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. . . . . .
Section3.5InverseTrigonometric
FunctionsV63.0121.006/016, CalculusI
March11, 2010
Announcements
I ExamsreturnedinrecitationI ThereisWebAssigndueTuesdayMarch23andwrittenHWdueThursdayMarch25
I nextquizisFridayApril2
. . . . . .
Announcements
I ExamsreturnedinrecitationI ThereisWebAssigndueTuesdayMarch23andwrittenHWdueThursdayMarch25
I nextquizisFridayApril2
. . . . . .
Whatisaninversefunction?
DefinitionLet f beafunctionwithdomain D andrange E. The inverse of f isthefunction f−1 definedby:
f−1(b) = a,
where a ischosensothat f(a) = b.
Sof−1(f(x)) = x, f(f−1(x)) = x
. . . . . .
Whatisaninversefunction?
DefinitionLet f beafunctionwithdomain D andrange E. The inverse of f isthefunction f−1 definedby:
f−1(b) = a,
where a ischosensothat f(a) = b.
Sof−1(f(x)) = x, f(f−1(x)) = x
. . . . . .
Whatfunctionsareinvertible?
Inorderfor f−1 tobeafunction, theremustbeonlyone a in Dcorrespondingtoeach b in E.
I Suchafunctioniscalled one-to-oneI Thegraphofsuchafunctionpassesthe horizontallinetest:anyhorizontallineintersectsthegraphinexactlyonepointifatall.
I If f iscontinuous, then f−1 iscontinuous.
. . . . . .
Outline
InverseTrigonometricFunctions
DerivativesofInverseTrigonometricFunctionsArcsineArccosineArctangentArcsecant
Applications
. . . . . .
arcsin
Arcsinistheinverseofthesinefunctionafterrestrictionto[−π/2, π/2].
. .x
.y
.sin.
.−π
2
.
.π
2
.y = x
.
. .arcsin
I Thedomainof arcsin is [−1, 1]I Therangeof arcsin is
[−π
2,π
2
]
. . . . . .
arcsin
Arcsinistheinverseofthesinefunctionafterrestrictionto[−π/2, π/2].
. .x
.y
.sin.
.
.
.−π
2
.
.π
2
.y = x
.
. .arcsin
I Thedomainof arcsin is [−1, 1]I Therangeof arcsin is
[−π
2,π
2
]
. . . . . .
arcsin
Arcsinistheinverseofthesinefunctionafterrestrictionto[−π/2, π/2].
. .x
.y
.sin.
.
.
.−π
2
.
.π
2
.y = x
.
. .arcsin
I Thedomainof arcsin is [−1, 1]I Therangeof arcsin is
[−π
2,π
2
]
. . . . . .
arcsin
Arcsinistheinverseofthesinefunctionafterrestrictionto[−π/2, π/2].
. .x
.y
.sin.
.
.
.−π
2
.
.π
2
.y = x
.
. .arcsin
I Thedomainof arcsin is [−1, 1]I Therangeof arcsin is
[−π
2,π
2
]
. . . . . .
arccos
Arccosistheinverseofthecosinefunctionafterrestrictionto[0, π]
. .x
.y
.cos..0
..π
.y = x
.
. .arccos
I Thedomainof arccos is [−1,1]I Therangeof arccos is [0, π]
. . . . . .
arccos
Arccosistheinverseofthecosinefunctionafterrestrictionto[0, π]
. .x
.y
.cos.
.
..0
..π
.y = x
.
. .arccos
I Thedomainof arccos is [−1,1]I Therangeof arccos is [0, π]
. . . . . .
arccos
Arccosistheinverseofthecosinefunctionafterrestrictionto[0, π]
. .x
.y
.cos.
.
..0
..π
.y = x
.
. .arccos
I Thedomainof arccos is [−1,1]I Therangeof arccos is [0, π]
. . . . . .
arccos
Arccosistheinverseofthecosinefunctionafterrestrictionto[0, π]
. .x
.y
.cos.
.
..0
..π
.y = x
.
