Calculus Vector-Valued Functions
Outline 1 Vector-Valued Functions 2 The Calculus of Vector-Valued Functions 3 Motion in Space 4 Curvature 5 Tangent and Normal Vectors 6 Parametric Surfaces
Vector Valued Functions: An Example To specified the path of the plane indicated in the figure, we describe the plane s location at any given time by the endpoint of a vector called position vector Figure: [9.1a] Airplane s flight path. Figure: [9.1b] Vectors indicating plane s position at several times.
Definition of Vector-Valued Functions Definition (1.1) A vector-valued function r(t) is a mapping from its domain D R to its range R V 3, so that for each t in D, r(t) = v for only one vector v V 3. We can always write a vector-valued function as r(t) = f (t)i + g(t)j + h(t)k, for some scalar functions f, g and h (called the component functions of r). we can likewise define a vector-valued function r(t) in V 2 by r(t) = f (t)i + g(t)j for some scalar function f and g.
Sketching the Curve Defined by a Vector-Valued Function Example (1.1) Sketch a graph of the curve traced out by the endpoint of the two-dimensional vector-valued function r(t) = (t + 1)i + (t 2 2)j.
The Trace of a Vector-Valued Function In example 1.1, the curve traced out by the endpoint of the vector-valued function r(t) = (t + 1)i + (t 2 2)j is identical to the curve described by the parametric equations x(t) = t + 1 and y(t) = t 2 1 Figure: [9.2a] Some values of r(t) = (t + 1)i + (t 2 2)j. Figure: [9.2b] Curve defined by r(t) = (t + 1)i + (t 2 2)j.
A Vector-Valued Function Defining an Ellipse Example (1.2) Sketch a graph of the curve traced out by the endpoint of the vector-valued function r(t) = 4 cos t i 3 sin t j, t R. Figure: [9.3] Curve defined by r(t) = 4 cos t i 3 sin t j.
A Vector-Valued Function Defining an Elliptical Helix Example (1.3) Plot the curve traced out by the vector-valued function r(t) = sin t i 3 cos t j + 2t k, for t 0. Figure: [9.4a] Elliptical helix, r(t) = sin t i 3 cos t j + 2t k. Figure: [9.4b] Computer sketch: r(t) = sin t i 3 cos t j + 2t k.
A Vector-Valued Function Defining a Line Example (1.4) Plot the curve traced out by the vector-valued function r(t) = 3 + 2t, 5 3t, 2 4t, t R. Figure: [9.5] Straight line: r(t) = 3 + 2t, 5 3t, 2 4t.
Matching a Vector-Valued Function to Its Graph (I) Example (1.5) Match each of the vector-valued functions r 1 (t) = cos t, ln t, sin t, r 2 (t) = t cos t, t sin t, t, r 3 (t) = 3 sin 2t, t, t and r 4 (t) = 5 sin 3 t, 5 cos 3 t, t with the corresponding computer-generated graph.
Matching a Vector-Valued Function to Its Graph (II) r 1 (t) = cos t, ln t, sin t r 2 (t) = t cos t, t sin t, t r 3 (t) = 3 sin 2t, t, t r 4 (t) = 5 sin 3 t, 5 cos 3 t, t
Matching a Vector-Valued Function to Its Graph (III) r 1 (t) = cos t, ln t, sin t r 2 (t) = t cos t, t sin t, t r 3 (t) = 3 sin 2t, t, t r 4 (t) = 5 sin 3 t, 5 cos 3 t, t
Arc Length in R 2 Recall from section 5.3 that if f and f are continuous on the interval [a, b], then the arc length of the curve y = f (x) on that interval is given by s = b a 1 + (f (x)) 2 dx Consider a curve defined parametrically by x = f (t) and y = g(t), where f, f, g and g are all continuous for t [a, b]. If the curve is traversed exactly once as t increases from a to b, then the arc length is given s = b a (f (x)) 2 + (g (x)) 2 dt
Arc Length in R 3 Suppose that a curve in three dimensions is traced out by the end point of the vector-valued function r(t) =< f (t), g(t), h(t) >, where f, f, g, g, h and h are all continuous for t [a, b] and where the curve is traversed exactly once as t increases from a to b. For this situation, the arc length is given by s = b a (f (x)) 2 + (g (x)) 2 + (h (x)) 2 dt The integral can only rarely be computed exactly and we must typically be satisfied with a numerical approximation.
Approximating the Arc Length of a Curve in R 3 We approximate the curve by a number of line segments and use the distance formula to yield an approximation to the arc length. As the number of segments increases without bound, the sum of the distances approaches the actual arc length. Figure: [9.7a] Approximate arc length in R 2. Figure: [9.7b] Approximate arc length in R 3. Figure: [9.7c] Improved arc length approximation.
