Math 150 (62493) Spring 2019 Quiz 4A Solutions Instructor: C. Panza Quiz 4A Solutions: (20 points) Neatly show your work in the space provided, clearly mark and label your answers. Show proper equality, correctly using the equals sign (=). Answer with a fraction unless instructed to round. (5 pts ) 1. Use the limit definition to compute f (a) for f(x) = x + 3 at a = 2. f f(a + h) f(a) (a) h 0 h f f( 2 + h) f( 2) ( 2) h 0 h h + 1 1 h 0 h h + 1 1 h 0 h ( h + 1 + 1 ) h h 0 h ( h + 1 + 1 ) h 0 1 h + 1 + 1 = 1 2 h + 1 + 1 h + 1 + 1 (5 pts ) 2. Use the limit definition to compute f (a) and find an equation of the tangent line at x = a. f(x) = x 2 2x, a = 3 f (a) x a f(x) f(a) x a f (3) x 3 f(x) f(3) x 3 x 2 2x 3 x 3 x 3 (x 3)(x + 1) x 3 x 3 (x + 1) x 3 = 4 Equation of tangent line: y f(3) = f (3)(x 3) y 3 = 4(x 3) y = 4x 9
Math 150 (62493)/Quiz 4A Solutions Page 2 of 2 (5 pts ) 3. Use the power rule to compute the derivative of f(x) = 2x 2 6x + 9 and find the equation of the tangent line at x = 1. Derivative: f (x) = 4x 6 Tangent line: f(1) = 5 f (1) = 2 y 5 = 2(x 1) y = 2x + 7 (5 pts ) 4. Use the power rule to compute the derivative. g(x) = 2x5 x 3 + 5x 1 x 2 g(x) = 2x 3 x + 5x 3 g (x) = 6x 2 1 15x 4
Math 150 (62493) Spring 2019 Quiz 4B Solutions Instructor: C. Panza Quiz 4B Solutions: (20 points) Neatly show your work in the space provided, clearly mark and label your answers. Show proper equality, correctly using the equals sign (=). Answer with a fraction unless instructed to round. (5 pts ) 1. Use the limit definition to compute f (a) for f(x) = x 2 at a = 3. f f(a + h) f(a) (a) h 0 h f f(3 + h) f(3) ( 2) h 0 h 1 + h 1 h 0 h 1 + h 1 h 0 h ( 1 + h + 1 ) h h 0 h ( h + 1 + 1 ) h 0 1 h + 1 + 1 = 1 2 1 + h + 1 1 + h + 1 (5 pts ) 2. Use the limit definition to compute f (a) and find an equation of the tangent line at x = a. f(x) = x 2 + 3x, a = 1 f (a) x a f(x) f(a) x a f (1) x 1 f(x) f(1) x 1 x 2 + 3x 4 x 1 x 1 (x 1)(x + 4) x 1 x 1 (x + 4) x 1 = 5 Equation of tangent line: y f(1) = f (21)(x 1) y 4 = 5(x 1) y = 5x 1
Math 150 (62493)/Quiz 4B Solutions Page 2 of 2 (5 pts ) 3. Use the power rule to compute the derivative of f(x) = 2x 2 + 8x 2 and find the equation of the tangent line at x = 1. Derivative: f (x) = 4x + 8 Tangent line: f( 1) = 8 f ( 1) = 4 y ( 8) = 4(x ( 1)) y = 4x 4 (5 pts ) 4. Use the power rule to compute the derivative. g(x) = x6 5x 4 2x 1 x 3 g(x) = x 3 5x 2x 4 g (x) = 3x 2 5 + 8x 5
