Parametric Curves: Suppose a particle is moving around in a circle or any curve that fails the vertical line test, then we cannot describe the path of this particle using an equation of the form y fx) since y is not a function of x. Instead, we can imagine that, at any moment t in time, the location of the particle, as expressed in the x, y) coordinate, is a function of time. I.e. both x and y are functions of time, then we can describe the location of the particle using the variable t with the parametric equations: x xt), y yt)) Each ordered pair of values xt), yt)) can be interpreted as the location of a particle at any given moment in time t. As t varies, the points xt), yt)) traces out a curve C. This curve is called a parametric curve. While we used the example of xt), yt)) as location of a particle in time t, this does not have to be the case. t can be any variable and xt), yt)) can simply be interpreted as a curve given in this fashion. The variable t or any other variable used) is called the parameter. The direction of the movement of the particle is in the direction as t increases. The advantage of describing a curve in this fashion is that it allows a more flexible way to describe more curves in the x y plane, even the curves that cannot be expressed with y being a function of x. Example: The parabola y x can be described by the parametric curve t, t ). In this case, the particle moves along the curve from left to right as t increases. A different parametric curve t, t ) traces the same curve, but this time the par-
ticle is moving in the opposite direction, moving from right to left as t increases. The parabola x y does not have y as a function of x, but we can express this as the parametric curve t, t). Notice that the graph of any function y fx) can be represented by the parametric curve t, ft)) Example:
Describe the curve given by the parametric equation x cos t, y sin t, t π Ans: It would be easier to see if we square both sides of the two equations and use the pythegorean identity: x 4 cos t, y 4 sin t x + y 4 cos t + 4 sin t 4cos t + sin t) 4 We are looking at the equation of a semi-circle of radius. As t moves from to π radian), the parametric curve traces out the top half of the circle, starting at 1, ) and ends at 1, ). I.e., the particle moves in a counter clockwise direction. Example: A cycloid is a curve traced out a point P on the circumference of a circle as the circle rolls along a straing line. Using parametric curve, we want to find an expression for the equation of a cycloid. We let the circle has radius r and rolls along the x axis. We use the angle of rotation, θ in radian), as the independent parameter, and that at θ, the point P is at the origin,, ).
As seen in the picture, as the circle rolled an angle measure of θ radian, the point P will have a coordinate of x, y). The length of segment OT is equal to the length of the arc rolled, arc P T, which, by the definition of radian measure, is equal to rθ. OT P T rθ The coordinate of the center, C, will then be: rθ, r) The length of P Q and length of CQ can be found from trigonometry: P Q sin θ P Q r sin θ r cos θ CQ CQ r cos θ r The x and y coordinate of P can now be found: x OT P Q rθ r sin θ r θ sin θ) y QT CT CQ r r cos θ r1 cos θ) We have found the parametric equation for a cycloid: x r θ sin θ) y r 1 cos θ) where θ in radian treated as a real number) is the parameter and r is the radius of the circle.
Example: Describe the parametric equation: x 3 θ sin θ) y 3 1 cos θ) Answer: This is a cycloid generated by rotating a circle of radius 3 along the x axis. Its graph is given below: Example: Describe the parametric equation: x θ sin θ) y 1 cos θ) Answer: We can look at this as the graph of a cycloid formed when a circle of radius r is rolled along the x axis. Except this time, with the extra negative sign on y, the graph of the cycloid is inverted: Example: Describe the parametric equation: x θ sin θ) y 1 cos θ) Answer: This is a cycloid with the same shape as the one above, but the extra negative sign on x makes the cycloid travese from right to left.
When a curve is expressed by a parametric equation xt), yt)), it is still possible to find the slope of the line tangent to the curve at any given point xt), yt)). With substitution and application of the chain rule it can be shown that, if, then the slope of the tangent line to the parametrized curve is given by: y t) x t) / / Example: Given the parametrized curve t + 1, t 3 1 ), find the equation of the tangent line to the curve at t 1 Ans: In this example, xt) t +1 and yt) t 3 1, so x t) t and y t) 3t, and y t) x t) 3t t 3 t if t, At t 1, 3 1) 3 At t 1, the curve is at the point 1) + 1, 1) 3 1), ) Equation of the tangent line is: y 3 x ) y 3 x 3 Notice that if t, then has a sharp change. is undefined. At that point, the graph of the curve Example: Given the parametric curve 3 cos t, sin t) Find the values) of t where the curve has a horizontal tangent.
