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1 Tangent Vectors and Normal Vectors Unit Tangent Vector: Let be a smooth curve represented by on an open interval. The unit tangent vector at is defined to be
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2 Example: (Larson, Example-1,Page-807) Find the unit tangent vector to the curve given by, when Solution: Given, Thus, the unit tangent vector is When, the unit tangent vector is Ans.
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3 Principal Unit Normal Vectors: Let be a smooth curve represented by on an open interval. If then the principal unit normal vector at is defined to be
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4 Example: (Larson, Example-3,Page-809) Find principal unit normal vector for the curve represented by,when Solution: (Do)
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5 Example: (Larson, Example-4,Page-810) Find the principal unit normal vector for he helix given by, when Solution:
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6 Thus, the principal unit normal vector is When, the principal unit normal vector is Ans.
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7 Tangential and Normal Components of Acceleration: If is the position vector for a smooth curve [for which exists], then the tangential and normal components of acceleration are as follows. Example: Find the tangential and normal components of acceleration for the position function given by
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8 Solution: Given, The tangential component of acceleration is And the normal component of acceleration is
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9 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 12.4 Page no: 814 Problems: 14,17,18;29,30;
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10 Arc Length and Curvature Arc Length: The arc length of a curve given by is
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11 Example: (Larson, Example-3,Page-817) Find the arc length of one turn of the helix given by Solution: Here, Thus, the arc length of one turn of the helix is Ans.
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12 Curvature: The curvature measures how fast a curve is changing direction at a given point. Let be a smooth curve given by,where is the arc length parameter. The curvature at is given by [Formal definition] Let be a smooth curve given by,then the curvature at is given by [General definition]
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13 Example: (Larson, Example-7,Page-821) Find the curvature of the curve given by Solution: Given vector valued function We have, Tangent vector
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15 We know, Curvature Ans.
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16 Practice Problems Problem 1: Find the arc length of the curve on the interval Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 12.5 Page no: 825 Problems: 21, 23,29,32;
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17 Directional Derivatives and Gradients To determine the slope at a point on a surface, we define new type of derivative called a Directional Derivative. And to determine in which direction at that point on that surface the slope is maximum, we introduce Gradient.
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18 Definition of Directional derivative: Let be a function of two variables and in the direction of the unit vector Then the directional derivative of in the direction of denoted by and is defined by [difficult to use] Or [simpler than previous one] [for 3 dimensions]
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19 Example: (Larson, Example-1, Page-879) Find the directional derivative of at in the direction of Solution: Given function Given direction unit vector,
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20 Therefore the directional derivative at is -1.866. Ans. -1.866. We have,
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21 Example: (Larson, Example-2, Page-879) Find the directional derivative of at in the direction of Solution: Given function Given direction vector, direction unit vector
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22 We have,
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23 Definition of Gradient: Let be a function of two variables and such that and exist. Then the gradient of is denoted by and is defined by the vector [ for two variables] [ for three variables]
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24 Properties of the Gradient:
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25 Example: (Larson, Example-8, Page-885) Find the gradient of at the point Solution: Given function We have,
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26 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 13.6 Page no: 886 Problems: 1,5,7,12; 17,20; 25,27,30;
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27 Tangent Planes and Normal Lines One nice use of tangent planes is they give us a way to approximate a surface near a point. Normal lines are important in analyzing surfaces and solids. Surface S: Normal Line Tangent Plane
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28 Definition of Tangent Plane and Normal Line: Let be differentiable at the point on the surface given by such that 1.The plane through that is normal to is called the tangent plane to at. In particular the equation of the tangent plane is
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29 2. The line through having the direction of is called the normal line to at. In particular the equation of the normal line is
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30 Example: (Larson, Example-2, Page-891) Find an equation of the tangent plane to the hyperboloid given by at the point. Solution: Given equation is
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31 We have, equation of tangent line at is at the point, the partial derivatives are
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32 Example: (Larson, Example-3, Page-892) Find an equation of the tangent plane to the paraboloid given by at the point. Solution: Do
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33 Example: (Larson, Example-4, Page-893) Find the equation of the normal line to the surface given by at the point. Solution: Given equation of surface is Let
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34 at the point, we get we have, the parametric equation of normal line at is
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35 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 13.7 Page no: 895 Problems: 11,17,21;26,29;
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36 Vector Fields Functions that assign a vector to a point in the plane or to a point in space. Such functions are called vector fields, and they are useful in representing various types of force fields and velocity fields. Definition: Let and be functions of two variables and, defined on a plane region. The function defined by is called a vector field over and is called a vector field over Vector Field
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37 Conservative Vector Field: A vector field is called conservative if there exists a differentiable function such that. The function is called the potential function for. Example: (Larson, Example-4, Page-996) Show that the vector field given by is conservative if there exists Solution: Given Because it follows that is conservative. ( Showed)
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38 Test for Conservative Vector Field in Plane: Let and have continuous first partial derivatives on an open disc. The vector field given by is conservative if and only if Example: (Larson, Example-5, Page-997) Show that the vector field given by is not conservative. Solution: Here, and Since, it shows that is not conservative. (Showed).
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39 Example: (Larson, Example-6, Page-998) Find a potential function for Solution: Given function Since, so that is conservative comparing (i) and (ii), we get
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40 Integrating (iii) with respect to x, we obtain Integrating (iv) with respect to y, we obtain Therefore the required potential function is Ans. [ Where g(y) is constant with respect to x. ] [ Where h(x) is constant with respect to y. ]
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41 Example: (Larson, Example-8, Page-1000) Find a potential function for Solution: Do.
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42 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 15.1 Page no: 1002 Problems: 21,27;37,39;
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43 Curl of a Vector Field Curl of a vector field captures the idea of how a fluid may rotate. Imagine that the below vector field represents fluid flow. The vector field indicates that the fluid is circulating around a central axis.
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44 Definition of Curl of a vector field: The Curl of a vector field is Remark: If, then we say that is irrotational and conservative.
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45 Example: (Larson, Example-6, Page-998) Find the curl of a Vector Field Solution: The curl of a Vector Field is given by
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46 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 15.1 Page no: 1002 Problems: 29,31;
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47 Divergence of a Vector Field Imagine that the vector field F pictured below gives the velocity of some fluid flow. It appears that the fluid is exploding outward from the origin (fig-1).This expansion of fluid flowing with velocity field F is captured by the divergence of F, which we denote divF. The divergence of that vector field is positive since the flow is expanding. Since this compression of fluid is the opposite of expansion in (fig-2), the divergence of this vector field is negative. Fig-2Fig-1
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48 Definition of Divergence of a Vector Field The divergence of is denoted by and defined by If, then is said to be divergence free.
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49 Example: Find the divergence at for the vector field Solution: The divergence of at the point, the divergence is
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50 Practice Problems Book: Calculus with Analytical Geometry (5 th ed.) (Larson, Edward and Hostetler) Exercises for Section 15.1 Page no: 1002 Problems: 49,52,53;
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51 End (of Mid Term Syllabus)
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52 Discussed Topics Tangent Vectors and Normal Vectors Arc Length and Curvature Directional Derivatives and Gradients Tangent Planes and Normal Lines Vector Fields
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