MATHS N6 TUTORIAL

OPEN LEARNING UNIT MATHEMATICS N6 TUTORIAL GUIDE Downloaded from hlayiso.com

MATHEMATICS N6 TUTORIAL 1 1|Page Prepared by: Mr Sithole Z Downloaded from hlayiso.com

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MATHEMATICS N6 TUTORIAL 2 𝑑𝑦 Find 𝑑𝑥 if 4√𝑥 a) 𝑦 = tan−1 [1−4𝑥] SOLUTION 𝑑 1 Since 𝑑𝑥 (tan−1 𝑥) = 1+𝑥 2 4 (1 − 4𝑥) ( ) − 4√𝑥(−4) 𝑑 4√𝑥 1 2√𝑥 → (tan−1 [ ]) = 2∙ 𝑑𝑥 1 − 4𝑥 4√𝑥 (1 − 4𝑥)2 1+( ) 1 − 4𝑥 1 (2 − 8𝑥) + 16𝑥 (1 − 4𝑥)2 2 + 8𝑥 = ∙ = ∙ 16𝑥 √𝑥(1 − 4𝑥)2 (1 − 4𝑥)2 + 16𝑥 √𝑥(1 − 4𝑥)2 1+ 2 (1 − 4𝑥) 2 + 8𝑥 2(1 + 4𝑥) = = √ 𝑥[(1 − 4𝑥)2 + 16𝑥] √𝑥[1 − 8𝑥 + 16𝑥 2 + 16𝑥] 2(1 + 4𝑥) 2(1 + 4𝑥) 2 = = = √𝑥(1 + 16𝑥 + 16𝑥 2 ) √𝑥(1 + 4𝑥) 2 √𝑥(1 + 4𝑥) 1+tan 𝑥 b) 𝑦 = tan−1 [ ] 1−tan 𝑥 SOLUTION 𝑑𝑦 1 (1 − tan 𝑥) ∙ sec 2 𝑥 − (1 + tan 𝑥) ∙ (−sec 2 𝑥) = 2∙ 𝑑𝑥 1 + tan 𝑥 (1 − tan 𝑥)2 1 + (1 − tan 𝑥) (1 − tan 𝑥)2 sec 𝑥 [(1 − tan 𝑥) + (1 − tan 𝑥)] = 2 2 ∙ (1 − tan 𝑥) + (1 + tan 𝑥) (1 − tan 𝑥)2 2sec 2 𝑥 = (1 − tan 𝑥)2 + (1 + tan 𝑥)2 2sec 2 𝑥 2sec 2 𝑥 = = 1 − 2 tan 𝑥 + tan2 𝑥 + 1 + 2 tan 𝑥 + tan2 𝑥 2(1 + tan2 𝑥) 2sec 2 𝑥 = =1 2sec 2 𝑥 4|Page Prepared by: Mr Sithole Z Downloaded from hlayiso.com

2) Find the coordinates of the points of inflexion of the curve 𝑦 = 4𝑥 3 + 3𝑥 2 − 18𝑥 − 9 SOLUTION 𝑑𝑦 = 12𝑥 2 + 6𝑥 − 18 𝑑𝑥 𝑑2𝑦 = 24𝑥 + 6 𝑑𝑥 2 At the point of inflexion, 𝑑2𝑦 = 24𝑥 + 6 = 0 𝑑𝑥 2 → 24𝑥 + 6 = 0 1 →𝑥=− 4 1 3 1 2 1 35 → 𝑦 = 4 (− ) + 3 (− ) − 18 (− ) − 9 = − 4 4 4 8 𝑑2 𝑦 𝑑2 𝑦 Since 𝑑𝑥 2 > 0 for x > -1/4 and 𝑑𝑥 2 < 0 for x < -1/4, it implies the point (-1/4;-35/8) is a point of inflexion. sin−1 𝑥 3) Given that 𝑦= , prove that (1 − 𝑥 2 ) 𝑑𝑦 = 𝑥𝑦 + 1 √1−𝑥 2 𝑑𝑥 SOLUTION sin−1 𝑥 𝑦= √1 − 𝑥 2 1 1 1 √1 − 𝑥 2 ∙ − sin −1 𝑥 ∙ (1 − 𝑥 2 )−2 ∙ (−2𝑥) 𝑑𝑦 √1 − 𝑥 2 2 → = 𝑑𝑥 1 − 𝑥2 𝑑𝑦 𝑥 sin−1 𝑥 → (1 − 𝑥 2 ) =1+ 𝑑𝑥 √1 − 𝑥 2 𝑑𝑦 → (1 − 𝑥 2 ) = 1 + 𝑥𝑦 𝑑𝑥 5|Page Prepared by: Mr Sithole Z Downloaded from hlayiso.com

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MATHEMATICS N6 TUTORIAL 3 7|Page Prepared by: Mr Sithole Z Downloaded from hlayiso.com

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MATHEMATICS N6 TUTORIAL 4 Solve the following DE’s using separation of variables: a). SOLUTION b) . given that SOLUTION 15 | P a g e Prepared by: Mr Sithole Z Downloaded from hlayiso.com

Now, c) . given y(1) = 2. SOLUTION Now 16 | P a g e Prepared by: Mr Sithole Z Downloaded from hlayiso.com

6) Solve the following homogenous equations: a) . SOLUTION Let But NB: The solutions to differential equations are not unique as the final form of presentation depends on the approach, especially the choice of the constant of integration. You can use any letter for the integration constant b). 17 | P a g e Prepared by: Mr Sithole Z Downloaded from hlayiso.com

SOLUTION We proceed as in (a) by letting y = vx Now, v = y/x, c) . 18 | P a g e Prepared by: Mr Sithole Z Downloaded from hlayiso.com

SOLUTION To evaluate , we use partial fractions. Comparing coefficients, we have And so the integral becomes 19 | P a g e Prepared by: Mr Sithole Z Downloaded from hlayiso.com

Squaring both sides, we have v = y/x, 7) Solve the following linear differential equations: a) . given that . SOLUTION The first step is rewriting the equation in the form: 20 | P a g e Prepared by: Mr Sithole Z Downloaded from hlayiso.com

The general solution of these differential equations is where IF is the integration factor, given by And so by the general solution, we have Since , b) . SOLUTION And so from the general solution, we have 21 | P a g e Prepared by: Mr Sithole Z Downloaded from hlayiso.com

TBC... Contact Details: Engineering Department Tel: 031 301 0649 Email: admin@maritimebi.co.za Website: www.maritimebusinessinstitute.co.za 22 | P a g e Prepared by: Mr Sithole Z Downloaded from hlayiso.com