Example (1.6) Computing Arc Length in R 3 Find the arc length of the curve traced out by the endpoint of the vector-valued function r(t) = 2t, ln t, t 2, for 1 t e. Figure: [9.8] The curve defined by r(t) = 2t, ln t, t 2.
Approximating Arc Length in R 3 Example (1.7) Find the arc length of the curve traced out by the endpoint of the vector-valued function r(t) = e 2t, sin t, t, for 0 t 2.
Finding Parametric Equations for an Intersection of Surfaces Example (1.8) Find parametric equations for the curve determined by the intersection of the cone z = x 2 + y 2 and the plane y + z = 2. Figure: [9.9a] z = x 2 + y 2 and y + z = 2. Figure: [9.9b] x = t, y = 1 t2 4, z = 1 + t2 4.
Limit of a Vector-Valued Function (I) For a vector-valued function If we write r(t) =< f (t), g(t), h(t) >, lim r(t) = u, t a we mean that as t gets closer and closer to a, the vector r(t) is getting closer and closer to the vector u. If we write u =< u 1, u 2, u 3 >, this means that lim r(t) = lim < f (t), g(t), h(t) >= u =< u 1, u 2, u 3 > t a t a Notice that for this to occur, we must have that f (t) u 1, g(t) u 2, and h(t) u 3.
Limit of a Vector-Valued Function (II) Definition (2.1) For a vector-valued function r(t) = f (t), g(t), h(t), the limit of r(t) as t approaches a is given by lim r(t) = lim f (t), g(t), h(t) = lim f (t), lim g(t), lim h(t), t a t a t a t a t a provided all of the indicated limits exist. If any one of the limits indicated on the right-hand side of (2.1) fails to exist, then r(t) does not exist. lim t a
Finding the Limit of a Vector-Valued Function Example (2.1) Find lim t 0 t 2 + 1, 5 cos t, sin t.
A Limit That Does Not Exist Example (2.2) Find lim t 0 e 2t + 5, t 2 + 2t 3, 1 t.
Continuity of a Vector-Valued Function (I) Recall that for a scalar function f, we say that f is continuous at a if and only if lim f (t) = f (a) t a We defined the continuity of vector-valued functions in the same way. Definition (2.2) The vector-valued function r(t) = f (t), g(t), h(t), is continuous at t = a whenever lim r(t) = r(a) t a (i.e., whenever the limit and the value of the vector-valued function are the same).
Continuity of a Vector-Valued Function (III) Notice that in terms of the components of r, the definition says that t is continuous at a whenever Further since lim t a lim < f (t), g(t), h(t) >=< f (a), g(a), h(a) > t a < f (t), g(t), h(t) >=< lim t a f (a), lim g(a), lim h(a) > t a t a It follows that r(t) is continuous at t = a if and only if < lim t a f (t), lim t a g(t), lim t a h(t) >=< f (a), g(a), h(a) > Note that this occurs if and only if lim t a f (t) = f (a), lim t a g(t) = g(a), and lim h(t) = f (a). t a The above result is summarized in Theorem 2.1.
Continuity of a Vector-Valued Function (IV) Theorem (2.1) A vector-valued function r(t) = f (t), g(t), h(t) is continuous at t = a if and only if all of f, g and h are continuous at t = a. Notice that Theorem 2.1 says that if you want to determine whether or not a vector-valued function is continuous, you need only check the continuity of each component function.
Determining Where a Vector-Valued Function Is Continuous Example (2.3) Determine for what values of t the vector-valued function r(t) = e 5t, ln(t + 1), cos t is continuous.
A Vector-Valued Function with Infinitely Many Discontinuities Example (2.4) Determine for what values of t the vector-valued function 1 r(t) = tan t, t + 3, t 2 is continuous.
The Derivative of a Vector-Valued Function (I) Recall that we defined the derivative of a scalar function f to be f f (t + t) f (t) (t) = lim t 0 t Similarly, we define the derivative of a vector-valued function as follows. Definition (2.3) The derivative r (t) of the vector-valued function r(t) is defined by r (t) = lim t 0 r(t + t) r(t), t for any values of t for which the limit exists. When the limit exists for t = a, we say that r is differentiable at t = a.
The Derivative of a Vector-Valued Function (II) Theorem (2.2) Let r(t) = f (t), g(t), h(t) and suppose that the components f, g and h are all differentiable for some value of t. Then r is also differentiable at that value of t and its derivative is given by r (t) = f (t), g (t), h (t). Theorem 2.2 says that the derivative of a vector-valued function is found directly from the derivatives of the individual components.
