Parametric equationsParametric equation: x and y expressed in terms of aparameter t, for example, x = a cost, y = bsintA curve can be described by parametric equations x=x(t)y-y(t). Each value of t determines a point (x,y). So theparametric equations define a function.Typical parametric equations:circle: x=rcost,y=rsint, 0≤t≤2元.ellipse: x=acost,y=bsint, 0≤t≤2π
Parametric equations ◼ Parametric equation: x and y expressed in terms of a parameter t, for example, ◼ A curve can be described by parametric equations x=x(t), y=y(t). Each value of t determines a point (x,y). So the parametric equations define a function. ◼ Typical parametric equations: circle: ellipse: x a t y b t = = cos , sin x r t y r t t = = cos , sin , 0 2 . x a t y b t t = = cos , sin , 0 2
Example Ex. Sketch the curve with parametric equationsx =sint, y= sin? t.Sol. Observe that y = x and -1≤ x ≤1 so the curve is partof the parabola. Since sint is periodic, the point (x,y) movesback and forth infinitely often along the parabola, as tchanges.Ex. A cycloid is defined byx=r(0-sinO),y=r(1-cosO), e R
Example ◼ Ex. Sketch the curve with parametric equations ◼ Sol. Observe that and so the curve is part of the parabola. Since sint is periodic, the point (x,y) moves back and forth infinitely often along the parabola, as t changes. ◼ Ex. A cycloid is defined by 2 x t y t = = sin , sin . 2 y x = − 1 1 x x r y r R = − = − ( sin ), (1 cos ),
Derivative of functions definedby parametric equationsSuppose y=y(x) is defined by the parametric equationdyx= p(t), y = y(t). Thendy_ y'(t)- dtdxdxq'(t)dtEx. Find an equation of the tangent line to the curveTx = ln(1+t?)at the point(ln 2,1- "y = 1-arctant1Sol.11dyy'(t)1+t?2t2dxx'(t)2t1+t?
Derivative of functions defined by parametric equations Suppose y=y(x) is defined by the parametric equation Then Ex. Find an equation of the tangent line to the curve at the point Sol. x t y t = = ( ), ( ). ( ) . ( ) = = = dy dy t dt y dx t dx dt 2 ln(1 ) 1 arctan = + = − x t y t 2 2 1 ( ) 1 1 1 (1) . ( ) 2 2 2 1 − + = = = − = − + dy y t t y dx x t t t t (ln 2,1 ). 4 −
Questiondyx = ln(sint)findSupposey-e'sint=ldxdydye"costdnSol.cost=0=Sindtdtdt1-e' sintdye"coste' sinty-1dydt1-e’ sintdxdxcostI-e'sint2-ydtsint
Question Suppose find Sol. ln(sin ) , sin 1 = − = y x t y e t . dy dx cos (sin ) cos 0 . 1 sin − − = = − y y y y dy dy dy e t t e e t dt dt dt e t cos sin 1 1 sin . cos 1 sin 2 sin y y y y dy e t dy e t y dt e t dx e t y dx t dt t − − = = = = − −
Questionx =a(t-sint)d'yifFinddx?(y=a(1 -cost)dydydtasintTSol.cotdxdx2a(1-cost)dt1dCSCd'y1ddydtdx22dxdx?Tdxdxa(1-cost)4asin2dt
Question Find if Sol. 2 2 d y dx ( sin ) . (1 cos ) = − = − x a t t y a t sin cot , (1 cos ) 2 = = = − dy dy a t t dt dx a t dx dt 2 2 2 4 1 ( ) csc 1 2 2 ( ) . (1 cos ) 4 sin 2 − = = = = − − d dy t d y d dy dt dx dx dx dx a t dx t a dt