Math 150 (62493) Spring 2019 Quiz 5A Solutions Instructor: C. Panza Quiz 5A Solutions: (20 points) Neatly show your work in the space provided, clearly mark and label your answers. Show proper equality, correctly using the equals sign (=). Answer word problems with a sentence. (5 pts ) 1. Use the product rule to compute the derivative of f(x) = x 4 e x. f (x) = e x d dx x4 + x 4 d dx ex = 4x 3 e x + x 4 e x = e x x 3 (x + 4) (5 pts ) 2. Use the quotient rule to compute the derivative of f(t) = t t 2 3. f (t) = (t2 3) d dt t t d dt (t2 3) (t 2 3) 2 = (t2 3) 1 t (2t) (t 2 3) 2 = t2 3 (t 2 3) 2
Math 150 (62493)/Quiz 5A Solutions Page 2 of 2 (5 pts ) 3. A ball tossed in the air vertically from ground level returns to the earth 4 seconds later. Find the maximum height (in meters) and the initial velocity (in m/sec) of the ball. [Use g = 9.8 m/s 2.] At 2 seconds the ball is at its maximum height with a velocity of 0 m/sec. v(2) = v 0 9.8(2) 0 = v 0 19.6 19.6 = v 0 s(t) = 0 + 19.6t 4.9t 2 s(2) = 19.6(2) 4.9(2) 2 = 19.6 The ball had a maximum height of 19.6 m an initial velocity of 19.6 m/sec. (5 pts ) 4. Given f(x) = x 3 e x, find f (x) and evaluate f (1). Round your answer to two decimal places. f (x) = 3x 2 e x + x 3 e x f (x) = 6xe x + 3x 2 e x + 3x 2 e x + x 3 e x = 6xe x + 6x 2 e x + x 3 e x f (1) = 6(1)e 1 + 6(1) 2 e 1 + 1 3 e 1 = 13e 35.34
Math 150 (62493) Spring 2019 Quiz 5B Solutions Instructor: C. Panza Quiz 5B Solutions: (20 points) Neatly show your work in the space provided, clearly mark and label your answers. Show proper equality, correctly using the equals sign (=). Answer word problems with a sentence. (5 pts ) 1. Use the product rule to compute the derivative of f(x) = x 3 e x. f (x) = e x d dx x3 + x 3 d dx ex = 3x 2 e x + x 3 e x = e x x 2 (x + 3) (5 pts ) 2. Use the quotient rule to compute the derivative of f(t) = t t 2 + 2. f (t) = (t2 + 2) d dt t t d dt (t2 + 2) (t 2 + 2) 2 = (t2 + 2) 1 t (2t) (t 2 + 2) 2 = 2 t2 (t 2 + 2) 2
Math 150 (62493)/Quiz 5B Solutions Page 2 of 2 (5 pts ) 3. A ball tossed in the air vertically from ground level returns to the earth 6 seconds later. Find the maximum height (in meters) and the initial velocity (in m/sec) of the ball. [Use g = 9.8 m/s 2.] At 3 seconds the ball is at its maximum height with a velocity of 0 m/sec. v(3) = v 0 9.8(3) 0 = v 0 29.4 29.4 = v 0 s(t) = 0 + 29.4t 4.9t 2 s(3) = 29.4(3) 4.9(3) 2 = 44.1 The ball had a maximum height of 44.1 m and an initial velocity of 29.4 m/sec. (5 pts ) 4. Given f(x) = x 1 e x, find f (x) and evaluate f (1). Round your answer to two decimal places. f (x) = x 2 e x + x 1 e x f (x) = 2x 3 e x + ( x 2 e x) + ( x 2 e x) + x 1 e x = 2x 3 e x 2x 2 e x + x 1 e x f (1) = 2(1) 3 e 1 2(1) 2 e 1 + 1 1 e 1 = e 2.72
