University:
California State University, Northridge
Course:
MATH 351 | Differential Equations
Academic year:

2024
Views:

199

Pages:

3
Author:

Javon Neal

Question 1 What is the instantaneous rate of change of 𝑓 ( 𝑥 ) = 𝑥 𝑒 𝑥 − 𝑥 2 𝑒 𝑥 f(x)=xe x −x 2 e x at 𝑥 = − 1 x=−1? Answer Differentiate 𝑓 ( 𝑥 ) f(x): 𝑓 ′ ( 𝑥 ) = 𝑑 𝑑 𝑥 [ 𝑥 𝑒 𝑥 − 𝑥 2 𝑒 𝑥 ] . f ′ (x)= dx d [xe x −x 2 e x ]. Using the product rule: 𝑓 ′ ( 𝑥 ) = 𝑒 𝑥 + 𝑥 𝑒 𝑥 − 2 𝑥 𝑒 𝑥 − 𝑥 2 𝑒 𝑥 . f ′ (x)=e x +xe x −2xe x −x 2 e x . Simplify: 𝑓 ′ ( 𝑥 ) = 𝑒 𝑥 ( 1 + 𝑥 − 2 𝑥 − 𝑥 2 ) = − 𝑒 𝑥 ( 𝑥 2 + 𝑥 − 1 ) . f ′ (x)=e x (1+x−2x−x 2 )=−e x (x 2 +x−1). Evaluate at 𝑥 = − 1 x=−1: 𝑓 ′ ( − 1 ) = − 𝑒 − 1 ( ( − 1 ) 2 + ( − 1 ) − 1 ) . f ′ (−1)=−e −1 ((−1) 2 +(−1)−1). Simplify: 𝑓 ′ ( − 1 ) = − 𝑒 − 1 ( 1 − 1 − 1 ) = − 𝑒 − 1 ( − 1 ) = 1 𝑒 . f ′ (−1)=−e −1 (1−1−1)=−e −1 (−1)= e 1 . Result: The instantaneous rate of change at 𝑥 = − 1 x=−1 is 1 𝑒 e 1 . Question 2 What is the instantaneous rate of change of 𝑓 ( 𝑥 ) = 𝑒 𝑥 f(x)=e x when 𝑥 = 0 x=0? Answer The instantaneous rate of change is the derivative of 𝑓 ( 𝑥 ) f(x): 𝑓 ′ ( 𝑥 ) = 𝑑 𝑑 𝑥 ( 𝑒 𝑥 ) = 𝑒 𝑥 . f ′ (x)= dx d (e x )=e x . At 𝑥 = 0 x=0: 𝑓 ′ ( 0 ) = 𝑒 0 = 1. f ′ (0)=e 0 =1. Result: The instantaneous rate of change at 𝑥 = 0 x=0 is 1 1. Question 3 Find the directional derivative of 𝑓 f at the given point in the direction indicated by the angle 𝜃 θ. Given 𝑓 ( 𝑥 , 𝑦 ) = 𝑒 𝑥 cos ⁡ 𝑦 f(x,y)=e x cosy, point ( 0 , 0 ) (0,0), 𝜃 = 𝜋 4 θ= 4 π . Answer The formula for the directional derivative is: 𝐷 𝑢 𝑓 ( 𝑥 , 𝑦 ) = 𝑓 𝑥 ( 𝑥 , 𝑦 ) cos ⁡ 𝜃 + 𝑓 𝑦 ( 𝑥 , 𝑦 ) sin ⁡ 𝜃 . D u f(x,y)=f x (x,y)cosθ+f y (x,y)sinθ. Compute the partial derivatives: 𝑓 𝑥 = ∂ ∂ 𝑥 ( 𝑒 𝑥 cos ⁡ 𝑦 ) = 𝑒 𝑥 cos ⁡ 𝑦 f x = ∂x ∂ (e x cosy)=e x cosy, 𝑓 𝑦 = ∂ ∂ 𝑦 ( 𝑒 𝑥 cos ⁡ 𝑦 ) = − 𝑒 𝑥 sin ⁡ 𝑦 f y = ∂y ∂ (e x cosy)=−e x siny. Substitute the point ( 0 , 0 ) (0,0) and 𝜃 = 𝜋 4 θ= 4 π : At ( 0 , 0 ) (0,0): 𝑓 𝑥 ( 0 , 0 ) = 𝑒 0 cos ⁡ 0 = 1 , 𝑓 𝑦 ( 0 , 0 ) = − 𝑒 0 sin ⁡ 0 = 0. f x (0,0)=e 0 cos0=1,f y (0,0)=−e 0 sin0=0. Plug into the formula: 𝐷 𝑢 𝑓 ( 0 , 0 ) = 𝑓 𝑥 ( 0 , 0 ) cos ⁡ 𝜋 4 + 𝑓 𝑦 ( 0 , 0 ) sin ⁡ 𝜋 4 . D u f(0,0)=f x (0,0)cos 4 π +f y (0,0)sin 4 π . Simplify: 𝐷 𝑢 𝑓 ( 0 , 0 ) = 1 ⋅ 2 2 + 0 ⋅ 2 2 = 2 2 . D u f(0,0)=1⋅ 2 2 +0⋅ 2 2 = 2 2 . Result: The directional derivative is 𝐷 𝑢 𝑓 ( 0 , 0 ) = 2 2 D u f(0,0)= 2 2 . Question 4 Find 𝑦 ′ y ′ and 𝑦 ′ ′ y ′′ for 𝑦 = 𝑥 ln ⁡ ( 𝑥 ) y= x ln(x). Answer First derivative 𝑦 ′ y ′ : Using the product rule: 𝑦 = 𝑥 ln ⁡ ( 𝑥 ) = 𝑥 1 / 2 ln ⁡ ( 𝑥 ) . y= x ln(x)=x 1/2 ln(x). Differentiate: 𝑦 ′ = 𝑑 𝑑 𝑥 [ 𝑥 1 / 2 ] ln ⁡ ( 𝑥 ) + 𝑥 1 / 2 𝑑 𝑑 𝑥 [ ln ⁡ ( 𝑥 ) ] . y ′ = dx d [x 1/2 ]ln(x)+x 1/2 dx d [ln(x)]. Simplify: 𝑦 ′ = 1 2 𝑥 ln ⁡ ( 𝑥 ) + 𝑥 𝑥 . y ′ = 2 x 1 ln(x)+ x x . Combine terms: 𝑦 ′ = ln ⁡ ( 𝑥 ) 2 𝑥 + 1 𝑥 . y ′ = 2 x ln(x) + x 1 . Second derivative 𝑦 ′ ′ y ′′ : Differentiate 𝑦 ′ y ′ again: 𝑦 ′ = ln ⁡ ( 𝑥 ) 2 𝑥 + 1 𝑥 . y ′ = 2 x ln(x) + x 1 . The second derivative involves applying the product rule and chain rule on both terms. Result: The detailed solution simplifies 𝑦 ′ y ′ and 𝑦 ′ ′ y ′′ to: 𝑦 ′ = ln ⁡ ( 𝑥 ) 2 𝑥 + 1 𝑥 , 𝑦 ′ ′ = − ln ⁡ ( 𝑥 ) 4 𝑥 𝑥 + 1 4 𝑥 𝑥 . y ′ = 2 x ln(x) + x 1 ,y ′′ =− 4x x ln(x) + 4x x 1 .