Section 3.1 Derivatives of Polynomials and Exponential Functions 86. Find numbers

Section 3.1 Derivatives of Polynomials and Exponential Functions

86. Find numbers a and b such that the given function g is differentiable at 1.

g(x) =







ax³− 3x if x ≤ 1 bx²+ 2 if x > 1 Solution:

SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 189

so  +  + 4 = −1 or  +  = −5 (1). Also, ⁰(1) = 4 + 3 + 2 +  = 3 + 2 + 6and the slope of the tangent line

 = 2 − 3 at  = 1 is −3, so 3 + 2 + 6 = −3 or 3 + 2 = −9 (2). Adding −2 times (1) to (2) gives us  = 1 and hence,  = −6. The curve has equation  = ⁴+ ³− 6²+ 2 + 1.

81.  = () = ² ⇒ ⁰() = 2. So the slope of the tangent to the parabola at  = 2 is  = 2(2) = 4. The slope of the given line, 2 +  =  ⇔  = −2 + , is seen to be −2, so we must have 4 = −2 ⇔  = −¹2. So when

 = 2, the point in question has ­coordinate −¹₂· 2² = −2. Now we simply require that the given line, whose equation is 2 +  = , pass through the point (2 −2): 2(2) + (−2) =  ⇔  = 2. So we must have  = −¹2 and  = 2.

82. The slope of the curve  = √

is ⁰=  2√

and the slope of the tangent line  = ³₂ + 6is³₂. These must be equal at the point of tangency

 √

 , so 

2√

 = 3

2 ⇒  = 3√

. The ­coordinates must be equal at  = , so

√

 = ³₂ + 6 ⇒  3√

 √

 = ³₂ + 6 ⇒ 3 = ³2 + 6 ⇒ ³2 = 6 ⇒  = 4. Since  = 3√

, we have

 = 3√ 4 = 6.

83. The line  = 2 + 3 has slope 2. The parabola  = ² ⇒ ⁰= 2has slope 2 at  = . Equating slopes gives us 2 = 2, or  = 1. Equating ­coordinates at  =  gives us ²= 2 + 3 ⇔ () = 2 + 3 ⇔ 1 = 2 + 3 ⇔

 = −3. Thus,  = 1

 = −1 3.

84. () = ²+  +  ⇒ ⁰() = 2 + . The slope of the tangent line at  =  is 2 + , the slope of the tangent line at  =  is 2 + , and the average of those slopes is (2 + ) + (2 + )

2 =  +  + . The midpoint of the interval [ ]is  + 

2 and the slope of the tangent line at the midpoint is 2  +  2

+  = ( + ) + . This is equal to

 +  + , as required.

85. is clearly differentiable for   2 and for   2. For   2, ⁰() = 2, so ₋⁰(2) = 4. For   2, ⁰() = , so

+⁰(2) = . For  to be differentiable at  = 2, we need 4 = ₋⁰(2) = +⁰(2) = . So () = 4 + . We must also have continuity at  = 2, so 4 = (2) = lim

→2⁺ () = lim

→2⁺(4 + ) = 8 + . Hence,  = −4.

86. We have  () =

³− 3 if  ≤ 1

²+ 2 if   1

For   1, ⁰() = (3²) − 3(1) = 3²− 3, so ⁰−(1) = 3(1)²− 3 = 3 − 3. For   1, ⁰() = (2) + 0 = 2, so

+⁰ (1) = 2(1) = 2. For  to be differentiable at  = 1, we need ₋⁰ (1) = ⁰+(1), so 3 − 3 = 2, or  = 3 − 3

2 . For  to be continuous at  = 1, we need _(1) =  − 3 equal to ⁺(1) =  + 2. So we have the system of two equations:

 − 3 =  + 2,  = 3 − 3

2 . Substituting the second equation into the first equation we have  − 3 = 3 − 3

2 + 2 ⇒

2 − 6 = 3 − 3 + 4 ⇒  = −7 and  = 3(−7) − 3

2 = −12.

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88. A tangent line is drawn to the hyperbola xy = c at a point P as shown in the figure.

(a) Show that the midpoint of the line segment cut from this tangent line by the coordinate axes is P .

(b) Show that the triangle formed by the tangent line and the coordinate axes always has the same area, no matter where P is located on the hyperbola.

(b) Show that the triang1e formed by the tangent line and the coordinate axes a1ways has the same area, no matter where P is 10cated on the hyperb01a.

