Yvette Granillo, Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the derivative of f when 2 f (x) = 2x sin 5x + cos 5x. 5 1. f (x) = 10x cos 5x correct 2. f (x) = 10x cos 5x + 4 sin 5x 3. f (x) = 10x cos 5x - 4 sin 5x 4. f (x) = 10 cos 5x 5. f (x) = -10x cos 5x Explanation: Since d sin x = cos x, dx it follows that f (x) = 2 sin 5x + 10x cos 5x - 2 sin 5x. Consequently, f (x) = 10x cos 5x . d cos x = - sin x, dx keywords: Stewart5e, 003 (part 1 of 1) 10 points Determine f (x) when f (x) = 1. f (x) = - 2. f (x) = - 3. f (x) = 1+x . 1 - 2x2 4(x - 2) (x + 1)3
1
2. f (x) = - 3. f (x) = 4. f (x) =
6(x - 2) (x + 1)3 4(x - 1) correct (x + 1)3 6(x + 2) (x - 1)3
5. f (x) = - 6. f (x) =
6(x + 1) (x - 1)3
Explanation: By the Chain and Quotient Rules, f (x) = 2 Consequently, f (x) = 4(x - 1) . (x + 1)3 x - 1 (x + 1) - (x - 1) . x+1 (x + 1)2
1 - 2x (1 - 2x2 )1/2 1 + 2x (1 - 2x2 )1/2
keywords: Stewart5e, 002 (part 1 of 1) 10 points Find f (x) when f (x) = x-1 x+1
2
1 + 2x (1 - 2x2 )1/2
4. f (x) = - . 5. f (x) = 6. f (x) =
1. f (x) = -
4(x + 1) (x - 1)3
1 + 2x correct (1 - 2x2 )3/2 1 - 2x (1 - 2x2 )3/2
1 + 2x (1 - 2x2 )3/2
Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell Explanation: By the Product and Chain Rules, f (x) = 1 4x(1 + x) + 2 )1/2 (1 - 2x 2(1 - 2x2 )3/2 = Consequently, f (x) = 1 + 2x . (1 - 2x2 )3/2 (1 - 2x2 ) + 2x(1 + x) . (1 - 2x2 )3/2 keywords: Stewart5e, 005 (part 1 of 1) 10 points Find the value of f (1) when f (x) = (x2 + 8)1/2 + 1 3 1 . x
2
1. f (1) = - 2. f (1) = 2 3
(Note: the Quotient Rule could have been used, but it's simpler to use the Product Rule.) keywords: Stewart5e, 004 (part 1 of 1) 10 points Determine f (x) when f (x) = 2 cos2 x + 3 sin2 x . 1. f (x) = -4 sin x - 6 cos x 2. f (x) = 2 sin x cos x correct 3. f (x) = -10 sin x cos x 4. f (x) = 4 cos x + 6 sin x
3. f (1) = 0 2 4. f (1) = - correct 3 5. f (1) = 1 3
Explanation: Using the Chain Rule and the fact that d r x = r xr-1 dx holds for all values of r, we see that f (x) = x 1 - 2. 1/2 x + 8)
5. f (x) = -2 sin x cos x 6. f (x) = -4 sin x + 6 cos x 7. f (x) = 10 sin x cos x Explanation: Since d d sin x = cos x , cos x = - sin x , dx dx the Chain Rule ensures that f (x) = -4 cos x sin x + 6 sin x cos x . Consequently, f (x) = 2 sin x cos x .
(x2
At x = 1, therefore, f (1) = - 2 . 3
keywords: Stewart5e, 006 (part 1 of 1) 10 points Find the derivative of f when f (x) = 1 (3 - 4x) 3 + 8x . 4
Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell 1. f (x) = - 2. f (x) = - 3. f (x) = - 4. f (x) = 5. f (x) = 4x 3 + 8x 12x correct 3 + 8x 12x 3 + 4x 4 f (x + 8)2 x+4 x+8
3
6. F (x) = -
Explanation: By the Chain Rule, F (x) = f x+4 x+8 d dx x+4 x+8 .
