Module 13: Parametric Equations and Partial Derivatives PDF

MODULE 13: DERIVATIVES OF PARAMETRIC EQUATIONS AND PARTIAL DERIVATIVES PARAMETRIC EQUATIONS Suppose that 𝑥 and 𝑦 are both given as functions of a third variable 𝑡 (called a parameter) by the equations 𝑥 = 𝑓(𝑡) and 𝑦 = 𝑔(𝑡) then, these equations are called parametric equations. T...

MODULE 13: DERIVATIVES OF PARAMETRIC EQUATIONS AND PARTIAL DERIVATIVES PARAMETRIC EQUATIONS Suppose that 𝑥 and 𝑦 are both given as functions of a third variable 𝑡 (called a parameter) by the equations 𝑥 = 𝑓(𝑡) and 𝑦 = 𝑔(𝑡) then, these equations are called parametric equations. The derivatives of parametric equations can be obtained by 𝑑𝑦 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑦 𝑑𝑡 = or = ⋅ 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑡 𝑑𝑥 𝑑𝑡 and 𝑑 𝑑𝑦 𝑑2 𝑦 𝑑 𝑑𝑦 ( ) 𝑑 𝑑𝑦 𝑑𝑡 𝑑𝑡 𝑑𝑥 = ( ) = = ( )⋅ 𝑑𝑥 2 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑡 𝑑𝑥 𝑑𝑥 𝑑𝑡 EXERCISES 13.1 𝑑𝑦 𝑑2 𝑦 Find the and of the given parametric equations. 𝑑𝑥 𝑑𝑥 2 𝑑𝑦 3 1 𝑑2 𝑦 3(𝑡 2 +1) 1. 𝑥 = 𝑡 2 , 𝑦 = 𝑡 3 − 3𝑡 ans: = (𝑡 − ) ; = 𝑑𝑥 2 𝑡 𝑑𝑥 2 4𝑡 3 𝑑𝑦 3 𝑑2 𝑦 3 2. 𝑥 = 4 + 𝑡 2 , 𝑦 = 𝑡 2 + 𝑡 3 ans: 𝑑𝑥 =1+2 𝑡 ; 𝑑𝑥 2 = 4𝑡 𝑑𝑦 3 𝑑2 𝑦 3 3. 𝑥 = 2 sin 𝑡 , 𝑦 = 3 cos 𝑡 ans: 𝑑𝑥 = − 2 tan 𝑡 ; 𝑑𝑥 2 = − 4 sec 3 𝑡 APPLICATIONS OF THE DERIVATIVES OF PARAMETRIC EQUATIONS (Assignment) EXERCISES 13.2 1. Find an equation of the tangent to the curve defined by the parametric equations 𝑥 = 𝑡 4 + 1, 𝑦 = 𝑡 3 + 𝑡 at 𝑡 = −1 2. Find an equation of the tangent to the curve defined by the parametric equations 𝑥 = 1 + ln 𝑡 , 𝑦 = 𝑡 2 + 2 at (1, 3) 3. At what points on the curve 𝑥 = 2𝑡 3 , 𝑦 = 1 + 4𝑡 − 𝑡 2 does the tangent line have slope 1? PARTIAL DERIVATIVES In general, if 𝑧 is a function of two variables 𝑥 and 𝑦, that is 𝑧 = 𝑓(𝑥, 𝑦), and suppose we let only 𝑥 vary while keeping 𝑦 fixed, say 𝑦 = 𝑏, where 𝑏 is a constant. Then we are really considering a function of a single variable 𝑥, that is 𝑧 = 𝑓(𝑥, 𝑏). If 𝑧 has a derivative at 𝑥 = 𝑎, then we call it the partial derivative of 𝑓 with respect to 𝑥 at (𝑎, 𝑏) and denoted by 𝑓𝑥 (𝑎, 𝑏). The same principle applies if we let only 𝑦 vary while keeping 𝑥 fixed. NOTATIONS FOR PARTIAL DERIVATIVES If 𝑧 = 𝑓(𝑥, 𝑦), we write 𝜕𝑓 𝜕 𝜕𝑧 𝑓𝑥 (𝑥, 𝑦) = 𝑓𝑥 = = 𝑓(𝑥, 𝑦) = = 𝐷𝑥 𝑓 𝜕𝑥 𝜕𝑥 𝜕𝑥 and 𝜕𝑓 𝜕 𝜕𝑧 𝑓𝑦 (𝑥, 𝑦) = 𝑓𝑦 = = 𝑓(𝑥, 𝑦) = = 𝐷𝑦 𝑓 𝜕𝑦 𝜕𝑦 𝜕𝑦 RULE FOR FINDING PARTIAL DERIVATIVES OF 𝑧 = 𝑓(𝑥, 𝑦) 1. To find 𝑓𝑥 , regard 𝑦 as constant and differentiate 𝑓(𝑥, 𝑦) with respect to 𝑥. 2. To find 𝑓𝑦 , regard 𝑥 as constant and differentiate 𝑓(𝑥, 𝑦) with respect to 𝑦. EXERCISES 13.3 𝜕𝑧 𝜕𝑧 𝝏𝒛 𝝏𝒛 4. If 𝑧 = 2𝑥 + 3𝑦 − 6 , find and. ans: =𝟐 ; =𝟑 𝜕𝑥 𝜕𝑦 𝝏𝒙 𝝏𝒚 5. If 𝑓(𝑥, 𝑦) = 4 − 𝑥 2 − 2𝑦 2 , find 𝑓𝑥 (1, 1) and 𝑓𝑦 (1, 1). ans: 𝒇𝒙 (𝟏, 𝟏) = −𝟐 ; 𝒇𝒚 (𝟏, 𝟏) = −𝟒 𝜕𝑓 𝜕𝑓 𝝏𝒇 𝝏𝒇 6. If 𝑓(𝑥, 𝑦) = 𝑥 3 + 𝑥 2 𝑦 3 − 2𝑦 2 , find 𝜕𝑥 and 𝜕𝑦. ans: 𝝏𝒙 = 𝟑𝒙𝟐 + 𝟐𝒙𝒚𝟑 ; 𝝏𝒚 = 𝟑𝒙𝟐 𝒚𝟐 − 𝟒𝒚 7. If 𝑓(𝑥, 𝑦) = 𝑦 5 − 3𝑥 𝑦 , find 𝑓𝑥 and 𝑓𝑦. Assignment 𝜕𝑧 𝜕𝑧 8. If 𝑧 = (2𝑥 + 3𝑦)10 , find and. Assignment 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑧 9. If 𝑧 = tan 𝑥𝑦 , find 𝜕𝑥 and 𝜕𝑦. Assignment