Calculus with Parametric Curves

Transcription

1 Calculus with Parametric Curves Suppose f and g are differentiable functions and we want to find the tangent line at a point on the parametric curve x f(t), y g(t) where y is also a differentiable function of x. Then the Chain Rule gives If /, we can solve for /: if (1) It can be seen from (1) that the curve has a horizontal tangent when / (provided that / ) and it has a vertical tangent when / (provided that / ). EXAMPLE 1: A curve C is defined by the parametric equations x t, y t 3 3t. (a) Show that C has two tangents at the point (3, ) and find their equations. (b) Find the points on C where the tangent is horizontal or vertical. (c) Determine where the curve is concave upward or downward. (d) Sketch the curve. 1

2 EXAMPLE 1: A curve C is defined by the parametric equations x t, y t 3 3t. (a) Show that C has two tangents at the point (3, ) and find their equations. (b) Find the points on C where the tangent is horizontal or vertical. (c) Determine where the curve is concave upward or downward. (d) Sketch the curve. Solution: (a) Notice that x t 3 when t ± 3. Therefore, the point (3, ) on C arises from two values of the parameter, t 3 and t 3. This indicates that the curve C crosses itself at (3, ). Since / / 3t 3 3 ( t 1 ) t t the slope of the tangent when t ± 3 is / ±6/( 3) ± 3, so the equations of the tangents at (3, ) are y 3(x 3) and y 3(x 3) (b) C has a horizontal tangent when /, that is, when / and /. Since / 3t 3, this happens when t 1, that is, t ±1. The corresponding points on C are (1, ) and (1, ). C has a vertical tangent when / t, that is, t. (Note that / there.) The corresponding point on C is (, ). (c) To determine concavity we calculate the second derivative: ( ) d 3 (1 + 1t ) d y t 3(t + 1) 4t 3 Thus the curve is concave upward when t > and concave downward when t <. (d) Using the information from parts (b) and (c), we sketch C: EXAMPLE : (a) Find the tangent to the cycloid x r(θ sin θ), y r(1 cos θ) at the point where θ π/3. (b) At what points is the tangent horizontal? When is it vertical?

3 EXAMPLE : (a) Find the tangent to the cycloid x r(θ sin θ), y r(1 cos θ) at the point where θ π/3. (b) At what points is the tangent horizontal? When is it vertical? Solution: (a) The slope of the tangent line is When θ π/3, we have and / / ( π x r 3 sin π r 3) r sin θ r(1 cosθ) sin θ 1 cosθ ( π 3 ) 3 sin(π/3) 3/ 1 cos(π/3) (, y r 1 cos π ) r 3 Therefore the slope of the tangent is 3 and its equation is y r ( ) 3 x rπ 3 + r ( ) 3 π3 or 3x y r (b) The tangent is horizontal when /, which occurs when sin θ and 1 cosθ, that is, θ (n 1)π, n an integer. The corresponding point on the cycloid is ((n 1)πr, r). When θ nπ, both / and / are. It appears from the graph that there are vertical tangents at those points. We can verify this by using l Hospital s Rule as follows: lim θ nπ + lim θ nπ + sin θ 1 cosθ lim cos θ θ nπ + sin θ A similar computation shows that / as θ nπ, so indeed there are vertical tangents when θ nπ, that is, when x nπr. 3

4 Areas We know that the area under a curve y F(x) from a to b is A b a F(x), where F(x). If the curve is given by the parametric equations x f(t) and y g(t), α t β, then we can calculate an area formula by using the Substitution Rule for Definite Integrals as follows: [ or α β A b a y β α g(t)f (t) ] g(t)f (t) if (f(β), g(β)) is the leftmost endpoint EXAMPLE 3: Find the area under one arch of the cycloid x r(θ sin θ), y r(1 cosθ) 4

5 EXAMPLE 3: Find the area under one arch of the cycloid x r(θ sin θ), y r(1 cosθ) Solution: One arch of the cycloid is given by θ π. Using the Substitution Rule with y r(1 cos θ) and r(1 cosθ), we have A r y r(1 cosθ)r(1 cosθ) r (1 cosθ) r (1 cosθ + cos θ) r [1 cosθ + 1 (1 + cos θ) ] r [ 3 θ sin θ sin θ ] π r ( 3 π ) 3πr Arc Length We alrea know how to find the length L of a curve C given in the form y F(x), a x b. In fact, if F is continuous, then b ( ) L 1 + () a Suppose that C can also be described by the parametric equations x f(t) and y g(t), α t β, where / f (t) >. This means that C is traversed once, from left to right, as t increases from α to β and f(α) a, f(β) b. Putting Formula 1 into Formula and using the Substitution Rule, we obtain Since / >, we have L b a 1 + L ( ) β α β α ( ) ( ) / / ( ) (3) 5

