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

2024
Views:

134

Pages:

11
Author:

Javon Neal

Question 1 Find the derivative 𝑑 𝑑 𝑥 [ arctan ⁡ ( 𝑥 2 + 1 ) ] dx d [arctan(x 2 +1)]. Solution: Let 𝑓 ( 𝑥 ) = arctan ⁡ ( 𝑥 2 + 1 ) f(x)=arctan(x 2 +1). The derivative of arctan ⁡ ( 𝑢 ) arctan(u) is 1 1 + 𝑢 2 1+u 2 1 , where 𝑢 = 𝑥 2 + 1 u=x 2 +1. Use the chain rule to differentiate: 𝑑 𝑑 𝑥 [ arctan ⁡ ( 𝑥 2 + 1 ) ] = 1 1 + ( 𝑥 2 + 1 ) 2 ⋅ 𝑑 𝑑 𝑥 ( 𝑥 2 + 1 ) . dx d [arctan(x 2 +1)]= 1+(x 2 +1) 2 1 ⋅ dx d (x 2 +1). Differentiate 𝑥 2 + 1 x 2 +1: 𝑑 𝑑 𝑥 ( 𝑥 2 + 1 ) = 2 𝑥 . dx d (x 2 +1)=2x. Substitute back into the expression: 𝑑 𝑑 𝑥 [ arctan ⁡ ( 𝑥 2 + 1 ) ] = 2 𝑥 1 + ( 𝑥 2 + 1 ) 2 . dx d [arctan(x 2 +1)]= 1+(x 2 +1) 2 2x . Correct Answer: (d) 2 𝑥 1 + ( 𝑥 2 + 1 ) 2 1+(x 2 +1) 2 2x Question 2 At what point does the curve 𝑦 = 9 ln ⁡ ( 𝑥 ) y=9ln(x) have maximum curvature? What happens to the curvature as 𝑥 → ∞ x→∞? Solution: Curvature Formula: The curvature of a curve 𝑦 = 𝑓 ( 𝑥 ) y=f(x) is: 𝑘 ( 𝑥 ) = ∣ 𝑦 ′ ′ ∣ ( 1 + ( 𝑦 ′ ) 2 ) 3 / 2 . k(x)= (1+(y ′ ) 2 ) 3/2 ∣y ′′ ∣ . First and Second Derivatives: First derivative 𝑦 ′ y ′ : 𝑦 ′ = 9 𝑥 . y ′ = x 9 . Second derivative 𝑦 ′ ′ y ′′ : 𝑦 ′ ′ = − 9 𝑥 2 . y ′′ =− x 2 9 . Substitute into Curvature Formula: 𝑘 ( 𝑥 ) = ∣ − 9 𝑥 2 ∣ ( 1 + ( 9 𝑥 ) 2 ) 3 / 2 . k(x)= (1+( x 9 ) 2 ) 3/2 − x 2 9 . Simplify the denominator: 1 + ( 9 𝑥 ) 2 = 𝑥 2 + 81 𝑥 2 . 1+( x 9 ) 2 = x 2 x 2 +81 . Thus, 𝑘 ( 𝑥 ) = 9 𝑥 2 ( 𝑥 2 + 81 𝑥 2 ) 3 / 2 . k(x)= ( x 2 x 2 +81 ) 3/2 x 2 9 . Simplify further: 𝑘 ( 𝑥 ) = 9 𝑥 2 ⋅ 𝑥 3 ( 𝑥 2 + 81 ) 3 / 2 = 9 𝑥 ( 𝑥 2 + 81 ) 3 / 2 . k(x)= x 2 9 ⋅ (x 2 +81) 3/2 x 3 = (x 2 +81) 3/2 9x . Maximize the Curvature 𝑘 ( 𝑥 ) k(x): Take the derivative of 𝑘 ( 𝑥 ) k(x), set it equal to 0, and solve for 𝑥 x. The result is: 𝑥 = 9 2 2 . x= 2 9 2 . Find the 𝑦 y-value at 𝑥 = 9 2 2 x= 2 9 2 : 𝑦 = 9 ln ⁡ ( 9 2 2 ) . y=9ln( 2 9 2 ). Thus, the point of maximum curvature is: ( 𝑥 , 𝑦 ) = ( 9 2 2 , 9 ln ⁡ ( 9 2 2 ) ) . (x,y)=( 2 9 2 ,9ln( 2 9 2 )). Behavior as 𝑥 → ∞ x→∞: As 𝑥 → ∞ x→∞, the denominator ( 𝑥 2 + 81 ) 3 / 2 (x 2 +81) 3/2 dominates, and: lim ⁡ 𝑥 → ∞ 𝑘 ( 𝑥 ) = 0. x→∞ lim k(x)=0. Final Results: Maximum curvature occurs at 𝑥 = 9 2 2 x= 2 9 2 . As 𝑥 → ∞ x→∞, the curvature 𝑘 ( 𝑥 ) → 0 k(x)→0. Question 3 Find the derivative of 𝑦 = ln ⁡ ( 𝑥 ) 𝑥 4 y= x 4 ln(x) . Solution: Use the quotient rule: For 𝑦 = 𝑢 𝑣 y= v u , 𝑑 𝑦 𝑑 𝑥 = 𝑣 𝑑 𝑢 𝑑 𝑥 − 𝑢 𝑑 𝑣 𝑑 𝑥 𝑣 2 , dx dy = v 2 v dx du −u dx dv , where 𝑢 = ln ⁡ ( 𝑥 ) u=ln(x) and 𝑣 = 𝑥 4 v=x 4 . Differentiate 𝑢 u and 𝑣 v: 𝑑 𝑢 𝑑 𝑥 = 1 𝑥 , 𝑑 𝑣 𝑑 𝑥 = 4 𝑥 3 . dx du = x 1 , dx dv =4x 3 . Substitute into the quotient rule: 𝑑 𝑦 𝑑 𝑥 = 𝑥 4 ⋅ 1 𝑥 − ln ⁡ ( 𝑥 ) ⋅ 4 𝑥 3 ( 𝑥 4 ) 2 . dx dy = (x 4 ) 2 x 4 ⋅ x 1 −ln(x)⋅4x 3 . Simplify: 𝑑 𝑦 𝑑 𝑥 = 𝑥 3 − 4 𝑥 3 ln ⁡ ( 𝑥 ) 𝑥 8 . dx dy = x 8 x 3 −4x 3 ln(x) . Factorize the numerator: 𝑑 𝑦 𝑑 𝑥 = 𝑥 3 ( 1 − 4 ln ⁡ ( 𝑥 ) ) 𝑥 8 . dx dy = x 8 x 3 (1−4ln(x)) . Simplify further: 𝑑 𝑦 𝑑 𝑥 = 1 − 4 ln ⁡ ( 𝑥 ) 𝑥 5 . dx dy = x 5 1−4ln(x) . Final Answer: 𝑑 𝑦 𝑑 𝑥 = 1 − 4 ln ⁡ ( 𝑥 ) 𝑥 5 . dx dy = x 5 1−4ln(x) .