Differentiation Ru les CHAPTER 3

184

85. Let

if x:::三2

if x > 2

'。+

x

而/-哼，LV

〈自『fit-IL xn

--

、、‘.. E.'' ,

/

x

.. '

、

fJ n

y Find the values of m and b that make

f

differentiable

everywhere.

86. Find numbers a and b such that the given function 9 is differentiable at 1

x

89. Draw a diagram showing two pe叩endicularlines that inter- sect on the y-axis and are both tangent to the parab01a y = χ2 Where do these lines intersect?

。

if x 三 l

if x > 1

90. Sketch the parab01as y = x² and y = x^{2 -} 2x + 2. Do you think there is a line that is tangent to both curves? If so, fìnd its equation. If not, why not?

X 1000 - 1 87. Evaluate lim

x• 1 X - 1

91. If c > ~， how many lines through the point (0, c) are norma1

lines to 伽 parabola^{y =} x²? What if c 三;7 88. A tangent line is drawn to the hyperb01a xy = c at a

point P as shown in the fìgure.

(a) Show that the midpoint of the line segment cut from this tangent line by the coordinate axes is P.

Suppose you are asked to design the fìrst ascent and drop for a new roller coaster. By study- ing photographs of your favorite coasters, you decide to make the slope of the ascent 0.8 and the slope of the drop -1.6. You decide to connect these two straight stretches y = LI(x) and y = L2(x) with part of a parabola y = f(x) = ax² + bx + c, where x and f(x) are measured in meters. For the track to be smooth there can't be abrupt changes in direction, so you want the linear segments LI and L2 to be tangent to the parabola at the transition points P and Q.

(See the fìgure.) To simplify the equations, you decide to place the origin at P.

A叫 ED PROJ 的 |叫叫 NG

A BETTER ROLLER COASTER

1. (a) Suppose the horizontal distance between P and Q is 30 m. Write equations in a, b, and c that will ensure that the track is smooth at the transition points.

(b) Solve the equations in part (a) for a, b, and c to fìnd a formula for f(x).

囝 (c) Plot LI, f, and L2 to veri勾 graphically that the transitions are smooth (d) Find the difference in elevation between P and Q.

2. The s01ution in Problem 1 might look smooth, but it might notfeel smooth because the piecewise defìned function [consisting of LI(x) for x < 0, f(x) for 0 :::三 x:::三 30 ，and L2(x) for x

>

30] doesn't have a continuous second derivative. So you decide to improve the design by using a quadratic function q(x) = αx² + bx + c only on the interval 3 三三X :::三 27 and connecting it to the linear functions by means of two cubic functions:

g(x) = kx³ + Ix² + mx + n 0 :::三 x< 3

h(x) = px³ + qx² + rx + s

EDutuD】的」血口=平的\個白白亡。由E咱的=的

27 < ^{x :::三 30}

(a) Write a system of equations in 11 unknowns that ensures that the functions and their fìrst two derivatives agree at the transition points.

囝 (b) S01ve t伽hes叮咖圳y戶州s泣t岫emofe叩q叩叩u叫1

(扣c) P刊10叫tLJ, g, q, h, and L幻2， andc∞ompa缸rewith the p抖10叫tin Problem 1 (收c).

Solution:

SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 177

80. () = ²+  +  ⇒ ⁰() = 2 + . The slope of the tangent line at  =  is 2 + , the slope of the tangent line at  =  is 2 + , and the average of those slopes is (2 + ) + (2 + )

2 =  +  + . The midpoint of the interval [ ]is  + 

2 and the slope of the tangent line at the midpoint is 2  +  2

+  = ( + ) + . This is equal to

 +  + , as required.

81. is clearly differentiable for   2 and for   2. For   2, ⁰() = 2, so ₋⁰(2) = 4. For   2, ⁰() = , so

+⁰(2) = . For  to be differentiable at  = 2, we need 4 = ₋⁰(2) = +⁰(2) = . So () = 4 + . We must also have continuity at  = 2, so 4 = (2) = lim

→2⁺ () = lim

→2⁺(4 + ) = 8 + . Hence,  = −4.

82. (a)  =  ⇒  = 

. Let  =

 



. The slope of the tangent line at  =  is ⁰() = − 

². Its equation is

 − 

 = − 

²( − ) or  = − 

² +2

, so its -intercept is 2

. Setting  = 0 gives  = 2, so the -intercept is 2.

The midpoint of the line segment joining

 02





and (2 0) is







= .

(b) We know the - and -intercepts of the tangent line from part (a), so the area of the triangle bounded by the axes and the tangent is ¹₂(base)(height) =¹₂ = ¹₂(2)(2) = 2, a constant.