Finding the Derivative of a Vector-Valued Function Example (2.5) Find the derivative of r(t) = sin(t 2 ), e cos t, t ln t.
Rules for Computing Derivatives of Vector-Valued Functions Theorem (2.3) Suppose that r(t) and s(t) are differentiable vector-valued functions, f (t) is a differentiable scalar function and c is any scalar constant. Then d 1 dt [r(t) + s(t)] = r (t) + s (t) d 2 dt [cr(t)] = cr (t) d 3 dt [ f (t)r(t)] = f (t)r(t) + f (t)r (t) d 4 dt [r(t) s(t)] = r (t) s(t) + r(t) s (t) and d 5 dt [r(t) s(t)] = r (t) s(t) + r(t) s (t).
A Graphical Interpretation of the Derivative of a Vector-Valued Function (I) Recall that the derivative of r(t) at t = a is given by r r(a + t) a (a) = lim t 0 t Notice that the endpoint of the vector-valued function r(t) traces out a curve C in R 3 as shown in the figure. Observe that the vector r(a + t) r(t) points in the same t direction as r(a + t) r(t).
A Graphical Interpretation of the Derivative of a Vector-Valued Function (II) If we take smaller and smaller value of t, approach r (t). r(a + t) r(t) t will
A Graphical Interpretation of the Derivative of a Vector-Valued Function (III) As t 0, notice that the vector r(a + t) r(t) approaches a t vector that is tangent to the curve C at the terminal point of r(t). We refer to r (a) as the tangent vector to the curve C at the point corresponding to t = a. Observe that r (a) lies along the tangent line to the curve at t = a and points in the direction of the orientation of C.
A Graphical Interpretation of the Derivative of a Vector-Valued Function (IV)
Drawing Position and Tangent Vectors Example (2.6) For r(t) = cos 2t, sin 2t, plot the curve traced out by the endpoint of r(t) and draw the position vector and tangent vector at t = π 4. Figure: [9.11] Position and tangent vectors.
r(t) = constant r (t) r(t) Theorem (2.4) r(t) = constant if and only if r(t) and r (t) are orthogonal, for all t. Theorem 2.4 implies that: 1 in two (three) dimensions, if r(t) = c, then the curved traced out by the position r(t) must lie on the circle (sphere) of radius c, centered at the origin, and 2 the path traced out by r(t) lies on a circle (sphere) centered at the origin if and only if the tangent vector is orthogonal to the position vector at every point on the curve.
Antiderivatives of the Vector-Valued Functions Definition (2.4) The vector-valued function R(t) is an antiderivative of the vector-valued function r(t) whenever R (t) = r(t). Notice that if r(t) =< f (t), g(t), h(t) > and f, g and h have antiderivatives F, G and H, respectively, then d dt < F(t), G(t), H(t) >=< F (t), G (t), H (t) >=< f (t), g(t), h(t) > That is, < F(t), G(t), H(t) > is an antiderivative of r(t). In fact, < F(t) + c 1, G(t) + c 2, H(t) + c 3 > is also an antiderivative of r(t), for any choice of constants, c 1, c 2 and c 3.
Indefinite Integral of a Vector-Valued Function Definition (2.5) If R(t) is any antiderivative of r(t), the indefinite integral of r(t) is defined to be r(t)dt = R(t) + c, where c is an arbitrary constant vector. As in the scalar case, R(t) + c is the most general antiderivative of r(t). Notice that this says that r(t) dt = < f (t), g(t), h(t) >= f (t) dt, g(t) dt, h(t) dt That is, you integrate a vector-valued function by integrating each of the individual components.
Evaluating the Indefinite Integral of a Vector-Valued Function Example (2.7) Evaluate the indefinite integral t 2 + 2, sin 2t, 4te t2 dt.
Definite Integral of a Vector-Valued Function Similarly, we defined the definite integral of a vector-valued function in the obvious way. Definition (2.6) For the vector-valued function r(t) = f (t), g(t), h(t), we define the definite integral of r(t) by b b b b b r(t)dt = f (t), g(t), h(t) dt = f (t)dt, g(t)dt, h(t)dt. a a a a a Notice that this says that the definite integral of a vector-valued function r(t) is simply the vector whose components are the definite integrals of the corresponding components of r(t).
Fundamental Theorem of Calculus for Vector-Valued Functions Theorem (2.5) Suppose that R(t) is an antiderivative of r(t) on the interval [a, b]. Then, b a r(t)dt = R(b) R(a).
Evaluating the Definite Integral of a Vector-Valued Function Example (2.8) Evaluate 1 0 sin πt, 6t 2 + 4t dt.
