Question 4 From the following. Find the approximations (i) Sol: Let f(x) = , where x = 1000 and ∆x =– 1 f’ (x) = Approximate value is f (x + ∆x) = f(x) + f’ (x) ∆x = 10 – 0. 0033 = 9.9967 (ii) Sol: (iii) Sol: (iv) Sin 620 Sol: Let f(x) = sin x, where x = 600 and ∆x =20 Approximate value is f (x + ∆x) = f(x) + f’ (x) ∆x = sin 600 + cos x (20) = sin 600+ cos 600 (0.0348) = 0.8660 + 0.5 × 0.0348 = 0.8660 + 0.0174 =0.8834 Question 5 The radius of a sphere is measured as 14cm. Later it was found that there is an error of 0.02cm in measuring the radius. Find the approximate error in the surface area of the sphere. Sol: Given r = 14 cm and ∆r =0.02cm Surface area of sphere =A = 4π r2 ∆A = 8π r ∆r = 8 ×3.14× 14 × 0.02 = 7.0336 Visit my Youtube Channel: Click on Below Logo

TS Inter Maths 1B Concept

Education, TS Inter Maths, TS Inter Maths1B / Nayini Satyanarayana Reddy

TS Inter Maths 1B Concept Ts Inter Maths 1B Concept: designed by the ‘Basics in Maths’ team. These notes to do help the TS intermediate first-year Maths students fall in love with mathematics and overcome the fear. These notes cover all the topics covered in the TS I.P.E first year maths 1B syllabus and include plenty of formulae and concept to help you solve all the types of Inter Math problems asked in the I.P.E and entrance examinations. 0.COORDINATE GEOMETRY( BASICS) Distance between two points A(x1, y1), B(x2, y2) is distance between a point A(x1, y1) to the origin is The midpoint of two points A(x1, y1), B(x2, y2) is If P divides the line segment joining the points A(x1, y1), B(x2, y2) in the ratio m:n then the coordinates of P are Area of the triangle formed by the vertices A (x1, y1), B (x2, y2) and C (x3, y3) is 1. LOCUS Locus: The set of points that are satisfying a given condition or property is called the locus of the point. Ex:- If a point P is equidistant from the points A and B, then AP =BP Ex 2: – set of points that are at a constant distance from a fixed point. here the locus of a point is a circle. • In a right-angled triangle PAB, the right angle at P and P is the locus of the point, then AB2 = PA2 + PB2 •Area of the triangle formed by the vertices A (x1, y1), B (x2, y2), and C (x3, y3) is 2.CHANGE OF AXES Transformation of axes: When the origin is shifted to (h, k), without changing the direction of axes then •To remove the first degree terms of the equation ax2 + 2hxy + by2 +2gx +2fy+ c = 0, origin should be shifted to the point •If the equation ax2 + by2 +2gx +2fy+ c = 0, origin should be shifted to the point Rotation of axes: When the axes are rotated through an angle θ then •To remove the xy term of the equation ax2 + 2hxy + by2 = 0, axes should be rotated through an angle θ is given by 3.STRAIGHT LINES Slope:- A-line makes an angle θ with the positive direction of the X-axis, then tan θ is called the slope of the line. It is denoted by “m”. m= tan θ • The slope of the x-axis is zero. • Slope of any line parallel to the x-axis is zero. • The y-axis slope is undefined. • The slope of any line parallel to the y-axis is also undefined. • The slope of the line joining the points A (x1, y1) and B (x2, y2) is Slope of the line ax + by + c = 0 is Types of the equation of a straight line: Equation of x- axis is y = 0. Equation of any line parallel to the x-axis is y = k, where k is the distance from above or below the x-axis. Equation of y- axis is x = 0. Equation of any line parallel to y-axis is x = k, where k is the distance from the left or right side of the y-axis. Slope- intercept form The equation of the line with slope m and y-intercept c is y = mx + c. Slope point form: The equation of the line passing through the point (x1, y1) with slope m is y – y1 = m (x – x1) Two points form: The equation of the line passing through the points (x1, y1) and (x2, y2) ’ is Intercept form: The equation of the line with x-intercept a, y-intercept b is • The equation of the line ∥ el to ax +by + c = 0 is ax +by + k = 0. • The equation of the line ⊥ler to ax +by + c = 0 is bx −ay + k = 0. Note: – If two lines are parallel then their slopes are equal m1 = m2 If two lines are perpendicular then product of their slopes is – 1 m1 × m2 = – 1 The area of the triangle formed by the line ax + by + c = 0 with the coordinate axes is The area of the triangle formed by the line with the coordinate axes is Perpendicular distance (Length of the perpendicular): The perpendicular distance from a point P (x1, y1) to the line ax + by + c = 0 is • The perpendicular distance from origin to the line ax + by + c = 0 is Distance between two parallel lines: •The distance between the parallel lines ax1 + by1 + c1 = 0 and ax2 + by2 + c2 = 0 is Perpendicular form or Normal form: The equation of the line which is at a distance of ‘p’ from the origin and α (0≤ α ≤ 3600) is the angle made by the perpendicular with the positive direction of the x-axis is x cosα + y sinα = p. Symmetric form: The equation of the line passing through point P (x1, y1) and having inclination θ is Parametric form: if P (x, y) is any point on the line passing through A (x1, y1) and making inclination θ, then x = x1 + r cos θ, y = y1 + r sin θ where ‘r’ is the distance from P to A. • The ratio in which the line L ≡ ax + by + c = 0 divide the line segment joining the points A (x1, y1), B (x2, y2) is

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