Worksheet 1.8: Geometry of Vector Derivatives

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Vincent Richards
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1 Boise State Math 275 (Ultman) Worksheet 1.8: Geometry of Vector Derivatives From the Toolbox (what you need from previous classes): Calc I: Computing derivatives of single-variable functions y = f (t). Geometrically: these derivatives represent slopes of tangent lines. In real world applications, these derivatives represent instantaneous rates of change. Vectors: Vector operations (vector addition, scalar multiplication, the dot and cross products); computing the magnitude of a vector; know what a vector function is, and be able to evaluate vector functions at specific parameter (input) values. The derivative r (t) is vector function, computed by taking the derivatives of each of the coordinate functions of the vector function r(t). Since these coordinate functions are functions of a single variable, these are Calc I derivatives. The vector r (t) is tangent to r(t), and points in the direction along the curve in which t is increasing. The unit tangent vector ˆT (t) and the vector differential dr are related to the vector derivative r (t). In this worksheet, you will: Use vector derivatives to find vector equations of tangent lines and angles of intersection between curves. Determine when position vectors and their derivatives are orthogonal. Compute the unit tangent vector ˆT and the vector differential dr.

2 Boise State Math 275 (Ultman) Worksheet 1.8: Geometry of Vector Derivatives 1 Model 1: CALC I DERIVATIVES vs. VECTOR DERIVATIVES Diagram 1A: Derivative of a function y = f (t) (Calc I) Secant Line Tangent Line Diagram 1B: Derivative of a vector function r(t) = x(t), y(t) (Calc III) Displacement Vector Tangent Vector Critical Thinking Questions In this section, you will review the derivative of a function y = f (t) from Calc I, and see how the vector derivative r (t) is similar. (Q1) Review of Derivatives from Calc I (see Diagram 1A) (a) A secant line is a line between two points on the graph of a function. On the left side of Diagram 1A is a secant line between the point ( t, f (t) ) and a nearby point ( ) t + t, f (t + t). The slope of a secant line gives the average instantaneous rate of change of the function between the two points. This slope is: m sec = f (t + t) f (t) (t + t) t = f

3 Boise State Math 275 (Ultman) Worksheet 1.8: Geometry of Vector Derivatives 2 (b) The symbol means a change in, so the slope of this secant line is the change in the output f (t) relative to the change in the input. (c) The tangent line at a point is the line that best approximates the graph at that point; this means, as you zoom in on the point, the graph and the line look more and more similar different. If such a line exists, we say that the function is differentiable at that point, and the derivative is the of the tangent line: f (t) = df dt = m tan The slope of a tangent line gives the average instantaneous rate of change of the function at a point. (d) How do tangent lines relate to secant lines? On the graph on the left of Diagram 1A, draw a third point on the graph y = f (t) between the two points shown. Sketch the secant line between the point ( t, f (t) ) and the point you just drew. Do this again, with a fourth point between ( t, f (t) ) and the third point you just drew. Notice that, as the increment t gets smaller, the secant line begins to look more and more similar to different from the tangent line (we say that the secant line converges to the tangent line as t goes to zero). As this happens, the slope of the secant line converges to the slope of the tangent line. This process is called taking a limit: f lim t 0 t = df dt = f (t). (Please do not freak out. I will not make you compute limits. I just want you to know what it means in the context of derivatives.) (Q2) Derivatives of Vector Functions (see Diagram 1B) The left side of Diagram 1B shows three vectors. r(t) and r(t + t) are position vectors along the curve C. The third vector, r(t), is a displacement vector. r(t) gives the change in position along the curve between the parameter values t and t + t. (a) The right side of Diagram 1B shows the position vector r(t) and the vector derivative. This vector derivative is parallel to the line from Diagram 1A. This is why the vector derivative r (t) is sometimes called a tangent vector.

4 Boise State Math 275 (Ultman) Worksheet 1.8: Geometry of Vector Derivatives 3 (b) How do vector derivatives relate to displacement vectors? On the graph on the left of Diagram 1B, draw a third point on the curve C between the two points shown. Sketch the position vector for this new point, and the displacement vector between r(t) and the position vector for the new point you just drew. Do this again, with a fourth point between r(t) and the third point you just drew. Notice that, as the increment t gets smaller, the direction of the displacement vector r begins to look more and more similar to different from the direction of the tangent vector, but the magnitude of r(t) gets very large small. To get the vector derivative, multiply the displacement vector by 1/ t, and let t get very small. As t gets smaller, 1/ t gets smaller larger. The vector derivative is: r lim t 0 t = dr dt = r (t) (Q3) Parameterize the parabola y = 3 t 2 using the vector function r(t) = t, 3 t 2. (a) Evaluate r(1) and sketch it on the graph below. Remember, r(1) starts at the origin. (b) Compute r (1) and sketch it on the graph below, starting at the point (1, 2). ( c) Sketch the tangent line to the parabola above at the point (1, 2). A vector equation of this tangent line is r T (s) = sv + r 0, where the djrection vector v = r(1) r (1), and the fixed vector r 0 = r(1) r (1). So, using the vectors r(1) and r (1) that you computed in parts (a) and (b), a vector equation for the tangent line to this parabola at the point (1, 2) is: r T (s) = s, +,