6 Even if C can t be expressed in the form y F(x), Formula 3 is still valid but we obtain it by polygonal approximations. We divide the parameter interval [α, β] into n subintervals of equal wih t. If t, t 1, t,...,t n are the endpoints of these subintervals, then x i f(t i ) and y i g(t i ) are the coordinates of points P i (x i, y i ) that lie on C and the polygon with vertices P, P 1,...,P n approximates C. As in Section 7.4, we define the length L of C to be the limit of the lengths of these approximating polygons as n : n L lim P i 1 P i n i1 The Mean Value Theorem, when applied to f on the interval [t i 1, t i ], gives a number t i in (t i 1, t i ) such that f(t i ) f(t i 1 ) f (t i )(t i t i 1 ) If we let x i x i x i 1 and y i y i y i 1, this equation becomes x i f (t i) t Similarly, when applied to g, the Mean Value Theorem gives a number t i in (t i 1, t i ) such that y i g (t i ) t Therefore and so P i 1 P i ( x i ) + ( y i ) [f (t i ) t] + [g (t i ) t] [f (t i )] + [g (t i )] t L lim n n i1 [f (t i )] + [g (t i )] t (4) The sum in (4) resembles a Riemann sum for the function [f (t)] + [g (t)] but it is not exactly a Riemann sum because t i t i in general. Nevertheless, if f and g are continuous, it can be shown that the limit in (4) is the same as if t i and t i were equal, namely, L β α [f (t)] + [g (t)] Thus, using Leibniz notation, we have the following result, which has the same form as (3). 6

7 THEOREM: If a curve C is described by the parametric equations x f(t), y g(t), α t β, where f and g are continuous on [α, β] and C is traversed exactly once as t increases from α to β, then the length of C is β ( ) ( ) L + Notice that the formula in this Theorem is consistent with the general formulas L ds and (ds) () + () of Section 7.4. EXAMPLE 4: If we use the representation of the unit circle α x cost, y sin t t π then / sin t and / cos t, so the above Theorem gives ( ) ( ) π L + sin t + cos t π as expected. If, on the other hand, we use the representation x sin t, y cos t t π then / cost, / sin t, and the integral in the above Theorem gives ( ) ( ) π + 4 cos t + 4 sin t 4π REMARK: Notice that the integral gives twice the arc length of the circle because as t increases from to π, the point (sin t, cos t) traverses the circle twice. In general, when finding the length of a curve C from a parametric representation, we have to be careful to ensure that C is traversed only once as t increases from α to β. EXAMPLE 5: Find the length of one arch of the cycloid x r(θ sin θ), y r(1 cos θ). 7

8 EXAMPLE 5: Find the length of one arch of the cycloid x r(θ sin θ), y r(1 cos θ). Solution: From Example 3 we see that one arch is described by the parameter interval θ π. Since we have L r(1 cosθ) and r sin θ ( ) + ( ) r (1 cosθ) + r sin θ r (1 cosθ + cos θ + sin θ) r (1 cosθ) To evaluate this integral we use the identity sin x 1 (1 cos x) with θ x, which gives 1 cosθ sin (θ/) Since θ π, we have θ/ π and so sin(θ/). Therefore (1 cos θ) 4 sin (θ/) sin(θ/) sin(θ/) and so L r sin(θ/) r[ cos(θ/)] π r[ + ] 8r EXAMPLE 6: Find the length of the astroid x cos 3 t, y sin 3 t, t π. 8

9 EXAMPLE 6: Find the length of the astroid x cos 3 t, y sin 3 t, t π. Solution: Since we have 3 cos t( sin t) and 3 sin t cost π/ It follows that ( ) + ( ) π/ π/ π/ π/ π/ 9 cos 4 t sin t + 9 sin 4 t cos t 9 sin t cos t(cos t + sin t) 9 sin t cos t 3 sin t cost 3 sin t cost Length of first-quadrant portion π/ Therefore the length of the astroid is four times this: sin t u d(sin t) du cos t du ( ) + 1 ( ) 3 ] 1 3udu 3u 3 9