83. Solution 1: Let () = ¹⁰⁰⁰. Then, by the definition of a derivative, ⁰(1) = lim

→1

 () − (1)

 − 1 = lim

→1

¹⁰⁰⁰− 1

 − 1 . But this is just the limit we want to find, and we know (from the Power Rule) that ⁰() = 1000⁹⁹⁹, so

⁰(1) = 1000(1)⁹⁹⁹= 1000. So lim

→1

¹⁰⁰⁰− 1

 − 1 = 1000.

Solution 2: Note that (¹⁰⁰⁰− 1) = ( − 1)(⁹⁹⁹+ ⁹⁹⁸+ ⁹⁹⁷+ · · · + ²+  + 1). So

lim→1

¹⁰⁰⁰− 1

 − 1 = lim

→1

( − 1)(⁹⁹⁹+ ⁹⁹⁸+ ⁹⁹⁷+ · · · + ²+  + 1)

 − 1 = lim

→1(⁹⁹⁹+ ⁹⁹⁸+ ⁹⁹⁷+ · · · + ²+  + 1)

= 1 + 1 + 1 + · · · + 1 + 1 + 1  = 1000, as above.

1000 ones

84. In order for the two tangents to intersect on the -axis, the points of tangency must be at equal distances from the -axis, since the parabola  = ²is symmetric about the -axis.

Say the points of tangency are

 ²and

− ², for some   0. Then since the derivative of  = ²is  = 2, the left-hand tangent has slope −2 and equation

 − ²= −2( + ), or  = −2 − ², and similarly the right-hand tangent line has equation  − ²= 2( − ), or  = 2 − ². So the two lines intersect at

0 −². Now if the lines are perpendicular, then the product of their slopes is −1, so (−2)(2) = −1 ⇔ ²= ¹₄ ⇔  = ¹2. So the lines intersect at

0 −¹4

.

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1

90. Sketch the parabolas y = x² and y = x²− 2x + 2. Do you think there is a line that is tangent to both curves? If so, find its equation. If not, why not?

Solution:

178 ¤ CHAPTER 3 DIFFERENTIATION RULES

85.  = ² ⇒ ⁰= 2, so the slope of a tangent line at the point ( ²)is ⁰= 2and the slope of a normal line is −1(2), for  6= 0. The slope of the normal line through the points ( ²)and (0 ) is²− 

 − 0, so ²− 

 = − 1

2 ⇒

²−  = −¹2 ⇒ ²=  −¹2. The last equation has two solutions if   ¹₂, one solution if  = ¹₂, and no solution if

  ¹₂. Since the -axis is normal to  = ²regardless of the value of  (this is the case for  = 0), we have three normal lines if   ¹₂ and one normal line if  ≤ ¹2.

86. From the sketch, it appears that there may be a line that is tangent to both curves. The slope of the line through the points  ( ²)and

( ²− 2 + 2) is ²− 2 + 2 − ²

 −  . The slope of the tangent line at  is 2 [⁰= 2] and at  is 2 − 2 [⁰= 2 − 2]. All three slopes are equal, so 2 = 2 − 2 ⇔  =  − 1.

Also, 2 − 2 = ²− 2 + 2 − ²

 −  ⇒ 2 − 2 = ²− 2 + 2 − ( − 1)²

 − ( − 1) ⇒ 2 − 2 = ²− 2 + 2 − ²+ 2 − 1 ⇒ 2 = 3 ⇒  = ³2and  = ³₂ − 1 = ¹2. Thus, an equation of the tangent line at  is  −1

2

2

= 21 2

 −¹2

or

 =  − ¹4.

APPLIED PROJECT Building a Better Roller Coaster

1. (a) () = ²+  +  ⇒ ⁰() = 2 + .

The origin is at  :  (0) = 0 ⇒  = 0

The slope of the ascent is 08: ⁰(0) = 08 ⇒  = 08 The slope of the drop is −16: ⁰(30) = −16 ⇒ 60 +  = −16

(b)  = 08, so 60 +  = −16 ⇒ 60 + 08 = −16 ⇒ 60 = −24 ⇒  = −24

60 = −004.

Thus, () = −004²+ 08.

(c) Since 1passes through the origin with slope 08, it has equation  = 08.

The horizontal distance between  and  is 30, so the -coordinate at  is

 (30) = −004(30)²+ 08(30) = −12. Since ²passes through the point (30 −12) and has slope −16, it has equation  + 12 = −16( − 30) or  = −16 + 36.

(d) The difference in elevation between  (0 0) and (30 −12) is 0 − (−12) = 12 meters.

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