. .arccos
I Thedomainof arccos is [−1,1]I Therangeof arccos is [0, π]
. . . . . .
arctanArctanistheinverseofthetangentfunctionafterrestrictionto[−π/2, π/2].
. .x
.y
.tan
.−3π2
.−π
2.π
2 .3π2
.y = x
.arctan
.−π
2
.π
2
I Thedomainof arctan is (−∞,∞)
I Therangeof arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, limx→−∞
arctan x = −π
2
. . . . . .
arctanArctanistheinverseofthetangentfunctionafterrestrictionto[−π/2, π/2].
. .x
.y
.tan
.−3π2
.−π
2.π
2 .3π2
.y = x
.arctan
.−π
2
.π
2
I Thedomainof arctan is (−∞,∞)
I Therangeof arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, limx→−∞
arctan x = −π
2
. . . . . .
arctanArctanistheinverseofthetangentfunctionafterrestrictionto[−π/2, π/2].
. .x
.y
.tan
.−3π2
.−π
2.π
2 .3π2
.y = x
.arctan
.−π
2
.π
2
I Thedomainof arctan is (−∞,∞)
I Therangeof arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, limx→−∞
arctan x = −π
2
. . . . . .
arctanArctanistheinverseofthetangentfunctionafterrestrictionto[−π/2, π/2].
. .x
.y
.tan
.−3π2
.−π
2.π
2 .3π2
.y = x
.arctan
.−π
2
.π
2
I Thedomainof arctan is (−∞,∞)
I Therangeof arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, limx→−∞
arctan x = −π
2
. . . . . .
arcsecArcsecantistheinverseofsecantafterrestrictionto[0, π/2) ∪ (π, 3π/2].
. .x
.y
.sec
.−3π2
.−π
2.π
2 .3π2
.y = x
.
.
.π
2
.3π2
I Thedomainof arcsec is (−∞,−1] ∪ [1,∞)
I Therangeof arcsec is[0,
π
2
)∪(π2, π]
I limx→∞
arcsec x =π
2, limx→−∞
arcsec x =3π2
. . . . . .
arcsecArcsecantistheinverseofsecantafterrestrictionto[0, π/2) ∪ (π, 3π/2].
. .x
.y
.sec
.−3π2
.−π
2.π
2 .3π2
.
.
.y = x
.
.
.π
2
.3π2
I Thedomainof arcsec is (−∞,−1] ∪ [1,∞)
I Therangeof arcsec is[0,
π
2
)∪(π2, π]
I limx→∞
arcsec x =π
2, limx→−∞
arcsec x =3π2
. . . . . .
arcsecArcsecantistheinverseofsecantafterrestrictionto[0, π/2) ∪ (π, 3π/2].
. .x
.y
.sec
.−3π2
.−π
2.π
2 .3π2
.
.
.y = x
.
.
.π
2
.3π2
I Thedomainof arcsec is (−∞,−1] ∪ [1,∞)
I Therangeof arcsec is[0,
π
2
)∪(π2, π]
I limx→∞
arcsec x =π
2, limx→−∞
arcsec x =3π2
. . . . . .
arcsecArcsecantistheinverseofsecantafterrestrictionto[0, π/2) ∪ (π, 3π/2].
. .x
.y
.sec
.−3π2
.−π
2.π
2 .3π2
.
.
.y = x
.
.
.π
2
.3π2
I Thedomainof arcsec is (−∞,−1] ∪ [1,∞)
I Therangeof arcsec is[0,
π
2
)∪(π2, π]
I limx→∞
arcsec x =π
2, limx→−∞
arcsec x =3π2
. . . . . .
ValuesofTrigonometricFunctions
x 0π
6π
4π
3π
2
sin x 0 12
√22
√32
1
cos x 1√32
√22
12
0
tan x 01√3
1√3 undef
cot x undef√3 1
1√3
0
sec x 12√3
2√2
2 undef
csc x undef 22√2
2√3
1
. . . . . .