12x 3 + 8x 4x 3 + 8x
But by the Quotient Rule, dx dx x+4 x+8 = (x + 8) - (x + 4) . (8 + x)2
Explanation: By the Product and Chain Rules, 3 - 4x f (x) = - 3 + 8x + . 3 + 8x Bringing the right hand side to a common denominator, we thus see that: (3 - 4x) - (3 + 8x) 12x f (x) = = - . 3 + 8x 3 + 8x
Consequently, F (x) = 4 f (x + 8)2 x+4 x+8 .
keywords: Stewart5e, 008 (part 1 of 1) 10 points Find the derivative of f when f (x) = sin 5+x 2-x .
keywords: Stewart5e, 007 (part 1 of 1) 10 points Find the derivative of F when F (x) = f and f is differentiable. 1. F (x) = f 2. F (x) = 3. F (x) = 4 (x + 8)2 x+4 x+8 x+4 x+8 correct x+4 x+8
1. f (x) =
7 sin (2 - x)2 7 cos (2 - x)2 7 (2 - x)2 7 (2 - x)2
5+x 2-x 5+x 2-x
2. f (x) = -
3. f (x) = sin 4. f (x) = cos 5. f (x) = - 6. f (x) = x+4 x+8
4 f (x + 8)2 4 f (x + 8)2 4 (x + 8)2 4 f (x + 8)2
7 sin (2 - x)2
5+x 2-x 5+x 2-x correct
4. F (x) = f 5. F (x) = -
7 cos (2 - x)2
Explanation:
Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell By the Chain and Quotient rules, f (x) = cos 5+x 2-x 5+x 2-x d dx 5+x 2-x . keywords: Stewart5e, 010 (part 1 of 1) 10 points . Find the value of F (2) when F (x) = f (g(x)) keywords: Stewart5e, 009 (part 1 of 1) 10 points Find an equation for the tangent line to the graph of f (x) = 8 sin(sin(x)) at the point P (, f ()). 1. y + 8x = 8 correct 2. y + = 8x 3. y + 8x + = 0 4. y = 8x + 5. y = 8x - 8 Explanation: Since f () = sin(sin()) = sin 0 = 0 , we have to find the equation of the tangent line to the graph of f (x) = 8 sin(sin(x)) at P (, 0). Now f (x) = (8 cos x) cos(sin x)) , so at P (, 0), the tangent line has slope = 8(cos ) cos 0 = -8 . Thus and g(2) = 6, f (2) = 5, 1. F (2) = 10 2. F (2) = 14 3. F (2) = 12 correct 4. F (2) = 11 5. F (2) = 13 Explanation: By the Chain Rule, F (x) = f (g(x))g (x) . F (2) = f (g(2))g (2) = f (6)g (2) . Consequently, when g(2) = 6, f (2) = 5, we see that F (2) = 12 . g (2) = 4 , f (6) = 3 , g (2) = 4 , f (6) = 3 .
4
By the point-slope formula, therefore, the tangent line is y = -8(x - ), i.e., y + 8x = 8 .
= cos Thus
(2 - x) + (5 + x) (2 - x)2 5+x 2-x
7 f (x) = cos (2 - x)2
Notice that the value of f (2) was not needed.
Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell keywords: Stewart5e, 011 (part 1 of 1) 10 points 3. Find dy/dx when 3x2 + 2y 2 = 1 . dy = -3xy dx 3x dy = - correct dx 2y 3x dy = dx 2y 3x dy = - dx y x dy = dx 2y dy = 2xy dx 4. 5. 2. dy = cot x cot y dx dy = tan x tan y correct dx dy = tan xy dx dy = cot x tan y dx
5
1. 2. 3. 4. 5. 6.
Explanation: Differentiating implicitly with respect to x we see that 2 cos x cos y Thus dy cos x cos y = sin x sin y . dx Consequently, dy sin x sin y = = tan x tan y . dx cos x cos y dy - sin y sin x dx = 0.