5 Boise State Math 275 (Ultman) Worksheet 1.8: Geometry of Vector Derivatives 4 Model 2: VECTOR DERIVATIVES and ANGLES Diagram 2: Curve y = 3 x 2 Diagram 2A: Vector Function (Position Vectors) Diagram 2B Vector Derivative (Tangent Vectors) Position: r(t) = t, 3 t 2 r( 2) = 2, 1 r(0) = 0, 3 r(1) = 1, 2 Derivative: r (t) = 1, 2t r ( 2) = 1, 4 r (0) = 1, 0 r (1) = 1, 2 Critical Thinking Questions In this section, you will use what you know about measuring angles between vectors to measure angles between position and tangent vectors, and to find angles of intersection. (Q4) Diagram 2 shows the parabola y = 3 x 2, parameterized by the vector function r(t) = t, 3 t 2. Three position vectors are shown: r( 2), r(0), and r(1). (On the left, they are shown as solid arrows; on the right, as dashed arrows.) Use these position arrows to determine the direction that an object with position r(t) moves along the parabola as t increases. Object moves from left to right. Object moves from right to left. (Q5) On Diagram 2B: The solid arrows represent the vector derivatives r (t) for t = 2, 0, 1. What is the relationship between the direction of the vector derivative r (t) and the direction of motion as an object travels along the parabola as t increases?

6 Boise State Math 275 (Ultman) Worksheet 1.8: Geometry of Vector Derivatives 5 (Q6) Finding points where r(t) and r (t) are orthogonal. Using the parabola r(t) = t, 3 t 2 from Model 2: (a) In Diagram 2, it appears that the vectors r(0) and r (0) are orthogonal at the point (0, 3). Use the dot product to confirm this. Note: you are given the components for these vectors at the bottom of the diagrams. ( b) There are two additional values of t (besides t = 0) where r(t) and r (t) are orthogonal on the parabola r(t) = t, 3 t 2. i. First, use the graph on the right to estimate where on the curve r(t) and r (t) are orthogonal, and sketch these three pairs of orthogonal vectors r(t) and r (t). You know that r(0) and r (0) are orthogonal from part (a) now estimate the other two places this happens. ii. Now, find the other two t-values exactly, and use them to find the exact coordinates of the two additional points. Hint: use the dot product.

7 Boise State Math 275 (Ultman) Worksheet 1.8: Geometry of Vector Derivatives 6 If two curves intersect, their angle of intersection is the angle between their tangent lines at the point of intersection. You can find this angle by finding the angle between the tangent vectors. (Q8) The circle r c (t) = cos t, sin t and the parabola r p (s) = s, s 2 1 intersect at the point (1, 0). Your goal is to find the angle of intersection θ at this point. (a) On the circle r c (t) = cos t, sin t, find t 0 so that r c (t 0 ) = 1, 0. (b) On the parabola r p (s) = s, s 2 1, find s 0 so that r p (s 0 ) = 1, 0. s 0 and t 0 are different! (c) Compute the vector derivatives r c(t 0 ) and r p(s 0 ) (using t 0 and s 0 from parts (a) and (b)). (d) Find the angle between these two tangent vectors. This is the angle of intersection between the circle and the parabola at the point (1, 0). ( Q9) The graphs of the exponential functions y = e αx and y = e βx intersect at the point (0, 1). Find the relationship between α and β that must hold so that the graphs intersect at right angles. Use this to choose two specific values for α and β, and graph the resulting curves. Hint: Use the parameterization r(t) = t, e ct, c = α, β for the curves. For this parameterization, t = 0 at the point of intersection.

8 Boise State Math 275 (Ultman) Worksheet 1.8: Geometry of Vector Derivatives 7 Model 3: Other Tangent Vectors: ˆT and dr The unit tangent vector ˆT is tangent to the curve, and has magnitude ˆT = 1. The vector differential (or infinitesimal displacement vector) dr is an infinitesimal (very very small) tangent vector. This vector represents an infinitesimal displacement along a curve. If a curve is parameterized by a vector function r(t), then: ˆT = r (t) r (t) (as long as r (t) 0) dr = r (t) dt Critical Thinking Questions In this section, you will practice computing the unit tangent vector ˆT and vector differential dr. (Q10) We know that ˆT = r (t)/ r (t) and dr = r (t) dt are tangent to the curve r(t), because 1/ r (t) and dt are both vectors scalars. This means ˆT and dr are both scalar multiples of r (t), so they are both parallel orthogonal to r (t). So, since r (t) is tangent to the curve, so are ˆT and dr. (Q11) Compute ˆT and dr for the following vector functions: (a) r(t) = 5 cos t, 5 sin t (circle) (b) r(t) = t, t 2 (parabola) (c) r(t) = 5 cos t, 5 sin t, t (helix) (d) r(t) = 2t + 3, t 6, 5t (line)

Worksheet 4.2: Introduction to Vector Fields and Line Integrals

Boise State Math 275 (Ultman) Worksheet 4.2: Introduction to Vector Fields and Line Integrals From the Toolbox (what you need from previous classes) Know what a vector is. Be able to sketch vectors. Be