Motion in Space (I) Suppose that an object moves along a curve described parametrically by C : x = f (t), y = g(t), z = h(t) where t [a, b]. We can think of the curve as being traced out by the endpoint of the vector-valued function r(t) =< f (t), g(t), h(t) >. Figure: [9.12] Position, velocity and acceleration vectors.
Motion in Space (II) Differentiating r(t), we have r (t) =< f (t), g (t), h (t) > and the magnitude of this vector-valued function is r(t) = [f (t)] 2 + [g (t)] 2 + [h (t)] 2 Figure: [9.12] Position, velocity and acceleration vectors.
Motion in Space (III) Recall that the arc length of the portion of the curve from u = t 0 up to u = t is given by s(t) = t t 0 [f (u)] 2 + [g (u)] 2 + [h (u)] 2 dt. If we differentiate both sides of the above equation, we get s (t) = [f (t)] 2 + [g (t)] 2 + [h (t)] 2 = r (t). Figure: [9.12] Position, velocity and acceleration vectors.
Motion in Space (IV) Since s(t) represent arc length, s (t) gives the instantaneous rate of change of arc length with respect to time, i.e., the speed of the object as it moves along the curve. Figure: [9.12] Position, velocity and acceleration vectors.
Motion in Space (V) For any given t, r (t) is a tangent vector pointing in the direction of the orientation of C and whose magnitude gives the speed of the object. We call r(t) the velocity vector, usually denoted v(t). Figure: [9.12] Position, velocity and acceleration vectors.
Motion in Space (VI) We refer to the derivative of the velocity vector v (t) = r (t) as the acceleration vector, denoted a(t). Figure: [9.12] Position, velocity and acceleration vectors.
Finding Velocity and Acceleration Vectors Example (3.1) Find the velocity and acceleration vectors if the position of an object moving in the xy-plane is given by r(t) = t 3, 2t 2. Figure: [9.13] Position, velocity and acceleration vectors.
Finding Velocity and Position from Acceleration Example (3.2) Find the velocity and position of an object at any time t, given that its acceleration is a(t) = 6t, 12t + 2, e t, its initial velocity is v(0) = 2, 0, 1 and its initial position is r(0) = 0, 3, 5. Figure: [9.14] Position, velocity and acceleration vectors.
Newton s Second Law of Motion The Newton s second law of motion states that the net force acting on an object equals the product of the mass and the acceleration. Using vector notation, we have the vector form of Newton s second law: F = ma Here, m is the mass, a is the acceleration vector and F is the vector representing the net force acting on the object.
Newton s Second Law of Motion and Linear Momentum When a force F(t) applies on an object of constant mass m, the motion of the object obeys the Newton s second law of motion, F(t) = ma(t) where a is the acceleration of the object. Now, integrating the Newton s second law with respect to time we have t2 t 1 F(t) dt = m t2 t 1 a dt = mv(t) t 2 t1 = mv(t 2 ) mv(t 1 ) In the last expression, the term mv(t) is referred to as the linear momentum of the object.
Rotational Version of Newton s Second Law of Motion In the case of an object rotating in two dimensions, the primary variable that we track is an angle of displacement, denoted by θ. For a rotating body, the angle measured from fixed ray changes with time t, so that the angle is a function θ(t). We define the angular velocity to be ω(t) = θ (t) and the angular acceleration to be α(t) = θ (t) = θ (t) The equation of rotation motion is then τ = Iα where I is the moment of initial of a body and τ is the torque causing the rotation.
Rotational Version of Newton s Second Law of Motion and Angular Momentum For rotational motion in three dimensions, the calculations are somewhat more complicated. Recall that we had defined the torque τ due to a force F applied at position r to be τ = r F Integrating the above equation with respect to time, we yield t2 t 1 τ dt = = t2 t2 t 1 r F dt = t2 t 1 v mv + r mv dt = = (r mv) t 2 t1 t 1 v mv + r ma dt t2 t 1 (r mv) dt The term r mv is referred to as the angular linear momentum
Finding the Force Acting on an Object Example (3.3) Find the force acting on an object moving along a circular path of radius b centered at the origin, with constant angular speed. Figure: [9.15a] Motion along a circle. Figure: [9.15b] Centripetal force.
Example (3.4) Analyzing the Motion of a Projectile A projectile is launched with an initial speed of 140 feet per second from ground level at an angle of π 4 to the horizontal. Assuming that the only force acting on the object is gravity (i.e., there is no air resistance, etc.), find the maximum altitude, the horizontal range and the speed at impact of the projectile. Figure: [9.16a] Initial velocity vector. Figure: [9.16b] Path of a projectile.
