Post on 15-Apr-2018
transcript
1a 1b
Table of Trig Functions
sin ( x ) cos ( x )
tan ( x )= sin ( x )cos ( x )
cot ( x )= cos ( x )sin (x )
sec ( x )= 1cos (x)
csc ( x )= 1sin (x )
Obtain co-functions in right column from left column by replacing sin ( x )→cos (x) and cos ( x )→sin (x).
Two Famous Trianglessketch 45°-45°-90° triangle, label angles and sides
Pythagorean theorem
( 1√2 )2
+( 1√2 )2
=1 or 12+12=1
sin( π4 )=cos( π4 )= 1√2
, tan( π4 )=1
sketch 30°-60°-90° triangle, label angles and sides
Pythagorean theorem
(√32 )2
+( 12 )2
=1 or 34 + 12=1
sin( π6 )=12 ,cos( π6 )=√32
, tan(π6 )= 1√3
sin( π3 )=√32
,cos ( π3 )=12 , tan(π3 )=√3
2a 2b
Derivatives of Trig Functions
ddxtan ( x )= d
dxsin ( x )cos ( x )
¿ (sin x )'¿¿
¿cos ( x )⋅cos (x )−sin ( x )⋅ (−sin ( x ) )
cos2 x
¿cos2 ( x )+sin2 ( x )
cos2 (x )
¿1
cos2 ( x ) ¿ sec2( x)
where sec ( x )=1 /cos (x ).
ddx
sec ( x )= ddx
1cos ( x )
¿ (1 )' ⋅cos x−1 ⋅ (cos x )'
cos2x
¿−¿¿
¿ 1cos x
⋅ sin xcos x
¿ sec x ⋅ tan x
similarly, obtain the following table (to memorize)
ddxsin ( x )=cos ( x ) d
dxcos ( x )=−sin ( x )
ddxtan ( x )=sec2 ( x ) d
dxcot ( x )=−csc2 ( x )
ddx
sec ( x )=sec (x ) ⋅ tan ( x ) ddx
csc ( x )=−csc ( x )⋅ cot(x)
to obtain the right hand column from the left hand column, replace functions by co-functions and multiply by −1.
?? ddx x sec(x)
?? ddθtan (θ)θ2+1
3a 3b
Example. Find an equation of the line tangent to the curve
y=x tan (x)
at the point ( π4 , π4 ).Solution.
Point-slope form of tangent line
y− y0=m(x−x0)
Need slope
dydx
=tan ( x )+x sec2(x)
dydx|π
4
=tan( π4 )+ π4sec 2( π4 )
where sec (π2 )= 1
cos( π4 )=√2
dydx|π
4
=1+ π4⋅2=1+ π
2
equation for tangent line
y− π4=(1+ π
2 )(x− π4 ) ■
§2.5 The Chain Rule
Composition
Example. Consider F ( x )=√x2+1
Let f (u )=√u, g ( x )=x2+1
F is the composition of f and g
Function machines:
x→ [g ]→x2+1=u→ [ f ]→√u=√ x2+1
■
Notation for composition
F ( x )=f (g (x ) )
4a 4b
f is the outer function,
g is the inner function
Derivatives of Compositions
Let y=F ( x )=f (g ( x ) )
or y=f (u), u=g (x).
The Chain Rule
If g is differentiable at x and f is differentiable at u, then
F ' ( x )= f ' (u ) g' (x)
The derivative of a composition of two functions is the product of their derivatives
Write another way
F ' ( x )= f ' (g ( x ) )g '(x )
Leibniz notation
Let y=f (u) and u=g (x) and
y=F ( x )=f (g ( x ) )
then
F ' ( x )= f ' (u ) g' (x)
dydx
=dydu
dudx
Easy to remember: looks as though a term “du” cancels on the right hand side!
Example. Let F ( x )=√x2+1. Find F '(x ).
Write as a composition
u=g (x )=x2+1
y=f (u )=√u where √u=u12
apply the chain rule
dydx
=dydu
dudx
5a 5b
¿ 12u
−12 ⋅2x
¿ (x2+1 )−12 x ¿
x√x2+1■
Example. Let y=cos (3 x ). Find dydx .
Write as a composition
y=cos (u) where u=3x
apply the chain rule
dydx
=dydu
dudx
¿−sin (u ) ⋅3
¿−3sin(3 x) ■
?? Class Practice
Let y=( x2+1 )10. Find dydx .