Explanation: Diferentiating 3x2 + 2y 2 = 1 implicitly with respect to x we see 6x + 4y Consequently, 6x 3x dy = - = - . dx 4y 2y dy = 0. dx
keywords: Stewart5e, 013 (part 1 of 1) 10 points dy when dx tan(x y) - = x . dy x2 1. = correct dx 1 + x2 2. 3. 4. dy x2 = dx 1 + y2 dy x2 = - dx 1 + y2 dy x2 = - dx 1 + x2
Find
keywords: Stewart5e, 012 (part 1 of 1) 10 points Determine dy/dx when 2 cos x sin y = 5 . dy = tan x 1. dx
Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell dy y2 = dx 1 + x2 dy y2 = - dx 1 + x2 1. 2. 3. 4. 5. = 1. dy dx dy dx dy dx dy dx dy dx = = 7 11 1 7
6
5. 6.
P
P
Explanation: Differentiating tan(x - y) = x implicitly with respect to x we see that sec2 (x - y) 1 - Thus - sec2 (x - y) in which case dy = dx sec2 (x - y) - 1 sec2 (x - y)
2
P
= 7 = -7 correct = - 1 7
P
dy dx
P
Explanation: Differentiating implicitly with respect to x we see that 15x2 + 9y 2 Consequently, dy dy = 9y + 9x . dx dx
dy = 1 - sec2 (x - y) , dx
. But at
2
But sec2 = 1 + tan2 , so sec (x - y) = 1 + tan (x - y) = 1 + x , while sec2 (x - y) - 1 = tan2 (x - y) = x2 . Consequently, dy x2 = . dx 1 + x2 keywords: Stewart5e, 014 (part 1 of 1) 10 points Find the slope of the tangent line to the graph of 5x3 + 3y 3 = 9xy at the point P = 9 9 , . 8 8
2
dy 3y - 5x2 = . dx 3y 2 - 3x 9 9 , , 8 8
P =
simple algebra shows that 3y - 5x2 = while 3y 2 - 3x = Hence at P , dy dx = -7. 243 27 27 - = . 64 8 64 189 27 405 - = - , 8 64 64
P
keywords: Stewart5e, 015 (part 1 of 1) 10 points Determine F (x) when F (x) = -2xf (x) + 3f (x)
Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell and f is a twice-differentiable function. 1. F (x) = -5 f (x) + 2 x f (x) 2. F (x) = f (x) - 3 x f (x) 3. F (x) = f (x) - 2 x f (x) correct 4. F (x) = f (x) + 3 x f (x) 5. F (x) = -5 f (x) - 2 x f (x) Explanation: After differentiation using the Product Rule we see that F (x) = -2f (x) - 2xf (x) + 3f (x) . Consequently, F (x) = f (x) - 2xf (x) . keywords: Stewart5e, 016 (part 1 of 1) 10 points The figure below shows the graphs of three functions of time t: 6. F (x) = -5 f (x) - 3 x f (x) 4. s : 5. s : 6. s : v: v: v: a: a: a: correct 1. s : 2. s : 3. s : v: v: v: a: a: a:
7
Explanation: Experience tells us that the car is (i) moving forwards when its velocity is positive, (ii) moving backwards when its velocity is negative, (iii) speeding up when its velocity is positive and increasing, i.e., when both velocity and acceleration are positive, (iv) slowing down when its velocity is positive but decreasing, i.e., when its velocity is positive but its acceleration is negative. Now ds dv v = a = , dt dt so we need to look at the slope of the tangent line to three graphs to determine which graph is that of position, that of velocity and that of acceleration. Inspection of the graphs thus shows that s: v: a: .
t
keywords: Stewart5e, One is the graph of the position function s of a car, one is its velocity v, and one is its acceleration a. Identify which graph goes with which function. 017 (part 1 of 1) 10 points Determine d2 y/dx2 when 4x2 + 2y 2 = 2.
Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell d2 y 1 1. = - 3 2 dx y 2. 3. 4. 5. 2 d2 y = 3 2 dx y 2 d2 y = - 2 2 dx y d2 y 2 = - 3 correct 2 dx y d2 y 2 = 2 2 dx y
8
Find the value of R (0) by first finding R (0). 1. R (0) = 2. R (0) = 3. R (0) = 4. R (0) = 5. R (0) = 3 14 1 4 3 28 1 7 5 correct 28
Explanation: Differentiating implicitly with respect to x we see that 8x + 4y so x dy = -2 . dx y But then d2 y d x = - 2 2 dx dx y = - Thus 1 d2 y = - 3 4x2 + 2y 2 2 dx y keywords: Stewart5e, 018 (part 1 of 1) 10 points A special function known as the Rivaldo function satisfies both the equation 3xR (x) + 4R (x) + R(x) = 0 and the equation R(0) = 5 . 2 = - 3. y = - 8y - 8x 4y 2 dy dx dy = 0, dx
Explanation: Setting x = 0 in the equation 3xR (x) + 4R (x) + R(x) = 0 , we see that 4R (0) + R(0) = 0 , so 5 R (0) = - . 4 ()
On the other hand, by differentiating () implicitly with respect to x we see also that 3R (x) + 3xR (x) + 4R (x) + R (x) = 0 . At x = 0, therefore, 3R (0) + 4R (0) + R (0) = 0 . Consequently, R (0) = 5 . . 28
4x2 1 2y + . y2 y
keywords: Stewart5e, 019 (part 1 of 1) 10 points Determine the third derivative, f (x), of f when f (x) = 4x + 1 .
Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell 1. f (x) = -24(4x + 1)
-5/2
9
2. f (x) = -24(4x + 1)-3/2 3. f (x) = 24(4x + 1)-5/2 correct 4. f (x) = -4(4x + 1)-3/2 5. f (x) = 4(4x + 1)-3/2 6. f (x) = 4(4x + 1)-5/2 Explanation: To use the Chain Rule successively it's more convenient to write f (x) = 4x + 1 = (4x + 1)1/2 . For then f (x) = 1 4 (4x + 1)-1/2 2 = 2(4x + 1) while 1 f (x) = - 4 2(4x + 1)-3/2 2 = -4(4x + 1)-3/2 , and f (x) = Consequently, f (x) = 24(4x + 1)-5/2 . 3 4 4(4x + 1)-5/2 . 2
-1/2
where A is the amplitude of its oscillations and is a constant. Find the velocity and acceleration as functions of time. 1. v(t) = - A cos t, a(t) = -A 2 cos t 2. v(t) = A sin t, a(t) = -A cos t 3. v(t) = - A sin t, a(t) = -A 2 cos t correct 4. v(t) = A sin t, a(t) = -A 2 cos t 5. v(t) = - A sin t, a(t) = -A cos t Explanation: y(t) = A cos t v(t) = y (t) = - A sin t , a(t) = v (t) = -A 2 cos t keywords: Stewart5e, 021 (part 1 of 1) 10 points If P is a second degree polynomial such that P (2) = 8, P (2) = 5, P (2) = 5,
find the value of P (1). 1. P (1) = 6 2. P (1) = 7 3. P (1) = 11 correct 2 13 2
keywords: Stewart5e, 020 (part 1 of 1) 10 points A mass attached to a vertical spring has position function given by y(t) = A cos t, 4. P (1) =
5. P (1) = 5 Explanation:
Granillo, Yvette Homework 7 Due: Oct 13 2005, 3:00 am Inst: Edward Odell The most general second degree polynomial has the form P (x) = ax2 + bx + c . But then P (x) = 2ax + b, Thus P (2) = 5 in which case P (x) = 5x + b . The condition on P (2) now ensures that P (2) = 10 + b = 5 , i.e., b = -5. Thus P (x) = in which case P (2) = 8 Consequently, P (x) = At x = 1, therefore, P (1) = 11 . 2 5 2 x - 5x + 8 . 2 = c = 8. 5 2 x - 5x + c , 2 = a = 5 , 2 P (x) = 2a .
10
keywords: Stewart5e,