The Rotational Motion of a Merry-Go-Round Example (3.5) A stationary merry-go-round of radius 5 feet is started in motion by a push consisting of a force of 10 pounds on the outside edge, tangent to the circular edge of the merry-goround, for 1 second. The moment of inertia of the merry-go-round is I = 25. Find the resulting angular velocity of the merry-go-round.
Relating Torque and Angular Momentum Example (3.6) Show that torque equals the derivative of angular momentum.
Analyzing the Motion of a Projectile in Three Dimensions (I) Example (3.7) A projectile of mass 1 kg is launched from ground level toward the east at 200 meters/second, at an angle of π 6 to the horizontal. If a gusting northerly wind applies a steady force of 2 newtons to the projectile, find the landing location of the projectile and its speed at impact. Figure: [9.17a] The initial velocity and wind velocity vectors.
Analyzing the Motion of a Projectile in Three Dimensions (II) Figure: [9.17b] Path of the projectile. Figure: [9.17c] Projection of path onto the xz-plane.
Arc Length of a Curve Recall that for the curve traced out by the endpoint of the vector-valued function r(t) =< f (t), g(t), h(t) >, for a t b, we define the arc length parameter s(t) to be the arc length of that portion of the curve from u = a up to u = t, i.e., t s(t) = [f (u)] 2 + g (u)] 2 + h (u)] 2 du 0 Recognizing that [f (u)] 2 + g (u)] 2 + h (u)] 2 = r (u), we can write this more simply as s(t) = t a r (u) du
Parameterizing a Curve in Terms of Arc Length Example (4.1) Find an arc length parameterization of the circle of radius 4 centered at the origin.
Unit Tangent Vector (I) Consider the curve C traced out by the endpoint of the vector-valued function r(t). Recall that for each t, v(t) = r (t) can be thought of as both the velocity vector and a tangent vector, pointing in the direction of motion (i.e., the orientation of C). Figure: [9.18] Unit tangent vectors.
Unit Tangent Vector (II) Notice that T(t) = r (t) r (t) is also a tangent vector, but has length one ( T(t) = 1). We call T is a tangent vector of length one pointing in the direction of the orientation of C. Figure: [9.18] Unit tangent vectors.
Example (4.2) Finding a Unit Tangent Vector Find the unit tangent vector to the curve determined by r(t) = t 2 + 1, t. Figure: [9.18] Unit tangent vectors.
Tangent Vectors and "Sharpness" of Curves (I) In the following figures, we show two curves both connecting the points A and B. The curve in Figure 9.19b indicates a much sharper turn than the curve in Figure 9.19a. The question before us is to see how to mathematically describe this degree of "sharpness". Figure: [9.19a] Gentle curve. Figure: [9.19b] Sharp curve.
Tangent Vectors and "Sharpness" of Curves (II) In the same figures, we have draw in a number of unit tangent vectors at equally spaced points on the curve. Notice that the unit tangent vectors change very slowly along the gentle curve in Figure 9.19c, but twist and turn quite rapidly in the vicinity of the sharp curve in Figure 9.19d. Figure: [9.19c] Unit tangent vectors. Figure: [9.19d] Unit tangent vectors.
Tangent Vectors and "Sharpness" of Curves (III) Based on the analysis, the rate of change of the unit tangent vectors with respect to the arc length along the curve will give us a measure of sharpness. Figure: [9.19c] Unit tangent vectors. Figure: [9.19d] Unit tangent vectors.
Curvature of a Curve (I) Definition (4.1) The curvature κ of a curve is the scalar quantity κ = dt ds. Computing the curvature κ of given curve by the definition is not a simple matter. To do so, we would need to first find the arc length parameter and the unit tangent vector T(t), rewrite T(t) in terms of the arc length parameter s and then differentiate with respect to s
Observe that by the chain rule so that Recall that Curvature of a Curve (II) T (t) = dt dt = dt ds ds dt, κ = dt ds = T (t) ds s(t) = t a dt r (u) du By the Fundamental Theorem of Calculus, Hence, we have ds dt = r (t) κ = T (t) r (t)
Finding the Curvature of a Straight Line Example (4.3) Find the curvature of a straight line.
Finding the Curvature of a Circle Example (4.4) Find the curvature for a circle of radius a.
Another Way of Computing the Curvature of a Curve Theorem (4.1) The curvature of the curve traced out by the vector-valued function r(t) is given by κ = r (t) r (t) r (t) 3. This is a relatively simple matter to use this formula to compute the curvature for nearly any three-dimensional curve.
Example (4.5) Finding the Curvature of a Helix Find the curvature of the helix traced out by r(t) = 2 sin t, 2 cos t, 4t. Figure: [9.20] Circular helix.