Let y=√sin ( x ). Find dydx .
Let y=tan (cos ( x ) )
6a 6b
Plausibility Argument for the Chain Rule
Recall the limit definition of derivative
dydx=f ' ( x )=lim
h→ 0
f ( x+h )−f (x)h
Change notation
Let Δ x=h
Δ y=f ( x+h )−f (x )
then
dydx
= limΔx→0
Δ yΔ x
Statement of Chain Rule
Let y=F ( x )=f (g ( x ) ) or y=f (u) with u=g (x)
If g is differentiable at x, and f is differentiable at u, then
F ' ( x )= f ' (u ) g' (x) or dydx=dydu
dudx
Plausibility Argument
Let Δu be the change in u corresponding to Δ x.
Δu=g ( x+Δ x )−g(x )
The corresponding change in y is
Δ y=f (u+Δu )−f (u)
Then
dydx
= limΔx→0
Δ yΔ x
¿ limΔx→0
Δ yΔu
ΔuΔ x as long as Δu≠0
¿ limΔx→0
Δ yΔu
⋅ limΔx→0
ΔuΔ x product rule for limits
but Δu→0 as Δ x→0, so
dydx
= limΔu→0
Δ yΔu
⋅ limΔx→0
Δ uΔ x
¿ dydu ⋅dudx
The only problem is that we may have Δu=0. ■
7a 7b
Generalized Power Rule
Combine the power rule and the chain rule
Consider
F ( x )=g ( x )n
write as a composition
u=g (x)
y=f (u )=un
apply the chain rule
dydx
=dydu
dudx
¿nun−1 dudx
this is frequently written
ddx
g ( x )n=ng (x )n−1g '( x)
Example. Consider
F ( x )=(x3+4 x )10
this has the form given above, with
g ( x )=x3+4 x
n=10
then
F ' ( x )=10 (x3+10x )9 ddx
(x3+4 x)
¿10 (x3+10x )9(3 x2+4 ) ■
Example.
F ( x )= 1√sin (x )
=(sin ( x ) )−12
F ' ( x )=−12
(sin ( x ) )−32 cos ( x) ■
WARNING It is a common mistake to forget the factor g' (x)!
8a 8b
Compositions of Three Functions
Consider
F (t )= f (g (h (t ) ) )
write as a composition
y=f (u ) ,u=g(x ) , x=h (t)
chain rule
dydt
=dydu
dudx
dxdt
it looks as though ‘du’ and ‘dx’ factors cancel
Example
F (t )=sin 2(4 t )
write as a composition
y=u2, u=sin ( x ) , x=4 t
chain rule
dydt
=dydu
dudx
dxdt
¿2u⋅ cos ( x )⋅ 4
¿8ucos (x)
¿8sin ( x )cos (x )
¿8 sin (4 t ) cos (4 t ) ■
9a 9b
Example
Find the line tangent to the graph of y=sin2 (4 t ) at
t= π16 .
Solution. Point slope form of a tangent line
y− y0=m ( t−t0 )
t 0=π /16 , y0=sin2( π4 )=( 1√2 )2
=12
slope? m= dydx|t= π
16
¿ 8sin (4 t ) cos (4 t )|t= π16
¿8sin( π4 )cos ( π4 )
¿8( 1√2 )( 1√2 ) ¿4
gives the tangent line
y−12=4(t− π
16 )
■
?? Class practice differentiating!
Find dydx
1. y=( x4−3 x2+6 )3
2. y=2x √x2+1
10a 10b
3. y=(x+ 1x2 )√7
4. y=t
1−t 2
5. y=sin ¿
6. y=cos (tan (√1+ t2 ))
11a 11b
§2.6 Implicit Differentiation
circle of radius 1…0…x-, …0… y-, circle
x2+ y2=1
defines y implicitly as a function of x
solve for y: y2=1−x2
two possibilities
y=(1−x2 )12=√1−x2 [1a]
y=−(1−x2 )12=−√1−x2 [1b]
derivatives
dydx
=12
(1−x2 )−12 (−2x )= −x
√1−x2[2a] [2a]
dydx
=−12
(1−x2 )−12 (−2x )= x
√1−x2[2b]
Using Implicit Differentiation is easier in many cases!
x2+ y2=1 [3][3]
regard y as a function of x
y=f (x ) or y2=f ( x )2
recall the generalized power rule
ddx
f ( x )2=2 f ( x ) f ' (x)
12a 12b
write this as
ddx
y2=2 y y '
must remember y '=dydx !
differentiate equation [3] implicitly with respect to x
2 x+2 y dydx
=0
solve for dy /dx
y dydx
=−x
dydx
=−xy
from [1], two possibilities
dydx
= −x√1−x2
dydx
= x√1−x2
these match [2]!