Math 150 (62493) Spring 2019 Quiz 6A Solutions Instructor: C. Panza Quiz 6A Solutions: (20 points) Neatly show your work in the space provided, clearly mark and label your answers. Show proper equality, correctly using the equals sign (=). (5 pts ) 1. Compute the derivative using the Chain rule. f(x) = e x4 5x f (x) = ( ) e x4 5x (4x 3 5 ) = e x4 5x ( 4x 3 5 ) (5 pts ) 2. Compute the derivative using the Chain Rule. y = ( ) 3 sin 1 (4x) 1 ( ) ( ) 2 y = 3 sin 1 1 (4x) 1 4 1 (4x) 2 = ( ) 2 12 sin 1 (4x) 1 1 16x 2
Math 150 (62493)/Quiz 6A Solutions Page 2 of 2 (5 pts ) 3. Compute the higher derivative y. y = sec x y = d ( ) 1 dx cos x 0 ( sin x) = cos 2 x = sec x tan x y = tan x d d sec x + sec x dx dx tan x = tan x sec x tan x + sec x sec 2 x = sec x tan 2 x + sec 3 x = sec x ( tan 2 x + sec 2 x ) (5 pts ) 4. Find the derivative with respect to x using implicit differentiation. Solve for y. 2x 3 + cos y = sin y d ( 2x 3 + cos y ) = d dx dx sin y 6x 2 sin yy = cos yy 6x 2 cos y + sin y = y 6x 2 = cos yy + sin yy 6x 2 = y (cos y + sin y)
Math 150 (62493) Spring 2019 Quiz 6B Solutions Instructor: C. Panza Quiz 6B Solutions: (20 points) Neatly show your work in the space provided, clearly mark and label your answers. Show proper equality, correctly using the equals sign (=). (5 pts ) 1. Compute the derivative using the Chain rule. f(x) = e 2x4 +3x f (x) = ( ) e 2x4 +3x (8x 3 + 3 ) = ( 8x 3 + 3 ) e 2x4 +3x (5 pts ) 2. Compute the derivative using the Chain Rule. y = ( ) 5 2 + sin 1 (3x) ( ) ( ) 4 y = 5 2 + sin 1 1 (3x) 3 1 (3x) 2 = ( 15 2 + sin 1 (3x) 1 9x 2 ) 4
Math 150 (62493)/Quiz 6B Solutions Page 2 of 2 (5 pts ) 3. Compute the higher derivative y. y = sec x y = d ( ) 1 dx cos x 0 ( sin x) = cos 2 x = sec x tan x y = tan x d d sec x + sec x dx dx tan x = tan x sec x tan x + sec x sec 2 x = sec x tan 2 x + sec 3 x = sec x ( tan 2 x + sec 2 x ) (5 pts ) 4. Find the derivative with respect to x using implicit differentiation. Solve for y. sin y = 4x 2 cos y d dx sin y = d ( 4x 2 cos y ) dx cos yy = 8x + sin yy cos yy sin yy = 8x y (cos y sin y) = 8x y 8x = cos y sin y
Math 150 (62493) Spring 2019 Quiz 7A Solutions Instructor: C. Panza Quiz 7A Solutions: (20 points) Neatly show your work in the space provided, clearly mark and label your answers. Show proper equality, correctly using the equals sign (=). (5 pts ) 1. Find the derivative using logarithmic differentiation. (2x + 5)3 y = (x 3 8) 2 [ ] (2x + 5) 3 ln y = ln (x 3 8) 2 ln y = 3 ln (2x + 5) 2 ln ( x 3 8 ) y y = 6 2x + 5 y = (2x + 5)3 (x 3 8) 2 6x2 x 3 8 ( 6 2x + 5 ) 6x2 x 3 8 (5 pts ) 2. Calculate the derivative of f(x) = x 2x4. y = x 2x4 ln y = ln x 2x4 ln y = 2x 4 ln x y y = 8x3 ln x + 2x 4 1 x y y = 8x3 ln x + 2x 3 y = y ( 8x 3 ln x + 2x 3) f (x) = x 2x4 ( 8x 3 ln x + 2x 3) f (x) = 2x 3 x 2x4 (4 ln x + 1) f (x) = 2x 2x4 +3 (4 ln x + 1)
