Derivatives #9 - Questions and Answers

Question 1 Find three positive numbers whose sum is 12 and the sum of whose squares is as small as possible. Answer Define the function to minimize: 𝐹 ( 𝑥 , 𝑦 , 𝑧 ) = 𝑥 2 + 𝑦 2 + 𝑧 2 , subject to the constraint:  𝑥 + 𝑦 + 𝑧 = 12. F(x,y,z)=x 2 +y 2 +z 2 ,subject to the constraint: x+y+z=12. Using the method of Lagrange multipliers: ∇ 𝐹 = 𝜆 ∇ 𝐺 , 𝐺 ( 𝑥 , 𝑦 , 𝑧 ) = 𝑥 + 𝑦 + 𝑧 − 12. ∇F=λ∇G,G(x,y,z)=x+y+z−12. Partial derivatives: ∂ 𝐹 ∂ 𝑥 = 2 𝑥 , ∂ 𝐹 ∂ 𝑦 = 2 𝑦 , ∂ 𝐹 ∂ 𝑧 = 2 𝑧 , and ∂ 𝐺 ∂ 𝑥 = ∂ 𝐺 ∂ 𝑦 = ∂ 𝐺 ∂ 𝑧 = 1. ∂x ∂F ​ =2x, ∂y ∂F ​ =2y, ∂z ∂F ​ =2z,and ∂x ∂G ​ = ∂y ∂G ​ = ∂z ∂G ​ =1. Equating gradients: 2 𝑥 = 𝜆 , 2 𝑦 = 𝜆 , 2 𝑧 = 𝜆 . 2x=λ,2y=λ,2z=λ. This implies 𝑥 = 𝑦 = 𝑧 x=y=z. Substituting 𝑥 = 𝑦 = 𝑧 x=y=z into the constraint 𝑥 + 𝑦 + 𝑧 = 12 x+y+z=12: 3 𝑥 = 12    ⟹    𝑥 = 𝑦 = 𝑧 = 4. 3x=12⟹x=y=z=4. Verify minimum: The minimum value of 𝐹 F is: 𝐹 ( 4 , 4 , 4 ) = 4 2 + 4 2 + 4 2 = 16 + 16 + 16 = 48. F(4,4,4)=4 2 +4 2 +4 2 =16+16+16=48. Question 2 Suppose 𝑓 ′ ′ f ′′ is continuous on ( − ∞ , ∞ ) (−∞,∞). If 𝑓 ′ ( 2 ) = 0 f ′ (2)=0 and 𝑓 ′ ′ ( 2 ) = − 5 f ′′ (2)=−5, what can you say about 𝑓 f? Answer 1 By the Second Derivative Test: If 𝑓 ′ ( 2 ) = 0 f ′ (2)=0 and 𝑓 ′ ′ ( 2 ) < 0 f ′′ (2)<0, then 𝑓 f has a local maximum at 𝑥 = 2 x=2. Answer 2 𝑓 ′ ( 2 ) = 0 f ′ (2)=0 indicates a critical point. 𝑓 ′ ′ ( 2 ) = − 5 < 0 f ′′ (2)=−5<0 confirms that the critical point is a local maximum. Question 3 Find the derivatives of the function 𝑦 = ( 1 + 2 𝑥 ) 𝑒 − 2 𝑥 y=(1+2x)e −2x . Answer Differentiate using the product rule: 𝑦 = 𝑢 ⋅ 𝑣 , 𝑢 = ( 1 + 2 𝑥 ) , 𝑣 = 𝑒 − 2 𝑥 . y=u⋅v,u=(1+2x),v=e −2x . First, find derivatives of 𝑢 u and 𝑣 v: 𝑑 𝑢 𝑑 𝑥 = 2 , 𝑑 𝑣 𝑑 𝑥 = − 2 𝑒 − 2 𝑥 . dx du ​ =2, dx dv ​ =−2e −2x . Apply the product rule: 𝑑 𝑦 𝑑 𝑥 = 𝑢 ′ 𝑣 + 𝑢 𝑣 ′ . dx dy ​ =u ′ v+uv ′ . Substituting: 𝑑 𝑦 𝑑 𝑥 = ( 2 ) ( 𝑒 − 2 𝑥 ) + ( 1 + 2 𝑥 ) ( − 2 𝑒 − 2 𝑥 ) . dx dy ​ =(2)(e −2x )+(1+2x)(−2e −2x ). Simplify: 𝑑 𝑦 𝑑 𝑥 = 2 𝑒 − 2 𝑥 − 2 ( 1 + 2 𝑥 ) 𝑒 − 2 𝑥 . dx dy ​ =2e −2x −2(1+2x)e −2x . Combine terms: 𝑑 𝑦 𝑑 𝑥 = 2 𝑒 − 2 𝑥 − 2 𝑒 − 2 𝑥 − 4 𝑥 𝑒 − 2 𝑥 . dx dy ​ =2e −2x −2e −2x −4xe −2x . Final result: 𝑑 𝑦 𝑑 𝑥 = − 4 𝑥 𝑒 − 2 𝑥 . dx dy ​ =−4xe −2x . Question 4 What must be true of 𝐹 ( 𝑥 ) F(x) and 𝐺 ( 𝑥 ) G(x) if both are antiderivatives of 𝑓 ( 𝑥 ) f(x)? Answer 1 If 𝐹 ( 𝑥 ) F(x) and 𝐺 ( 𝑥 ) G(x) are both antiderivatives of 𝑓 ( 𝑥 ) f(x), then they must differ by a constant: 𝐹 ( 𝑥 ) − 𝐺 ( 𝑥 ) = 𝐶 . F(x)−G(x)=C. Answer 2 𝐹 ( 𝑥 ) F(x) and 𝐺 ( 𝑥 ) G(x) represent families of functions differing by a constant 𝐶 C. For example: ∫ sin ⁡ 𝑥 cos ⁡ 𝑥   𝑑 𝑥 = sin ⁡ 2 𝑥 2 + 𝐶 . ∫sinxcosxdx= 2 sin 2 x ​ +C. Any two antiderivatives will differ by 𝐶 C. Question 5 How many tangent lines to the curve 𝑦 = 𝑥 𝑥 + 1 y= x+1 x ​ pass through the point ( 1 , 2 ) (1,2)? Answer Equation of the curve: 𝑦 = 𝑥 𝑥 + 1 . y= x+1 x ​ . Differentiate to find the slope: 𝑑 𝑦 𝑑 𝑥 = ( 𝑥 + 1 ) ( 1 ) − 𝑥 ( 1 ) ( 𝑥 + 1 ) 2 . dx dy ​ = (x+1) 2 (x+1)(1)−x(1) ​ . Simplify: 𝑑 𝑦 𝑑 𝑥 = 𝑥 + 1 − 𝑥 ( 𝑥 + 1 ) 2 = 1 ( 𝑥 + 1 ) 2 . dx dy ​ = (x+1) 2 x+1−x ​ = (x+1) 2 1 ​ . Tangent line equation: 𝑦 − 2 = 𝑚 ( 𝑥 − 1 ) , y−2=m(x−1), where 𝑚 = 1 ( 𝑥 + 1 ) 2 m= (x+1) 2 1 ​ . Substituting 𝑦 = 𝑥 𝑥 + 1 y= x+1 x ​ and solving for 𝑥 x, equate slopes and points to find the required tangent lines. Solving gives two values for 𝑥 x, indicating two tangent lines. Final Answer: There are two tangent lines to the curve passing through ( 1 , 2 ) (1,2).