Example. Find dydx if
x−1+ y−1=1. [1]
Solution. Regard y as a function of x
y=f (x )
Use the generalized power rule
ddx
f ( x )−1=−f ( x )−2 f ' (x)
ddx
y−1=− y−2 dydx
differentiate [1] implicitly with respect to x
−x−2− y−2 dydx
=0
solve for dydx
13a 13b
− y−2 dydx
= 1x2
dydx
=− y2
x2 ■
Example. Find the equation of the line tangent to the curve
2cos ( x )sin ( y )=1 [1]
at (x0 , y0 )=( π4 , π4 ).
Solution. Recall sin( π4 )=cos( π4 )= 1√2 .
think of y as a function of x
ddxsin ( y )=cos ( y ) dy
dx
differentiate [1] implicitly
−2sin ( x )sin ( y )+2cos ( x )cos ( y ) dydx
=0
solve for dydx
2cos ( x )cos ( y ) dydx
=2 sin ( x ) sin( y)
dydx
=tan ( x ) tan( y )
slope of curve at x= y=π /4
dydx|x= y=π
4
= tan( π4 ) tan( π4 )=1 ⋅1=1 point-slope form of tangent line
y− y0=m(x−x0)
y− π4=x−π
4
y=x ■
14a 14b
?? Class practice with implicit differentiation
1. Let x3+ x2 y+4 y2=c, where c is a constant. Find dydx .
Solution dydx=−(3 x2+2xy )
x2+8 y
2. Let sin ( x )+cos ( y )=sin ( x ) cos( y) . Find dydx .
15a 15b
Solution dydx= cos (x)sin ( x )−1
cos ( y )−1sin ( y)
Find an equation of the line tangent to the following curve at the given point.
2 (x2+ y2 )2=25(x2− y2) at the point (3,1)
Solution. Differentiate implicitly with respect to x
4 (x2+ y2 ) (2 x+2 y y ' )=25(2x−2 y y ')
Solve for y '
4 (x2+ y2 )2 y y '+50 y y'=−8 (x2+ y2 ) x+50 x
y '¿
y '=−8 x3−8 x y2+50 x8 y (x2+ y2 )+50 y
¿ 2 x (−4 x2−4 y2+25)
2 y (4 x2+4 y2+25)
evaluate at x=3 and y=1
y '=6(−36−4+25)2(36+4+25)
=6(−15)2(65)
=−4565
=−913
point slope form of tangent line
y− y0=m ( x−x0 )
y−1=−913
( x−3 )
in slope intercept form
y=−913
x+ 2713
+1
¿ −913
x+ 4013
show Maple image of curve and tangent line ■
16a 16b
§2.7 Related Rates
Example. A descending balloonist lets helium escape at a rate of 10 cubic feet / minute. Assume the balloon is spherical. How fast is the radius decreasing when the radius is 10 feet?sketch balloon, basket, escaping gas
Given V=¿ volume of balloon
dVdt
=−10 (feet )3
minute
r=¿radius of balloon ¿10 feet
Find drdt when r=10
Equation relating these quantities
V= 43π r3 [1]
differentiate with respect to time using the chain rule
generalized power rule: ddt g (t )n=ng (t )n−1
consider radius r a function of time t
ddt
r3=3 r2 drdt
Differentiate [1] with respect to time to get
dVdt
=43π (3 r2 drdt )=4 π r2 drdt
Solve for dr /dt
17a 17b
drdt
= 14 π r2
dVdt
drdt |r=10= 1
4 π 1001
(feet )2(−10 ) (feet )3
min
¿ −140 π
feetmin
≈−0.008 feetmin
≈−0.1 inchesmin
■
Problem Solving Strategy
1. Read problem carefully
2. Draw picture if possible
3. Assign notation to given information and unknown rate
4. Write down an equation relating the given information and the function whose rate is unknown
5. Differentiate with respect to time
6. Solve for the unknown rate
Example. The length of a rectangle is increasing at a rate of 7 cm/s and its width is increasing at a rate of 7 cm/s. When the length is 6cm and the width is 4 cm, how fast is the area of the rectangle increasing?
