Derivatives #29 - Questions and Answers

Question 1 How do you show that the point ( 𝑝 , 4 𝑝 2 ) (p,4p 2 ) lies on the curve 𝑦 = 4 𝑥 2 y=4x 2 for all real values of 𝑝 p, and then find the equation of the tangent to 𝑦 = 4 𝑥 2 y=4x 2 at ( 𝑝 , 4 𝑝 2 ) (p,4p 2 )? Answer Substitute 𝑥 = 𝑝 x=p into 𝑦 = 4 𝑥 2 y=4x 2 : 𝑦 = 4 𝑝 2 . y=4p 2 . Thus, the point ( 𝑝 , 4 𝑝 2 ) (p,4p 2 ) lies on the curve. To find the slope of the tangent, differentiate 𝑦 = 4 𝑥 2 y=4x 2 : 𝑑 𝑦 𝑑 𝑥 = 8 𝑥 . dx dy ​ =8x. At 𝑥 = 𝑝 x=p, the slope of the tangent is: slope = 8 𝑝 . slope=8p. Using the point-slope form of the equation of a line: 𝑦 − 𝑦 1 = 𝑚 ( 𝑥 − 𝑥 1 ) , y−y 1 ​ =m(x−x 1 ​ ), where 𝑚 = 8 𝑝 m=8p, 𝑥 1 = 𝑝 x 1 ​ =p, and 𝑦 1 = 4 𝑝 2 y 1 ​ =4p 2 : 𝑦 − 4 𝑝 2 = 8 𝑝 ( 𝑥 − 𝑝 ) . y−4p 2 =8p(x−p). Simplify: 𝑦 = 8 𝑝 𝑥 − 8 𝑝 2 + 4 𝑝 2    ⟹    𝑦 = 8 𝑝 𝑥 − 4 𝑝 2 . y=8px−8p 2 +4p 2 ⟹y=8px−4p 2 . Question 2 How do you find the slope of the tangent line to the graph of the given function 𝑦 = − 𝑥 3 + 𝑥 − 2 y=−x 3 +x −2 at 𝑥 = 2 x=2? Answer Differentiate 𝑦 = − 𝑥 3 + 𝑥 − 2 y=−x 3 +x −2 : 𝑑 𝑦 𝑑 𝑥 = − 3 𝑥 2 − 2 𝑥 − 3 . dx dy ​ =−3x 2 −2x −3 . Evaluate at 𝑥 = 2 x=2: 𝑑 𝑦 𝑑 𝑥 = − 3 ( 2 2 ) − 2 ( 2 − 3 ) . dx dy ​ =−3(2 2 )−2(2 −3 ). Simplify: 𝑑 𝑦 𝑑 𝑥 = − 3 ( 4 ) − 2 ( 1 8 ) = − 12 − 2 8 = − 12 − 1 4 . dx dy ​ =−3(4)−2( 8 1 ​ )=−12− 8 2 ​ =−12− 4 1 ​ . Convert to a single fraction: 𝑑 𝑦 𝑑 𝑥 = − 48 4 − 1 4 = − 49 4 . dx dy ​ =− 4 48 ​ − 4 1 ​ =− 4 49 ​ . Result: The slope of the tangent line is − 49 4 − 4 49 ​ . Question 3 How do you find the equation of the tangent line to the graph 𝑦 = 𝑥 2 𝑒 𝑥 − 2 𝑥 𝑒 𝑥 + 2 𝑒 𝑥 y=x 2 e x −2xe x +2e x through the point ( 1 , 𝑒 ) (1,e)? Answer 1 Differentiate 𝑦 = 𝑥 