Curvature for the plane curve y = f (x) For a plane curve y = f (x), we can derive a simple formula for the curvature. Notice that such a curve is traced out by the vector-valued function r(t) =< t, f (t), 0 > in the xy-plane. Further, r (t) =< 1, f (t), 0 > and r (t) =< 0, f (t), 0 > From Theorem 4.1, we have κ = r (t) r (t) r (t) 3 = f (t) {1 + [f (t)] 2 } 3/2 Replacing t by x, we write the curvature as κ = = < 1, f (t), 0 > < 0, f (t), 0 > < 1, f (t), 0 > 3 f (x) {1 + [f (x)] 2 } 3/2
Finding the Curvature of a Parabola Example (4.6) Find the curvature of the parabola y = ax 2 + bx + c. Also, find the limiting value of the curvature as x.
Stationary and Moving Frame of Reference Up to this point, we have used a stationary frame of reference for all of our work with vectors. That is, all the vector are in terms of the standard unit basis vectors i, j and k. In this section we introduce a moving frame of reference which is suitable for describing the motion of a moving object.
TNB Frame of Reference (I) Consider an object moving along the curved traced out by a vector-valued function r =< f (t), g(t), h(t) >. To define a reference frame that moves with the object, we will need to have three mutually orthogonal unit vectors at each point on the curve. Figure: [9.21] Principal unit normal vectors.
TNB Frame of Reference (II) One of these vectors should be pointing in the direction of motion, i.e., in the direction of the orientation of the curve. Since T = r r is a vector pointing in the direction of motion, T can be used as one of the three mutually orthogonal unit vectors. Figure: [9.21] Principal unit normal vectors.
TNB Frame of Reference (III) Recall that T is a unit vector, i.e., T = 1. Hence, from Theorem 2.4, we have T T = 0, implying that T and T are orthogonal. This gives us a second unit vector in our moving frame of reference, as in Definition 5.1. Figure: [9.21] Principal unit normal vectors.
TNB Frame of Reference (IV) Principle Unit Normal Vector Definition (5.1) The principal unit normal vector N(t) is a unit vector having the same direction as T (t) and is defined by N(t) = T (t) T (t).
TNB Frame of Reference (V) Recall that for a given curve traced out by r, the arc length parameter s(t) is given by Note that This implies that s(t) = t a r (u) du ds dt = r (t) > 0 ds dt = ds dt
TNB Frame of Reference (VI) From the chain, we have This gives us T (t) = dt dt = dt ds ds dt. N(t) = T (t) T (t) = dt ds dt ds ds dt ds dt = dt ds dt ds = 1 dt κ ds where we have used the definition of curvature.
TNB Frame of Reference (VII) The expression N(t) = 1 dt κ ds is not a practically useful formula for computing N. However, it can be used to determine the direction of N(t). Since κ > 0, N has the same direction as dt dt. Notice that ds ds is the instantaneous rate of change of the unit tangent vector with respect to the arc length. Figure: [9.21] Principal unit normal vectors.
TNB Frame of Reference (VIII) This says that dt points in the ds direction in which T is turning as arc length increases. That is, N(t) will always point to the concave side of the curve. Figure: [9.21] Principal unit normal vectors.
TNB Frame of Reference (IX) To get a third unit vector orthogonal to both T and N, we simply take their cross product. This leads to the following definition. Definition (5.2) We define the binormal vector B(t) to be B(t) = T(t) N(t). Notice that since T and N are unit vectors and orthogonal to each other, the magnitude of B is 1.
TNB Frame of Reference (X) This triple of three unit vectors T(t), N(t) and B(t) forms a frame of reference, called the TNB frame (or the moving trihedral), that moves along the curve defined as r(t). This moving frame of reference has particular importance in a branch of mathematics call differential geometry. Figure: [9.24] The TNB frame.
Finding Unit Tangent and Principal Unit Normal Vectors Example (5.1) Find the unit tangent and principal unit normal vectors to the curve defined by r(t) = t 2, t. Figure: [9.22] Unit tangent and principal unit normal vectors.
Finding Unit Tangent and Principal Unit Normal Vectors Example (5.2) Find the unit tangent and principal unit normal vectors to the curve determined by r(t) = sin 2t, cos 2t, t. Figure: [9.23] Unit tangent and principal unit normal vectors.
Example (5.3) Finding the Binormal Vector Find the binormal vector B(t) for the curve traced out by r(t) = sin 2t, cos 2t, t. Figure: [9.25] The TNB frame for r(t) = sin 2t, cos 2t, t.
