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

2024
Views:

184

Pages:

2
Author:

Javon Neal

Question 1 What is the equation of the line tangent to the curve 𝑦 = ( 𝑥 2 − 35 ) 7 y=(x 2 −35) 7 at 𝑥 = 6 x=6? Answer 1 The equation of the tangent line at a point ( 𝑥 0 , 𝑦 0 ) (x 0 ,y 0 ) is: 𝑦 − 𝑓 ( 𝑥 0 ) = 𝑓 ′ ( 𝑥 0 ) ( 𝑥 − 𝑥 0 ) . y−f(x 0 )=f ′ (x 0 )(x−x 0 ). First, evaluate 𝑓 ( 𝑥 0 ) f(x 0 ): 𝑓 ( 6 ) = ( 6 2 − 35 ) 7 = 1. f(6)=(6 2 −35) 7 =1. Find the derivative 𝑓 ′ ( 𝑥 ) f ′ (x): 𝑓 ′ ( 𝑥 ) = 7 ( 𝑥 2 − 35 ) 6 ⋅ 2 𝑥 . f ′ (x)=7(x 2 −35) 6 ⋅2x. Evaluate at 𝑥 = 6 x=6: 𝑓 ′ ( 6 ) = 7 ( 6 2 − 35 ) 6 ⋅ 2 ⋅ 6 = 84. f ′ (6)=7(6 2 −35) 6 ⋅2⋅6=84. The tangent line equation becomes: 𝑦 − 1 = 84 ( 𝑥 − 6 ) . y−1=84(x−6). Simplify: 𝑦 = 84 𝑥 − 503. y=84x−503. Question 2 How do you find the equation of the tangent line to the graph of 𝑓 ( 𝑥 ) = 𝑥 3 f(x)=x 3 at the given point ( 2 , 8 ) (2,8)? Answer The derivative of 𝑓 ( 𝑥 ) = 𝑥 3 f(x)=x 3 is: 𝑓 ′ ( 𝑥 ) = 3 𝑥 2 . f ′ (x)=3x 2 . At 𝑥 = 2 x=2: 𝑓 ′ ( 2 ) = 3 ( 2 ) 2 = 12. f ′ (2)=3(2) 2 =12. The equation of the tangent line in point-slope form is: 𝑦 − 𝑦 1 = 𝑚 ( 𝑥 − 𝑥 1 ) . y−y 1 =m(x−x 1 ). Substitute 𝑚 = 12 m=12, 𝑥 1 = 2 x 1 =2, and 𝑦 1 = 8 y 1 =8: 𝑦 − 8 = 12 ( 𝑥 − 2 ) . y−8=12(x−2). Simplify: 𝑦 = 12 𝑥 − 16. y=12x−16. Question 3 How do you find the equation of the tangent line and normal line to the curve at the given point 𝑦 = ( 1 + 2 𝑥 ) 2 y=(1+2x) 2 , ( 1 , 9 ) (1,9)? Answer Differentiate 𝑦 = ( 1 + 2 𝑥 ) 2 y=(1+2x) 2 : 𝑑 𝑦 𝑑 𝑥 = 2 ( 1 + 2 𝑥 ) ⋅ 2 = 4 ( 1 + 2 𝑥 ) . dx dy =2(1+2x)⋅2=4(1+2x). At 𝑥 = 1 x=1: 𝑑 𝑦 𝑑 𝑥 = 4 ( 1 + 2 ⋅ 1 ) = 12. dx dy =4(1+2⋅1)=12. The tangent line has slope 𝑚 = 12 m=12. Using point-slope form: 𝑦 − 9 = 12 ( 𝑥 − 1 ) . y−9=12(x−1). Simplify: 𝑦 = 12 𝑥 − 3. y=12x−3. The normal line has slope 𝑚 = − 1 12 m=− 12 1 . Using point-slope form: 𝑦 − 9 = − 1 12 ( 𝑥 − 1 ) . y−9=− 12 1 (x−1). Simplify: 𝑦 = − 1 12 𝑥 + 109 12 . y=− 12 1 x+ 12 109 . Question 4 How do you find an equation of the tangent line to the graph of 𝑦 = 𝑔 ( 𝑥 ) y=g(x) at 𝑥 = 5 x=5 if 𝑔 ( 5 ) = − 3 g(5)=−3 and 𝑔 ′ ( 5 ) = 4 g ′ (5)=4? Answer The tangent line equation is: 𝑦 − 𝑦 1 = 𝑚 ( 𝑥 − 𝑥 1 ) . y−y 1 =m(x−x 1 ). Here 𝑥 1 = 5 x 1 =5, 𝑦 1 = − 3 y 1 =−3, and 𝑚 = 𝑔 ′ ( 5 ) = 4 m=g ′ (5)=4. Substitute: 𝑦 − ( − 3 ) = 4 ( 𝑥 − 5 ) . y−(−3)=4(x−5). Simplify: 𝑦 + 3 = 4 𝑥 − 20 ⟹ 𝑦 = 4 𝑥 − 23. y+3=4x−20⟹y=4x−23. Question 5 How do you find an equation of the tangent line to the curve 𝑦 = 𝑥 ⋅ 𝑥 y=x⋅ x that is parallel to the line 𝑦 = 1 + 3 𝑥 y=1+3x? Answer First, find the derivative of 𝑦 = 𝑥 ⋅ 𝑥 = 𝑥 3 / 2 y=x⋅ x =x 3/2 : 𝑓 ′ ( 𝑥 ) = 3 2 𝑥 1 / 2 . f ′ (x)= 2 3 x 1/2 . The slope of the given line 𝑦 = 1 + 3 𝑥 y=1+3x is 3 3. Set 𝑓 ′ ( 𝑥 ) f ′ (x) equal to 3 3: 3 2 𝑥 1 / 2 = 3. 2 3 x 1/2 =3. Solve for 𝑥 x: 𝑥 1 / 2 = 2 ⟹ 𝑥 = 4. x 1/2 =2⟹x=4. At 𝑥 = 4 x=4, find 𝑦 y: 𝑦 = 4 ⋅ 4 = 4 ⋅ 2 = 8. y=4⋅ 4 =4⋅2=8. The tangent line has slope 𝑚 = 3 m=3, and passes through ( 4 , 8 ) (4,8): 𝑦 − 8 = 3 ( 𝑥 − 4 ) . y−8=3(x−4). Simplify: 𝑦 = 3 𝑥 − 4. y=3x−4.