Engineering Mathematics I Tutorial Sheet 5 PDF
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Jaypee Institute of Information Technology
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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.
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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€Μ + π₯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₯ + 7π¦ β 5π§ = 21. (v) the point (2,3,0) and perpendicular to π΄β = π€Μ + 2π₯Μ + 3π and π΅β = 3π€Μ + 4π₯Μ + 5π 2. Find the equation for the plane passing through: (i) the points (1,3,2), (3, β1, 6), (5,2,0). (ii) the point π (0,2, β1) & normal to πβ = 3π€Μ β 2π₯Μ β π. 3. Find the point of intersection of the lines π₯ = 2π‘ + 1, π¦ = 3π‘ + 2, π§ = 4π‘ + 3 and π₯ = π + 2, π¦ = 2π + 4, π§ = β4π β 1. Also find the plane determined by these lines. 4. Find the length of the curve defined by πβ(π‘) = 6π‘, 3β2π‘ , 2π‘ , 0 β€ π‘ β€ 1. 5. Find the directional derivative of the functions: (i) π(π₯, π¦, π§) = 2π¦π§ + π§ in the direction of the vector π€Μ + 2π₯Μ + 2π at the point(1, β1, 3). (ii) π(π₯, π¦, π§) = π₯ + π¦ + π§ at the point (1, β1, 2) in the direction of π€Μ + 2π₯Μ + π. (iii) π(π₯, π¦, π§) = π₯π§ β 3π₯π¦ + 2π₯π¦π§ β 3π₯ + 5π¦ β 17 from the point π(2, β6, 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₯ π¦ + π₯π§ = 2π¦ π§, π = (1,1,1). (ii) 2π§ β π₯ , π = (1,1,1). (iii) 4 β π₯ β 2π¦ , π = (1, β1,1). 7. Sketch the curve π(π₯, π¦) = π together with βπ and the tangent line at the given point. Also determine the equation of the tangent line. (i) π₯ + π¦ = 4, (β2, β2 ). (ii) π₯π¦ = β4, (2, β2). 8. Answer the following: (i) Find the gradient of π(π₯, π¦) = π¦ β 4π₯π¦, at (1,2). (ii) Find the gradient of π(π₯, π¦, π§) = π₯ π¦ + π₯π¦ β π§ , at (3, 1, 1). (iii) For a force field πΉβ(π₯, π¦) = π€Μ β π₯Μ applied on an object, find π(π₯, π¦) such that ππππ π = πΉβ(π₯, π¦). β 9. If πβ = π₯π€Μ + π¦π₯Μ + π§π, |πβ| = π and π = , show that ππππ = β. 10. Evaluate πππ£ ππ’ππ πΉβ whereπΉβ = π¦π§ π€Μ + π₯π¦π₯Μ + π¦π§π. *** Answers 1. (i) πβ = 5π‘ π€Μ β 2π‘π₯Μ + 3π‘π ; x=2t, y= -2t, z=3t (ii) πβ = (3 + π‘) π€Μ β (π‘ β 4)π₯Μ + (π‘ β 1)π; x=3+t, y= t-4, z= t-1 (iii) πβ = π€Μ + π₯Μ + (π‘ + 1)π; x = 1, y = 1, z = 1+t (iv) πβ = (2 + 3π‘) π€Μ β (4 + 7π‘)π₯Μ + (5 β 5π‘)π; x = 2+3t, y = 4+7t, z = 5-5t (v) πβ = (2 β 2π‘) π€Μ β (3 + 4π‘)π₯Μ + (β2π‘)π; x = 2 β 2t, y = 3 + 4t, z = -2t 2. (i) 6π₯ + 10π¦ + 7π§ = 50 (ii) 3π₯ β 2π¦ β π§ = β3 3. β20π₯ + 12π¦ + π§ = 7 4. 8 β 5. (i) (ii) (iii) 18 6. (i) π₯ β π¦ = 0, (1 + 3π‘, 1 β 3π‘, 1) (ii) 2π₯ β π§ β 2 = 0, (2 β 4π‘, 0, 2 + 2π‘) (iii) π₯ β 2π¦ β 3 = 0, (1 β π‘, β1 + 2π‘, 1) 7. (i) 2β2, 2β2 , π¦ = βπ₯ + 2β2 (ii) (β2,2), π₯ β π¦ = 4 8. (i) β8 π€Μ (ii) 7 π€Μ + 24π₯Μ β 2π (iii) π(π₯, π¦) = β + π; π ππ πππππ‘ππππ¦ ππππ π‘πππ‘. 9. N.A. 10. 0