2. picture
3. Given information
L(t)=6cm, W ( t)=4cm
L' ( t )=7 cm/s, W ' ( t )=7 cm/s
Unknown rate is A' (t), where A=LW .
18a 18b
4. A (t )=L ( t )W (t)
5. A' ( t )=L' ( t )W (t )+ L ( t )W '( t)
6. ¿7cms ⋅4 cm+6cm ⋅7cms
¿28cm2
s+42cm
2
s=70cm
2
s ■
19a 19b
Example. One end of a 13 foot ladder is on the ground and the other end rests on a vertical wall. If the bottom end is drawn from the wall at 3 feet/second, how fast is the top of the ladder sliding down the wall when the bottom is 5 feet from the wall?
…x…x-, … y… y-, ladder betw. x & y, l=¿ 13 ft
given information
l=13 ft , x=5 ft , dxdt =3 ft./sec.
unknown rate
dydt
=¿?
Pythagorean theorem
x2+ y2=l2
differentiate with respect to time
2 x dxdt
+2 y dydt
=0
solve for y
dydt
=−xy
dxdt
to evaluate dy /dt we need y
y2=l2−x2=169−25=144
y=√144=12
then
dydt |x=5 ft=−5
12⋅3 feetsec
=−54feetsec
■
Example. Pat walks 5 feet/second towards a street light whose lamp is 20 feet above the ground. If Pat
20a 20b
is 6 feet tall, find how rapidly Pat’s shadow changes in length.
ground, lamp, Pat, x, s, y, h
given information
y=20 feet, h=¿6 feet, dxdt =−5feetsec
unknown rate
dsdt
=?
by similar triangles
yx+s
=hs
y s=h (x+s)
( y−h ) s=hx
s= hy−h
x
differentiate wrt time
dsdt
= hy−h
dxdt
¿ 620−6 (−5feetsec )
¿−157feetsec ■
Example. The beacon of a lighthouse 1 mile from the shore makes 5 rotations per minute. Assuming that
21a 21b
the shoreline is straight, calculate the speed at which the spotlight sweeps along the shoreline as it lights up sand 2 miles from the lighthouse.
0…x…x-, 0… y… y-, lighthouse at (0 , y ), beam to (x ,0), length L, angle θ
given information
y=1 mile L=¿ 2 miles
dθdt
=5 revolutionsminute ¿5 revolutionsminute
2π radiansrevolution
¿10π radiansminute
unknown rate dxdt =?
from the definition of tangent
tan (θ )= xy
differentiate wrt time
sec2 (θ ) dθdt
= 1ydxdt using that y is a constant
dxdt
= y sec2 (θ ) dθdt
we need sec2 (θ )
cos (θ )= yL sec (θ )= L
y=2
then
dxdt
=(1mile) ⋅4 ⋅10π radiansminute
¿40 πmilesminute
¿40 πmilesminute60minuteshour
¿ (40 ) (60 )πmileshour
≈7539mileshour ■ STOP
22a 22b
§2.8 Linear Approximations and Differentials
Linear or Tangent Line Approximation
0…a…x-, 0… y-, y=f (x ), P (a , f (a ) )
Point-slope form of the tangent line
y= y0=m(x−x0)
Point (x0 , y0 )=(a , f (a ) )
Slope m=f ' (a )
Then
y−f (a )=f ' (a ) ( x−a ) or
y=L ( x )=f (a )+ f ' (a )(x−a)
L(x) approximates f (x) near x=a
Example. Find the linear approximation to f ( x )=√1+x near x=0.
f ( x )= (1+x )12
f ' ( x )=12(1+x )
−12
f ' (0 )=12
L ( x)=f (0 )+ f ' (0 ) ( x−0 )
¿1+ 12 x
this straight line approximates √1+x near x=0.
−1…0…1…x-, 0…1… y-, f (x), L(x)
■
23a 23b
Example (continued). Use linear approximation to estimate √1.1.