Normal Plane and Osculating Plane For each value of t, the plane determined by N and B is called the normal plane. By definition, the normal plane to a curve at a point contains all of the lines that are orthogonal to the tangent vector at the given point on the curve. For each value of t, the plane determined by T and N is called the osculating plane. For a two-dimensional curve, the osculating plane is simply the xy-plane.
Osculating Circle (I) For a given value of t, say t = t 0, if the curvature κ of the curve at the point P corresponding to t 0 is nonzero, then the circle of radius ρ = 1 lying completely in the κ osculating plane and whose center lies a distance of 1 from P along the κ normal N(t) is called the osculating circle (or the circle of curvature). Figure: [9.26] Osculating circle.
Osculating Circle (II) Since the curvature of a circle is the reciprocal of its radius, it implies that the osculating circle has the same tangent and curvature at P as the curve. Further, because the normal vector always points to the concave side of the curve, the osculating circle lies on the concave side of the curve. Figure: [9.26] Osculating circle.
Osculating Circle (III) In this sense, the osculating circle is the circle that "best fits" the curve at point P. The radius of the osculating circle is called the radius of curvature and the center of the circle is called the center of curvature. Figure: [9.26] Osculating circle.
Example (5.4) Finding the Osculating Circle Find the osculating circle for the parabola defined by r(t) = t 2, t at t = 0. Figure: [9.27] Osculating circle.
Tangential/Normal Component of Acceleration (I) Suppose that the position of an object at time t is given by the terminal point of the vector-valued function r(t). Recall that T = r (t) r (t) and r (t) = ds dt, where s represents arc length. Then the velocity of the object is given by v(t) = r (t) = r (t) T(t) = ds dt T(t). Further, we have the acceleration given by a(t) = v (t) = d dt ( ) ds dt T(t) = d2 s ds T(t) + dt2 dt T (t)
Tangential/Normal Component of Acceleration (II) Recall that N(t) = T (t) T (t) T (t) = T (t) N(t) Further by the chain rule, T (t) = dt dt = dt ds ds dt = ds dt dt ds = κds dt where we have used the definition of the curvature κ and the fact that ds > 0. So dt and T (t) = T (t) N(t) = κ ds dt N(t) ( ) a(t) = d2 s ds 2 dt 2 T(t) + κ N(t) dt
Tangential/Normal Component of Acceleration (III) We have obtained the acceleration vector as a(t) = ( ) d2 s ds 2 dt 2 + κ N(t) dt = a T T(t) + a N N(t) Notice that since a(t) is written as a sum of a vector parallel to T(t) and a vector parallel to N(t), the vector a(t) always lies in the osculating plane. The coefficient of T is the tangential component of acceleration a T and the coefficient of N(t) is the normal component of acceleration a N. Figure: [9.28] Tangential and normal components of acceleration.
A Strategy for Keeping the Car on the Road (I) From Newton s second law of motion, the net force acting on a car at any time t is F(t) = ma. Hence, F(t) = ma(t) ( ) = mκ d2 s ds 2 dt 2 + m N(t) dt where m is mass of the car. Figure: [9.29] Driving around a curve.
A Strategy for Keeping the Car on the Road (II) Observe that: 1 Since T points in the direction of the path of motion, we want the component of the force acting in the direction of T to be as large as possible compared to the component of the force acting in the direction of the normal N. 2 If the normal component of the force is too large, it may exceed the normal component of the force of friction between the tires and the highway, causing the car to skid off the road. Figure: [9.29] Driving around a curve.
A Strategy for Keeping the Car on the Road (III) 1 To minimize the force applied in the direction of N, we need to make ds/dt small (reducing speed). 2 To maximize the tangential component of the force, we need to make d 2 s/dt 2 as large as possible. It implies that we need to accelerate while in the curve. 3 To make a turn it is better to slow down before you enter the curve and then accelerate while in the curve. Figure: [9.29] Driving around a curve.
A Strategy for Keeping the Car on the Road (IV) If we wait until we are in the curve to slow down then d 2 s dt 2 < 0 in the curve and so the tangential component of the force is negative (acting in the opposite direction of T), making it harder to get through the curve. Figure: [9.30] Net force: d 2 s dt 2 < 0.
Finding Tangential and Normal Components of Acceleration Example (5.5) Find the tangential and normal components of acceleration for an object with position vector r(t) = 2 sin t, 2 cos t, 4t.
Computing a T and a N (I) 1 It is simple to compute a T = d2 s dt 2 We must only calculating ds/dt = r (t) and then differentiate the result. 2 Computing a N is a bit more complicated, since it requires you to first compute the curvature κ. Figure: [9.31] Components of a(t).