Ans: Since x t) 3 sin t, y t) cos t, we have: y t) x t) cos t 3 sin t 3 cot t The cotangent function is equal to at t π + πn, and cotangent is undefined at πn. The graph of the curve has horizontal tangents at t π + πn and vertical tangents at t πn. The graph of this is an ellipse. t π + πn corresponds to the two y-intercepts of the ellipse, Ç Ç å π Ç åå 3 cos + πn π, sin + πn, ) if n is even Ç 3 cos Ç π + πn å, sin Ç π + πn åå, ) if n is odd Example: Find the area under one arch of the cycloid generated by a circle of radius r: Answer: The area under a curve y fx) from a to b is given by b A y a We let y r1 cos θ), and x rθ sin θ), taking differentials we have: r1 cos θ). For one arch of the cycloid, θ goes from to π, we can set up the integral as:
π A π r π y π r1 cos θ) r1 cos θ) r 1 cos θ) r π 1 cos θ + cos θ) 1 cos θ + 1 1 + cos θ) r [ θ sin θ + 1 r [ θ sin θ + 1 θ + 1 4 sin θ ] π θ + )]π sin θ r π + 1 π ) r 3π) 3πr Arc-Length of a parametrized curve We know that if a curve C is defined by a continuous function y fx), then the arc-length of this curve from x a to x b is given by L b a 1 + ) Suppose that C can be parametrized by xt), yt)), where α t β. We know from previous that y t) x t) / / Since a fα) and b fβ), making a substitution into the arc-length formula gives: Ñ é b ) β / L 1 + 1 + a α / If we assume that simplifies to: β α β α 1 + Ñ é / / β α ) + ) >, then we can put Œ 1 + Ñ / / é inside the radical and the integral ) Using a method similar to the one we used in deriving the formula for arc-length of a function y fx), we can show that the above formula is true for the arc-length of any parametrized curve C, even if is not necessarily positive. Arc Length Formula for Parametrized Curve: When a curve C is expressed by a parametric equation xt), yt)), α t β, and C is traversed exactly once from t α to t β, then the arc-length of the curve is given by:
β L x t)) + y t)) β α α ) + ) Example: Find the arc-length of the given curve: x e t + e t, y 5 t, t 3 Answer: x t) e t e t, y t), so 3 L [e t e t ] 3 + [ ] e t e t e t + e t + 4 3 3 e t + e t + 4 e t + e t) e t e t 3 3 e t + + e t 3 [e t + e t ] e 3 e 3 ) e e ) e 3 1 e 3 Example: Find the arc-length of one arch of a cycloid generated by a circle of radius r:
Answer: Remember the parametric equation of a cycloid generated by a circle of radius r is given by: x rθ sin θ) y r1 cos θ) We have: r1 cos θ) r sin θ For one arch of the cycloid, the parameter θ goes from to π, so the arc-length formula gives: π ) ) π L + r 1 cos θ) + r sin θ) π r 1 cos θ + cos θ) + r sin π θ r 1 cos θ + cos θ + sin θ ) π π r 1 cos θ + 1) r 1 cos θ) π The power reduction identity for sin x gives: sin x 1 1 cos x) r cos θ) Making the substitution θ x x θ, the identity becomes: ) θ sin 1 θ 1 cos θ) 1 cos θ sin the above integral becomes: π π r 1 cos θ) r à 4 sin θ π ) π r For θ π, θ π. r à sin θ ), )) θ sin )
θ Inside this interval, sin, so ) sin θ the above integral further simplifies to: π ) r θ sin π θ r sin ) [ [ ) θ π r cos r cos )]π r [ 1) )] r[ + ] 8r ) ) θ sin, ] cos)) Using a similar method as the one for deriving the formula for arc-length, we can derive the formula for surface area of a surface formed by rotating a parametrized curve about the x axis: Suppose C is a curve parametrized by xt), yt)), α t β, and x t), y t) are continuous functions of t and yt), then the area of the surface obtained by rotating the curve C about the x axis is given by: β ) ) S πy + α Example: Find the surface area of a sphere with radius r. Ans: We can parametrize a semi-circle centered at the origin and rotate this semi-circle about the x axis to get a sphere: One way to parametriz this semi-circle is by using: r cost), r sint)), t π Since xt) r cost), yt) r sint), so r sint), r cost). According to the surface area formula, the surface area of the sphere obtained by rotating this curve about the x axis is given by: π S πr sin t) r sin t) + r cos t) π π r sin t) r sin t + r cos t )
