10.2 Calculus with Parametric Curves

Transcription

1 CHAPTER 1. PARAMETRIC AND POLAR Calculus with Parametric Curves Example 1. Return to the parametric equations in Example 2 from the previous section: x t + sin() y t + cos() (a) Find the Cartesian equation of the tangent line at t 7/4 (decimals ok). (b) Graph the original curve and the tangent line on your calculator. Solution. (a) As always, for any equation of a tangent line, our goal is to fill in y m(x x ) + y with x x-value of tangent point y y-value of tangent point m In this case, we calculate each of these as a function of t. x f(7/4) 7/4 + cos(7π/4) 7/4 2/2 1.4 y g(7/4) 7/4 + sin(7π/4) 7/4 + 2/ g (7/4) f (7/4) 1 π sin() ] 1 + π cos() 1 π sin(7π/4) 1 + π cos(7π/4) 1 π 2/2 1 π 2/2 1 t7/4 Combining the above steps we see that the tangent line is y (x 1.4) (b) To graph both the original curve, and the equation of the tangent line at the same time, we use parametric mode for both. To put the tangent line in parametric mode we need to make x and y both functions of t. Actually, y is alrea a function of x, so as soon as we figure out how to make x a function of t, we will implicitly alrea be done for y too. In other words, if x f(t) the y (f(t) 1.4) would give us y. In fact then, we could use just about any formula we want to calculate x, and then y is take care of. So, we could use something weird, like x e t or x ln(t), or x tan(t), but we will use the easiest possible function. x t

2 CHAPTER 1. PARAMETRIC AND POLAR 93 Example 2. [Based on Dwyer Gruenwald] A double Ferris wheel consists of two smaller Ferris wheels, attached to a single larger rotating bo, somewhat like pictured below: The position of a person riding on a double Ferris wheel is given ( ) ( ) 2 x(t) 2 sin + 1 sin 1 ( ) ( ) 2 y(t) 3 2 cos 1 cos 1 where t is in seconds, x and y are in feet. (a) Graph the position of a rider for t 2. (b) Take the derivative algebraically, but then use your calcualtor to find when the slope is. What does this mean in terms of the movement of the rider? Solution. (a) We enter the x- and y-equations into the calculator and get the following graph graphics/movement_of_double_ferris_wheel-eps-converted-to.pdf (b) 2 cos ( ) ( cos 2 ) 2 cos ( ) ( cos 2 )

3 CHAPTER 1. PARAMETRIC AND POLAR 9 Example 3. The shape defined by x (t sin(t)), y (1 cos(t)) is called the cycloid 2 (a) Use your calculator to sketch this shape for 2π t 4π. (b) Find the equation of the tangent line at t π/4. (c) Find the area under the first arch (you ll understand what this is after you graph it). Solution. (a) We enter this in our calculators and get the following: graphics/parametric_cycloid-eps-converted-to.pdf (b) The equation of a line is still given by y m(x x ) + y where m is the usual slope, and (x, y ) is a point on the curve. What s new in this problem is how we find the slope and the coordinates (x, y ). For the coordinates, we just plug in t π/4: x (π/4 sin(π/4)) (π/4 1/ 2).39 y (1 cos(π/4)) (1 1/ 2) 1.46 For the slope, we take the derivative as described above: m sin(t) (1 cos(t)) ] t π/4 sin(π/4) (1 cos(π/4)) / 2 (1 1/ 2) Putting it all together, we get the following equation: y 2 (x (π/4 1/ 2)) + (1 1/ 2) 2.414(x.39) (c) We integrate β α g(t)f (t) 2π (1 cos(t)) (1 cos(t)) 2 This shape was originally described in the 16s this way: the path traced out by a fixed point on the outside of a wheel as the wheel rolls along the ground.

4 CHAPTER 1. PARAMETRIC AND POLAR 97 Example 4. Find the arc-length, L, along the top part of a circle of radius 3, between θ and θ π/2, as shown, L 3 Solution. Before we do this using parametric equations, we pause to consider how difficult it would be using Cartesian equations. We would have to find the x-value corresponding to, we would have to take the derivative of y 3 2 x 2, square the derivative, add 1, put this in a square root, and hope for the best. We will find that parametric equations are much easier. We use sine and cosine for our parametric equations. In a circle of radius 3 we have sin(θ) y 3 and cos(θ) x. Solving these for x and y, and using t 3 instead of θ gives x 3 cos(t), y 3 sin(t) Then the arc-length is given by 3 cos(t), 3 sin(t) L ( 3 sin(t))2 + (3 cos(t)) 2 ] π/2 9 sin 2 (t) + 9 cos 2 (t) 3 sin 2 (t) + cos 2 (t) 3 1 3t 3(π/2 ) 3π/6 π/2

5 CHAPTER 1. PARAMETRIC AND POLAR 98 Example. Find the arc-length of one full arc of the following Archimedean spiral: x cos(t) + t sin(t) y sin(t) t cos(t) t 2π Solution. t sin(t) t cos(t) AL (t cos(t))2 + (t sin(t)) 2 t 2 cos 2 (t) + t 2 sin 2 (t) t 2 (cos 2 (t) + sin 2 (t)) t2 t 1 2 t2 ] 2π 1 2 4π2 2π 2