Math 150 (62493)/Quiz 7A Solutions Page 2 of 2 (5 pts ) 3. A sailboat 4 kilometers from shore passes by SBCC traveling parallel to the beach at 3 kilometers per hour. How fast is the distance between SBCC and the sailboat changing 30 minutes later? x = the sailboat s horizontal distance from SBCC z = the distance from SBCC to the sailboat 4 km x z z = 4 2 + x 2 dz dt = 2x 2 16 + x dx 2 dt At t = 30 minutes, ( ) 30 x = 3 = 3 60 2 km dx dt = 3 km/hr dz dt = 3/2 3 16 + (3/2) 2 1.05 The distance between SBCC and the sailboat is changing at about 1.05 km/hr. (5 pts ) 4. A man of height 1.8 m walks away from a 4-meter lamppost. After 2 seconds, his shadow is increasing at a rate of 1 meter per second. How fast is the man walking? x = the distance between the man and lamppost y = the length of the man s shadow 4 m x The man is walking about 1.2 meters per second. 1.8 y 4 x + y = 1.8 y 4y = 1.8(x + y) 4 dy ( dx dt = 1.8 dt + dy ) dt 2.2 dy dt = 1.8dx dt At t = 2 seconds, dy dt = 1 m/s 2.2(1) = 1.8 dx dt 1.2222 dx dt
Math 150 (62493) Spring 2019 Quiz 7B Solutions Instructor: C. Panza Quiz 7B Solutions: (20 points) Neatly show your work in the space provided, clearly mark and label your answers. Show proper equality, correctly using the equals sign (=). (5 pts ) 1. Find the derivative using logarithmic differentiation. (3x 7)4 y = (x 2 8) 3 [ ] (3x 7) 4 ln y = ln (x 2 8) 3 ln y = 4 ln (3x 7) 3 ln ( x 2 8 ) y y = 12 3x 7 y = (3x 7)4 (x 2 8) 3 6x x 2 8 ( 12 3x 7 6x ) x 2 8 (5 pts ) 2. Calculate the derivative of f(x) = x 4x3. y = x 4x3 ln y = ln x 4x3 ln y = 4x 3 ln x y y = 12x2 ln x + 4x 3 1 x y y = 12x2 ln x + 4x 2 y = y ( 12x 2 ln x + 4x 2) f (x) = x 4x3 ( 12x 2 ln x + 4x 2) f (x) = 4x 2 x 4x3 (3 ln x + 1) f (x) = 4x 4x3 +2 (3 ln x + 1)
Math 150 (62493)/Quiz 7B Solutions Page 2 of 2 (5 pts ) 3. A plane passes directly overhead SBCC at 800 kilometers per hour at a height of 3 kilometers above ground. How fast is the distance between SBCC and the plane changing one minute later? x = the plane s horizontal distance from SBCC z = the distance from SBCC to the plane 3 km x z z = 3 2 + x 2 dz dt = 2x 2 9 + x dx 2 dt At t = 1 minute, ( ) 1 x = 800 = 40 60 3 km dx = 800 km/hr dt dz dt = 40/3 800 9 + (40/3) 2 780.49 The distance between SBCC and the plane is changing at about 780.49 km/hr. (5 pts ) 4. A woman of height 1.6 m walks away from a 5-meter lamppost. After 2 seconds, her shadow is increasing at a rate of 1 meter per second. How fast is the woman walking? x = the distance between the woman and lamppost y = the length of the woman s shadow 5 m x The woman is walking about 2.2 meters per second. 1.6 y 5 x + y = 1.6 y 5y = 1.6(x + y) 5 dy ( dx dt = 1.6 dt + dy ) dt 3.4 dy dt = 1.6dx dt At t = 2 seconds, dy dt = 1 m/s 3.4(1) = 1.6 dx dt 2.125 = dx dt