π π r sin t) r sin + cos t ) π π r sin t) r π π π r sin t) r πr sin t πr cos t) πr [ cosπ) cos))] πr 1) 1)) πr 1 + 1) 4πr π
Given a polar equation r rθ) where r is a function of θ, since x r cos θ and y r sin θ, this means both x and y can be viewed as functions of θ. We can look at xθ), yθ)) as a parametrized curve with θ being the parameter. Differentiating using the produce rule, we have: cos θ r sin θ sin θ + r cos θ from the formula for of a parametrized curve, we have: sin θ + r cos θ cos θ r sin θ Example: Find the slope of the tangent line to the given polar equation at the given point: r sin θ, θ π/3 Ans: r θ) cos θ, therefore: cos θ)sin θ) + sin θ)cos θ) cos θ)cos θ) sin θ)sin θ) at θ π/3, cos θ 1/ and sin θ 3/, we have: 1 3 + 1 1 3 ) ã 3 4 + 1 3 4 3 1 4 3 + 3 4 1 3 1 3 Å + 3 cos θ sin θ + cos θ cos θ sin θ cos θ sin θ + sin θ 1 3 1 3 Å 1 ã 3 ) 1 3 ) )
Using the fact that the area of a sector of a circle a piece of pie) is proportional to the angle that the sector sustains, we know that the area of a sector of a circle with angle θ is given by: ) θ πr A ) 1 π r θ Using this fact, we can derive the following formula: θ i θ i 1 θ We divide the angle between a θ b into sub-angles of equal wihs, θ. Let θ θ i θ i 1 be a sub-angle of the bigger piece. Let θi be any number in the sub-interval [θ i 1, θ i ], let rθi ) be the radius of the sector, then the area of the region with the central angle θ can be approximated by the area of the sector with radius rθi ): A i 1 [rθ i )] θ Our intuition tells us that if we sum up the area of all these sectors we get a good approximation of the area of the region. Our approximation will be exact if take the limit as n the number of sectors) approaches infinity. This gives us a Riemann sum. lim n n i1 1 [rθ i )] θ b a 1 r) We have the following formula: Suppose r rθ) is a function of θ in polar coordinate, and A is the area of the region formed by r between a θ b, then: b 1 A a r
Example: Find the area formed by the curve between the given θ r e θ/, π θ π Ans: Plug in the formula we get: A π π 1 [ e θ/ ] π 1 e θ ) 1 π e θ ) π π 1 [ e π e π]
Example: Find the area of the region enclosed by one loop of the curve: r sin θ Ans: This is a four-leaved rose where one of the loop starts at θ and ends at θ π/, so we have the integral: π/ 1 A [sin π/ 1 θ] sin θ) ) Ñ é π/ 1 1 cos4θ) π/ 1 cos4θ) 4 1 4 θ 1 16 sin4θ) π/ π 8
Given a polar equation r rθ) where r is a function of θ, we know that we can treat the polar curve x r cos θ and y r sin θ, as a paramatrized curve with θ being the parameter, and cos θ r sin θ sin θ + r cos θ If follows that: ) ) + ) cos θ r cos θ sin θ + r sin θ [ ] [ ] cos θ r sin θ + sin θ + r cos θ + ) sin θ + r sin θ cos θ + r cos θ ) ) cos θ + sin θ + r sin θ + r cos θ ) [cos θ + sin θ ] + r [ sin θ + cos θ ] ) + r From the formula for arc-length of a parametrized curve, with θ being the parameter, we have: b ) ) L + a The algebra we just did gives us the formula for the arc-length of a curve with polar equation r rθ), a θ b: L b a r + ) Example: Find the arc-length of the polar curve: r θ, θ π Ans: Since rθ) θ, so r θ) 1, we have: L π θ) + 1 θ θ) + 1 + 1 Å ln x + ã π x + 1 π π) + 1 + 1 Å ln π + ã π) + 1 π 4π + 1 + 1 Åπ ln + ã 4π + 1