Engineering Mathematics I Tutorial Sheet 5 PDF

Summary

This document contains a tutorial sheet on engineering mathematics, covering topics including vector and Cartesian forms of the equation of a line, equations of planes, finding points of intersection, directional derivatives, and tangent planes/lines. The problems are focused on concepts in vector calculus and related topics in mathematics.

Full Transcript

Jaypee University of Information Technology Department of Mathematics B. Tech. Ist Semester Engineering Mathematics I (24B11MA111) Tutorial Sheet 5 1. Find the vector and Cartesian forms of the equation of line passing...

Jaypee University of Information Technology Department of Mathematics B. Tech. Ist Semester Engineering Mathematics I (24B11MA111) Tutorial Sheet 5 1. Find the vector and Cartesian forms of the equation of line passing through (i) the origin and the point (5,-2, 3). (ii) the point P(3,-4,-1) and parallel to the vector πš€Μ‚ + πš₯Μ‚ + π‘˜. (iii) the point (1, 1, 1) and parallel to the z-axis. (iv) the point (2, 4, 5) and perpendicular to the plane 3π‘₯ + 7𝑦 βˆ’ 5𝑧 = 21. (v) the point (2,3,0) and perpendicular to 𝐴⃗ = πš€Μ‚ + 2πš₯Μ‚ + 3π‘˜ and 𝐡⃗ = 3πš€Μ‚ + 4πš₯Μ‚ + 5π‘˜ 2. Find the equation for the plane passing through: (i) the points (1,3,2), (3, βˆ’1, 6), (5,2,0). (ii) the point 𝑃 (0,2, βˆ’1) & normal to 𝑛⃗ = 3πš€Μ‚ βˆ’ 2πš₯Μ‚ βˆ’ π‘˜. 3. Find the point of intersection of the lines π‘₯ = 2𝑑 + 1, 𝑦 = 3𝑑 + 2, 𝑧 = 4𝑑 + 3 and π‘₯ = πœ† + 2, 𝑦 = 2πœ† + 4, 𝑧 = βˆ’4πœ† βˆ’ 1. Also find the plane determined by these lines. 4. Find the length of the curve defined by π‘Ÿβƒ—(𝑑) = 6𝑑, 3√2𝑑 , 2𝑑 , 0 ≀ 𝑑 ≀ 1. 5. Find the directional derivative of the functions: (i) 𝑓(π‘₯, 𝑦, 𝑧) = 2𝑦𝑧 + 𝑧 in the direction of the vector πš€Μ‚ + 2πš₯Μ‚ + 2π‘˜ at the point(1, βˆ’1, 3). (ii) 𝑓(π‘₯, 𝑦, 𝑧) = π‘₯ + 𝑦 + 𝑧 at the point (1, βˆ’1, 2) in the direction of πš€Μ‚ + 2πš₯Μ‚ + π‘˜. (iii) 𝑓(π‘₯, 𝑦, 𝑧) = π‘₯𝑧 βˆ’ 3π‘₯𝑦 + 2π‘₯𝑦𝑧 βˆ’ 3π‘₯ + 5𝑦 βˆ’ 17 from the point 𝑃(2, βˆ’6, 3) in the direction of origin O. 6. Find the equation of the tangent plane and the normal line at the point P of the given surfaces: (i) π‘₯ 𝑦 + π‘₯𝑧 = 2𝑦 𝑧, 𝑃 = (1,1,1). (ii) 2𝑧 βˆ’ π‘₯ , 𝑃 = (1,1,1). (iii) 4 βˆ’ π‘₯ βˆ’ 2𝑦 , 𝑃 = (1, βˆ’1,1). 7. Sketch the curve 𝑓(π‘₯, 𝑦) = 𝑐 together with βˆ‡π‘“ and the tangent line at the given point. Also determine the equation of the tangent line. (i) π‘₯ + 𝑦 = 4, (√2, √2 ). (ii) π‘₯𝑦 = βˆ’4, (2, βˆ’2). 8. Answer the following: (i) Find the gradient of 𝑓(π‘₯, 𝑦) = 𝑦 βˆ’ 4π‘₯𝑦, at (1,2). (ii) Find the gradient of 𝑓(π‘₯, 𝑦, 𝑧) = π‘₯ 𝑦 + π‘₯𝑦 βˆ’ 𝑧 , at (3, 1, 1). (iii) For a force field 𝐹⃗(π‘₯, 𝑦) = πš€Μ‚ βˆ’ πš₯Μ‚ applied on an object, find 𝑓(π‘₯, 𝑦) such that π‘”π‘Ÿπ‘Žπ‘‘ 𝑓 = 𝐹⃗(π‘₯, 𝑦). βƒ— 9. If π‘Ÿβƒ— = π‘₯πš€Μ‚ + 𝑦πš₯Μ‚ + π‘§π‘˜, |π‘Ÿβƒ—| = π‘Ÿ and π‘Ÿ = , show that π‘”π‘Ÿπ‘Žπ‘‘ = βˆ’. 10. Evaluate 𝑑𝑖𝑣 π‘π‘’π‘Ÿπ‘™ 𝐹⃗ where𝐹⃗ = 𝑦𝑧 πš€Μ‚ + π‘₯𝑦πš₯Μ‚ + π‘¦π‘§π‘˜. *** Answers 1. (i) π‘Ÿβƒ— = 5𝑑 πš€Μ‚ βˆ’ 2𝑑πš₯Μ‚ + 3π‘‘π‘˜ ; x=2t, y= -2t, z=3t (ii) π‘Ÿβƒ— = (3 + 𝑑) πš€Μ‚ βˆ’ (𝑑 βˆ’ 4)πš₯Μ‚ + (𝑑 βˆ’ 1)π‘˜; x=3+t, y= t-4, z= t-1 (iii) π‘Ÿβƒ— = πš€Μ‚ + πš₯Μ‚ + (𝑑 + 1)π‘˜; x = 1, y = 1, z = 1+t (iv) π‘Ÿβƒ— = (2 + 3𝑑) πš€Μ‚ βˆ’ (4 + 7𝑑)πš₯Μ‚ + (5 βˆ’ 5𝑑)π‘˜; x = 2+3t, y = 4+7t, z = 5-5t (v) π‘Ÿβƒ— = (2 βˆ’ 2𝑑) πš€Μ‚ βˆ’ (3 + 4𝑑)πš₯Μ‚ + (βˆ’2𝑑)π‘˜; x = 2 – 2t, y = 3 + 4t, z = -2t 2. (i) 6π‘₯ + 10𝑦 + 7𝑧 = 50 (ii) 3π‘₯ βˆ’ 2𝑦 βˆ’ 𝑧 = βˆ’3 3. βˆ’20π‘₯ + 12𝑦 + 𝑧 = 7 4. 8 √ 5. (i) (ii) (iii) 18 6. (i) π‘₯ βˆ’ 𝑦 = 0, (1 + 3𝑑, 1 βˆ’ 3𝑑, 1) (ii) 2π‘₯ βˆ’ 𝑧 βˆ’ 2 = 0, (2 βˆ’ 4𝑑, 0, 2 + 2𝑑) (iii) π‘₯ βˆ’ 2𝑦 βˆ’ 3 = 0, (1 βˆ’ 𝑑, βˆ’1 + 2𝑑, 1) 7. (i) 2√2, 2√2 , 𝑦 = βˆ’π‘₯ + 2√2 (ii) (βˆ’2,2), π‘₯ βˆ’ 𝑦 = 4 8. (i) βˆ’8 πš€Μ‚ (ii) 7 πš€Μ‚ + 24πš₯Μ‚ βˆ’ 2π‘˜ (iii) 𝑓(π‘₯, 𝑦) = βˆ’ + π‘˜; π‘˜ 𝑖𝑠 π‘Žπ‘Ÿπ‘π‘–π‘‘π‘Ÿπ‘Žπ‘Ÿπ‘¦ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘. 9. N.A. 10. 0

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