2 𝑒 𝑥 − 2 𝑥 𝑒 𝑥 + 2 𝑒 𝑥 y=x 2 e x −2xe x +2e x : 𝑑 𝑦 𝑑 𝑥 = 𝑑 𝑑 𝑥 ( 𝑥 2 𝑒 𝑥 ) − 𝑑 𝑑 𝑥 ( 2 𝑥 𝑒 𝑥 ) + 𝑑 𝑑 𝑥 ( 2 𝑒 𝑥 ) . dx dy ​ = dx d ​ (x 2 e x )− dx d ​ (2xe x )+ dx d ​ (2e x ). Using the product rule: 𝑑 𝑑 𝑥 ( 𝑥 2 𝑒 𝑥 ) = 2 𝑥 𝑒 𝑥 + 𝑥 2 𝑒 𝑥 , 𝑑 𝑑 𝑥 ( 2 𝑥 𝑒 𝑥 ) = 2 𝑒 𝑥 + 2 𝑥 𝑒 𝑥 , 𝑑 𝑑 𝑥 ( 2 𝑒 𝑥 ) = 2 𝑒 𝑥 . dx d ​ (x 2 e x )=2xe x +x 2 e x , dx d ​ (2xe x )=2e x +2xe x , dx d ​ (2e x )=2e x . Combine terms: 𝑑 𝑦 𝑑 𝑥 = ( 2 𝑥 𝑒 𝑥 + 𝑥 2 𝑒 𝑥 ) − ( 2 𝑒 𝑥 + 2 𝑥 𝑒 𝑥 ) + 2 𝑒 𝑥 . dx dy ​ =(2xe x +x 2 e x )−(2e x +2xe x )+2e x . Simplify: 𝑑 𝑦 𝑑 𝑥 = 𝑥 2 𝑒 𝑥 . dx dy ​ =x 2 e x . Evaluate the slope at 𝑥 = 1 x=1: 𝑑 𝑦 𝑑 𝑥 = ( 1 ) 2 𝑒 1 = 𝑒 . dx dy ​ =(1) 2 e 1 =e. The tangent line equation at ( 1 , 𝑒 ) (1,e) with slope 𝑚 = 𝑒 m=e is: 𝑦 − 𝑒 = 𝑒 ( 𝑥 − 1 ) . y−e=e(x−1). Simplify: 𝑦 = 𝑒 𝑥 − 𝑒 + 𝑒    ⟹    𝑦 = 𝑒 𝑥 . y=ex−e+e⟹y=ex. Question 4 What is the instantaneous velocity of an object with position at time 𝑡 t equal to 𝑓 ( 𝑡 ) = ( 𝑡 𝑒 − 3 𝑡 , 𝑡 2 − 𝑒 𝑡 ) f(t)=(te −3t ,t 2 −e t ) at 𝑡 = 3 t=3? Answer Differentiate 𝑓 ( 𝑡 ) = ( 𝑡 𝑒 − 3 𝑡 , 𝑡 2 − 𝑒 𝑡 ) f(t)=(te −3t ,t 2 −e t ): For the first component 𝑡 𝑒 − 3 𝑡 te −3t , use the product rule: 𝑑 𝑑 𝑡 ( 𝑡 𝑒 − 3 𝑡 ) = 𝑒 − 3 𝑡 + 𝑡 ( − 3 𝑒 − 3 𝑡 ) = 𝑒 − 3 𝑡 ( 1 − 3 𝑡 ) . dt d ​ (te −3t )=e −3t +t(−3e −3t )=e −3t (1−3t). For the second component 𝑡 2 − 𝑒 𝑡 t 2 −e t : 𝑑 𝑑 𝑡 ( 𝑡 2 − 𝑒 𝑡 ) = 2 𝑡 − 𝑒 𝑡 . dt d ​ (t 2 −e t )=2t−e t . At 𝑡 = 3 t=3: First component: 𝑒 − 3 ( 3 ) ( 1 − 3 ( 3 ) ) = 𝑒 − 9 ( 1 − 9 ) = − 8 𝑒 − 9 . e −3(3) (1−3(3))=e −9 (1−9)=−8e −9 . Second component: 2 ( 3 ) − 𝑒 3 = 6 − 𝑒 3 . 2(3)−e 3 =6−e 3 . Result: The instantaneous velocity at 𝑡 = 3 t=3 is: 𝑓 ′ ( 3 ) = ( − 8 𝑒 − 9 , 6 − 𝑒 3 ) . f ′ (3)=(−8e −9 ,6−e 3 ).