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Parametric Formulas dy dy dt dx dx dt First derivative of a function in parametric form: d dy d y dt dx dx dx 2 dt 2 Second derivative of a function in parametric form : 2 2 dx dy Arc length of a function in parametric form: s dt a dt dt b Parametric Equations and Calculus Ex. 1 (Noncalculator) Given the parametric equations x 2 t and y 3t 2 2t , find dy d2 y and . dx dx 2 __________________________________________________________________________________ Ex. 2 (Noncalculator) Given the parametric equations x 4cost and y 3sin t , write an equation of the tangent line to the curve at the point where t 3 . 4 _________________________________________________________________________________ Ex 3 (Noncalculator) Find all points of horizontal and vertical tangency given the parametric equations x t 2 t, y t 2 3t 5. __________________________________________________________________________________ Ex. 4 (Noncalculator) Set up an integral expression for the arc length of the curve given by the parametric equations x t 2 1, y 4t 3 1, 0 t 1. Do not evaluate. Ex. 1 (Noncalculator) Given the parametric equations x 2 t and y 3t 2 2t , find dy d2 y and . dx dx 2 Solution: dy To find , we must differentiate both of the parametric equations with respect to t. dx 1 dy d dx d 1 12 2 3t 2t 6t 2 and 2 t 2 t t 2 so dt dt dt dt 2 dy 3 1 dy dt 6t 2 2 2t 2 6t 1 dx dx t 2 dt d2 y dy dy To find , we must differentiate with respect to x so [thinking of 2 dx dx dx as a function of t and t as a function of x] that 1 1 1 1 1 dt 1 dt d 2 y d dy 9t 12 t 12 dt 9t 2 t 2 9t 2 t 2 9t 1 2 2 9t t 1 dx dx dx dx dx 2 dx dx t 2 dt ________________________________________________________________________________ Ex. 2 (Noncalculator) Given the parametric equations x 4cost and y 3sin t , write an equation of the tangent line to the curve at the point where t 3 . 4 Solution: dy dy dt dx dx dt d 3sin t 3cost 3 dt cot t d 4sin t 4 4cost dt 2 3 2 3 dy 3 3 2 When t , 1 , x 4 2 2, and y 3 4 dx 4 4 2 2 2 3 2 3 x2 2 . 2 4 (Remind students that they may leave their tangent line equations in point-slope form.) Therefore the tangent line equation is: y Ex 3 (Noncalculator) Find all points of horizontal and vertical tangency given the parametric equations x t 2 t, y t 2 3t 5. dy d 2 t 3t 5 2t 3 dy dt dt Solution: d 2 dx dx 2t 1 t t dt dt dy A horizontal tangent will occur when 0 , which happens when 2t 3 0 (and 2t 1 0) dx 3 3 so a horizontal tangent occurs at t . Substituting t into the given equations, we find that a 2 2 15 11 dy horizontal tangent will occur at , . A vertical tangent will occur when is undefined, dx 4 4 1 . Substituting 2 1 27 1 t into the given equations, we find that a vertical tangent will occur at , . 2 4 4 _______________________________________________________________________________ Ex. 4 (Noncalculator) Set up an integral expression for the arc length of the curve given by the parametric equations x t 2 1, y 4t 3 1, 0 t 1. Do not evaluate. which happens when 2t 1 0 (and 2t 3 0) so a vertical tangent occurs at t Solution: For parametric equations, the formula for arc length is: 2 s b a 2 dx dy dt dt dt For our problem, dx d 2 dy d t 1 2t and 4t 3 1 12t 2 so the arc length is given by the dt dt dt dt integral expression s 1 0 2t 12t 2 dt 2 2 or s 1 0 4t 2 144t 4 dt . CALCULUS BC WORKSHEET ON PARAMETRICS AND CALCULUS Work the following on notebook paper. Do not use your calculator. dy d2 y and On problems 1 – 5, find . dx dx 2 1. x t 2 , y t 2 6t 5 2. x t 2 1, y 2t 3 t 2 3. x t , y 3t 2 2t 4. x ln t, y t 2 t 5. x 3sin t 2, y 4cost 1 _____________________________________________________________________________ 6. A curve C is defined by the parametric equations x t 2 t 1, y t 3 t 2 . dy in terms of t. dx (b) Find an equation of the tangent line to C at the point where t = 2. (a) Find 7. A curve C is defined by the parametric equations x 2cost, y 3sin t . (a) Find dy in terms of t. dx (b) Find an equation of the tangent line to C at the point where t = . 