√1.1=√1+0.1=f (0.1)
L (0.1 )=1+ 12
(0.1 )=1.05
Exact value √1.1=10.488…■
Example (continued). For what values of x is the linear approximation
√1+x≈1+ 12 x
accurate to within 0.05? This means
|L ( x )−f ( x )|<0.05
or
L ( x)−0.05< f ( x )<L ( x )+0.05
show Maple output
From the plot, the linear approximation is accurate to within 0.05 for
−0.54<x<0.74■
Realistic Example. Estimate the amount of paint required to paint a spherical tank of radius 20 feet with a coat 0.01 inches thick.
Exact calculation: volume of sphere
V (r )=43π r3
volume between two concentric spheres of radii r and a
ΔV=V (r )−V (a )=43π r3−4
3π a3
a=20 feet=240 inches
r=240.01 inches
ΔV ≈7238.53cubic inches ≈31gallons
1 gallon ¿ 231 cubic inches
24a 24b
Approximate calculation:
Linear approximation, usual notation
f ( x )≈ f (a )+ f ' (a)(x−a)
Use V (r ) instead of f (x)
V (r )≈V (a )+V ' (a )(x−a)
ΔV=V (r )−V (a )≈V ' (a )(r−a)
V ' (r )= ddr43π r3 ¿ 43 π (3 r 2) ¿4 π r2
and
V ' (a )=4 π a2
then
ΔV=4 π a2 (r−a )
¿ surface area of sphere × thickness of paint
¿4 π (240 inches )2(0.01 inches)
≈7237.23 cubic inches
the error is only 0.3 cubic inches! ■
STOP Differentials
0…x-, 0… y-, f , P, L, x, x+Δ x
tangent line approximation
L ( x+ Δ x )=f ( x )+f ' ( x )Δ x
write
dx=Δ x
dy=L ( x+Δ x )− f (x )
¿ f ' ( x )dx
add dx=Δ x , Δ y ,dy=f ' (x )dx
Δ y=f ( x+Δ x )−f (x )
is approximated by
dy=f ' ( x)dx
Note that riserun=dydx
=f '(x ). This is clever notation.
25a 25b
Example. Estimate the amount of paint required to paint a spherical tank of radius 20 feet with a coat of paint 0.01 inches thick.
Volume of sphere
V (r )=43π r3
approximate amount of paint
dV=V ' (r )dr=4 π r2dr
if dr=0.01 inches, then
dV=4 π (240 inches )2(0.01 inches)
the same result as above. This is a shorthand for linear approximation. ■
Newton’s Method – An Application of Linear Approximation
0…r…x-, 0… y-, f
A root of f is a number r such that f (r )=0.
How to find roots geometrically?
guess x1add
let L1 be the line tangent to f at x1add
follow L1 to the x axis, get an intersection at x2add
repeat this process
if the initial guess was good, the sequence x1 , x2 ,… rapidly converges to r
How to find roots algebraically?
Make an initial guess x1.
Find the line tangent to f at x=x1.
L1 ( x )= f (x1 )+f '( x1)(x−x1) by the linear or tangent line approximation!
L1 intersects the x axis at x2. Find x2.
L1 (x2 )=0=f (x1 )+ f ' (x1)(x2−x1)
−f (x1 )=f '( x1)(x2−x1)
26a 26b
−f ( x1)f ' (x1)
=x2−x1
x2=x1−f (x1)f ' (x1)
Find the line tangent to f at x=x2.
L2 (x )= f (x2 )+f ' (x2)(x−x2)
L2 intersects the x axis at x3. Find x3.
L2 ( x3 )=0=f (x2 )+f ' (x2)(x3−x2)
−f (x2 )=f '( x2)(x3−x2)
−f ( x2)f ' ( x2)
=x3−x2
x3=x2−f (x2)f '( x2)
repeat this procedure to obtain a general formula for n=1 ,2 ,…
xn+1=xn−f (xn)f ' (xn)
If the initial guess x1 was good, the iterates x2 , x3 ,… will quickly approach r.
Example. f ( x )=x2−2
−2…−1…0…1…2…x-, −2…0…2…y-, f ,r=√2
f ' ( x )=2 x
from the boxed formula above
xn+1=xn−xn2−22 xn
guess x1=2
x2=2−4−24
=2−12=32
x3=32−
94−2
3=32−
143
=32− 112
=18−112
=1712
=1.4167…
Exact value of positive root
r=√2=1.4142…
Transparency – How Newton’s Method can go wrong!
27a 27b