Computing a T and a N (II) An easier way to compute a N can be obtained by using the orthogonality of the two vectors T and N. Notice that a(t) = ( ) d2 s ds 2 dt 2 + κ N(t) dt = a T T(t) + a N N(t) Since we have T N = 0. a 2 = a 2 T + a 2 N followed by the Pythagorean Theorem. Figure: [9.31] Components of a(t).
Computing a T and a N (III) Solving for a N, we get a 2 = a 2 T + a 2 N a N = a 2 a 2 T where we have taken the positive root since ( ) ds 2 a N = κ dt Figure: [9.31] Components of a(t).
Finding Tangential and Normal Components of Acceleration Example (5.6) Find the tangential and normal components of acceleration for an object whose path is defined by r(t) = t, 2t, t 2. In particular, find these components at t = 1. Also, find the curvature. Figure: [9.32] Tangential and normal components of acceleration at t = 1.
Recall that Proving κ = r(t) r (t) r (t) 3 ( ) a(t) = d2 s ds 2 dt 2 T(t) + κ N(t) dt Taking the cross product of both sides of this equation with T gives us ( ) T a = d2 s ds 2 ( ) ds 2 dt 2 T T + κ T N = κ B dt dt Hence T a = κ ( ) ds 2 ( ) ds 2 B = κ dt dt Recalling that T = r / r, a = r and ds/dt = r gives us r(t) r (t) r (t) = κ r 2 κ = r(t) r (t) r (t) 3
KEPLER S LAWS OF PLANETARY MOTION Theorem 1 Each planet follows an elliptical orbit, with the sun at one focus. 2 The line segment joining the sun to a planet sweeps out equal areas in equal times. 3 If T is the time required for a given planet to make one orbit of the sun and if the length of the major axis of its elliptical orbit is 2a, then T 2 = ka 3, for some constant k (i.e., T 2 is proportional to a 3 ). Kepler s laws are based on a careful analysis of a massive number of astronomical observations. Using Newton s second law of motion and Newton s law of universal gravitation and Vector Calculus, one can derive the Kepler s Laws. A detail derivation can be found in the book.
Parametric Surfaces In this section, we extend the notion of parametric equations to those with two independent parameters. We will be working with the simple cases of functions of two variables, which are developed in more detail in Chapter 10.
Parametric Equations with One Variable: An Helix Consider the helix defined by the parametric equations x = cos t y = sin t and z = t This curve winds around the cylinder x 2 + y 2 = 1
Parametric Equations with Two Variables: An Cylinder Suppose that we wanted to obtain parametric equations that described the entire cylinder x 2 + y 2 = 1 we can use the parameters u and v to obtain the corresponding parametric equations x = cos u y = sin u and z = v Parametric equations with two independent parameters correspond to a two-dimensional surface.
Example (6.1) Graphing a Parametric Surface Identify and sketch a graph of the surface defined by the parametric equations x = 2 cos u sin v, y = 2 sin u sin v and z = 2 cos v. Figure: [9.35] x 2 + y 2 + z 2 = 4. Figure: [9.36] z = 4 x 2 y 2.
Parametric Equations for Hyperboloids and Hyperbolic Paraboloids For parametric equations of hyperboloids and hyperbolic paraboloids, it is convenient to use the hyperbolic functions Notice that cosh x = ex + e x 2 sinh x = ex e x 2 cosh 2 x sinh 2 x = 1
Example (6.2) Graphing a Parametric Surface Sketch the surface defined parametrically by x = 2 cos u cosh v, y = 2 sin u cosh v and z = 2 sinh v, 0 u 2π and < v <. Figure: [9.37] x 2 + y 2 z 2 = 4. Figure: [9.38] z = x 2 + y 2 4.
Parametric Equations in Two Dimensions In two dimensions, certain curves are more easily described in polar coordinates than in rectangular coordinates. For example, polar coordinates are essentially the parametric equations for circles. In particular, the polar coordinates r and θ are related to x and y by x = r cos θ, y = r sin θ and r = x 2 + y 2
Finding a Parametric Representation of a Hyperbolic Paraboloid Example (6.3) Find parametric equations for the hyperbolic paraboloid z = x 2 y 2.
Finding Parametric Representations of Surfaces Example (6.4) Find a parametric representation of each surface: 1 the portion of z = x 2 + y 2 inside x 2 + y 2 = 4 and 2 the portion of z = 9 x 2 y 2 above the xy-plane with y 0. Figure: [9.39a] Portion of z = x 2 + y 2 inside x 2 + y 2 = 4. Figure: [9.39b] Portion of z = 9 x 2 y 2 above the xy-plane, with y 0.