4 ______________________________________________________________________________ On problems 8 – 10, find: dy (a) in terms of t. dx (b) all points of horizontal and vertical tangency 8. x t 5, y t 2 4t 9. x t 2 t 1, y t 3 3t 10. x 3 2cost, y 1 4sint , 0 t 2 ______________________________________________________________________________ On problems 11 - 12, a curve C is defined by the parametric equations given. For each problem, write an integral expression that represents the length of the arc of the curve over the given interval. 11. x t 2 , y t 3 , 0 t 2 12. x e2t 1, y 3t 1, 2 t 2 dy d 2 dx d 2 t 6t 5 2t 6 and t 2t dt dt dt dt dy dy dt 2t 6 3 1 dx dx 2t t dt d2y dy To find , we must differentiate with respect to x so that 2 dx dx 3 3 2 2 d 2 y d dy d 3 3 dt t t 3 1 2t dx 2 dx dx dx t t 2 dx dx 2t 3 dt 1. ______________________________________________________________________________ dy d 3 2 dx d 2 2t t 6t 2 2t and t 1 2t dt dt dt dt dy dy dt 6t 2 2t 3t 1 dx dx 2t dt d2y dy To find , we must differentiate with respect to x so that 2 dx dx d 2 y d dy d dt 3 3 3t 1 3 2 dx dx dx dx dx 2t dx dt 2. ________________________________________________________________________________ dy d 2 dx d 1 12 3. 3t 2t 6t 2 and t t dt dt dt dt 2 dy 3 1 dy dt 6t 2 2 12t 4t 2 dx dx 1 12 dt 2 t d2y dy To find , we must differentiate with respect to x so that 2 dx dx 1 1 1 2 1 d 2 y d dy d 32 2 18t 2 2t 2 dt 18t 2t 12 t 4 t dx dx dx 2 dx dx dx dt 1 2 1 18t 2 2t 1 12 t 2 1 2 36t 4 dy d 2 dx d 1 t t 2t 1 and ln t dt dt dt dt t dy dy dt 2t 1 2t 2 t 1 dx dx t dt 2 d y dy To find , we must differentiate with respect to x so that 2 dx dx d 2 y d dy d 2 dt 4t 1 4t 1 2t t 4t 1 4t 2 t 2 1 dx dx dx dx dx dx t dt 4. ________________________________________________________________________________ dy d dx d 4cos t 1 4sin t and 3sin t 2 3cos t dt dt dt dt dy dy dt 4sin t 4 tan t dx dx 3cos t 3 dt d2y dy To find , we must differentiate with respect to x so that 2 dx dx 4 4 sec2 t sec2 t 4 d 2 y d dy d 4 dt 4 tan t sec2 t 3 3 sec3 t 2 dx dx dx dx 3 3cos t 9 dx 3 dx dt 5. _______________________________________________________________________________ 6. (a) dy 3t 2 2t dx 2t 1 (b) When t = 1, dy 3 12 2 1 8 , x 5, y 4 so the tangent line equation is dx 2 1 1 5 8 x 5 5 ______________________________________________________________________ dy 3cos t 3 7. (a) cot t dx 2sin t 2 dy 3 3 3 2 cot , x 2, y (b) When t , so the tangent line equation is 2 4 2 2 4 dx 3 2 3 y x 2 2 2 y4 8. (a) dy 2t 4 dx 1 dy dx 0 and 0 so a horizontal tangent dt dt occurs when 2t 4 0 which is at t= 2. When t = 2, x = 7 and y = 4 so a horizontal tangent occurs at the point 7, 4 . A vertical tangent occurs dy dx when 0 and 0. dt dt Since 1 0 , there is no point of vertical tangency on this curve. ______________________________________________________________________ dy 3t 2 3 9. (a) dx 2t 1 dy dx (b) A horizontal tangent occurs when 0 and 0 so a horizontal tangent dt dt occurs when 3t 2 3 0 which is at t 1 . When t = 1 , x = 1 and y = 2, and when t 1, x = 3 and y = 2 so a horizontal tangent occurs at the points 1, 2 and 3, 2 dy dx A vertical tangent occurs when 0 and 0 so a vertical tangent occurs dt dt 1 1 3 11 when 2t 1 0 so t . When t , x and y so a vertical tangent 2 2 4 8 3 11 occurs at the point , . 8 4 ______________________________________________________________________ dy 4cos t 10. (a) dx 2sin t dy dx (b) A horizontal tangent occurs when 0 and 0 so a horizontal tangent dt dt 3 occurs when 4cos t 0 which is at t and . When t = , x = 3 and y = 3, 2 2 2 3 and when t , x = 3 and y = 5 so a horizontal tangent occurs at the 2 points 3, 3 and 3, 5 . dy dx A vertical tangent occurs when 0 and 0 so a vertical tangent occurs dt dt when 2sin t 0 so t 0 and . When t 0 , x 5 and y 1 and when t , x 1 and y 1 so a vertical tangent occurs at the points 5, 1 and 1, 1 . (b) A horizontal tangent occurs when 2 2 2 dx dy 11. s dt a dt 0 dt b 2t 2 3t 2 2 dt or 2 0 4t 2 9t 4 dt ______________________________________________________________________ 2 2 2 dx dy 12. s dt a dt 2 dt b 2e2t 2 3 dt or 2 2 2 4e4t 9 dt