Check: Valuesofinversetrigonometricfunctions
ExampleFind
I arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6I −π
4
I3π4
. . . . . .
Check: Valuesofinversetrigonometricfunctions
ExampleFind
I arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6
I −π
4
I3π4
. . . . . .
Whatis arctan(−1)?
. .
.
..3π/4
..−π/4
.sin(3π/4) =
√22
.cos(3π/4) = −√22.sin(π/4) = −
√22
.cos(π/4) =
√22
I Yes, tan(3π4
)= −1
I But, therangeof arctan
is(−π
2,π
2
)I Anotheranglewhosetangentis −1 is −π
4, and
thisisintherightrange.
I So arctan(−1) = −π
4
. . . . . .
Whatis arctan(−1)?
. .
.
..3π/4
..−π/4
.sin(3π/4) =
√22
.cos(3π/4) = −√22
.sin(π/4) = −√22
.cos(π/4) =
√22
I Yes, tan(3π4
)= −1
I But, therangeof arctan
is(−π
2,π
2
)I Anotheranglewhosetangentis −1 is −π
4, and
thisisintherightrange.
I So arctan(−1) = −π
4
. . . . . .
Whatis arctan(−1)?
. .
.
..3π/4
..−π/4
.sin(3π/4) =
√22
.cos(3π/4) = −√22
.sin(π/4) = −√22
.cos(π/4) =
√22
I Yes, tan(3π4
)= −1
I But, therangeof arctan
is(−π
2,π
2
)
I Anotheranglewhosetangentis −1 is −π
4, and
thisisintherightrange.
I So arctan(−1) = −π
4
. . . . . .
Whatis arctan(−1)?
. .
.
..3π/4
..−π/4
.sin(3π/4) =
√22
.cos(3π/4) = −√22
.sin(π/4) = −√22
.cos(π/4) =
√22
I Yes, tan(3π4
)= −1
I But, therangeof arctan
is(−π
2,π
2
)I Anotheranglewhosetangentis −1 is −π
4, and
thisisintherightrange.
I So arctan(−1) = −π
4
. . . . . .
Whatis arctan(−1)?
. .
.
..3π/4
..−π/4
.sin(3π/4) =
√22
.cos(3π/4) = −√22
.sin(π/4) = −√22
.cos(π/4) =
√22
I Yes, tan(3π4
)= −1
I But, therangeof arctan
is(−π
2,π
2
)I Anotheranglewhosetangentis −1 is −π
4, and
thisisintherightrange.
I So arctan(−1) = −π
4
. . . . . .
Check: Valuesofinversetrigonometricfunctions
ExampleFind
I arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6I −π
4
I3π4
. . . . . .
Check: Valuesofinversetrigonometricfunctions
ExampleFind
I arcsin(1/2)I arctan(−1)
I arccos
(−√22
)
Solution
Iπ
6I −π
4
I3π4
. . . . . .
Caution: Notationalambiguity
..sin2 x = (sin x)2 .sin−1 x = (sin x)−1
I sinn x meansthe nthpowerof sin x, exceptwhen n = −1!I Thebookuses sin−1 x fortheinverseof sin x, andneverfor
(sin x)−1.
I I use csc x for1
sin xand arcsin x fortheinverseof sin x.
. . . . . .
Outline
InverseTrigonometricFunctions
DerivativesofInverseTrigonometricFunctionsArcsineArccosineArctangentArcsecant
Applications
. . . . . .
Theorem(TheInverseFunctionTheorem)Let f bedifferentiableat a, and f′(a) ̸= 0. Then f−1 isdefinedinanopenintervalcontaining b = f(a), and
(f−1)′(b) =1
f′(f−1(b))
“Proof”.If y = f−1(x), then
f(y) = x,
Sobyimplicitdifferentiation
f′(y)dydx
= 1 =⇒ dydx
=1
f′(y)=
1
f′(f−1(x))
. . . . . .
Theorem(TheInverseFunctionTheorem)Let f bedifferentiableat a, and f′(a) ̸= 0. Then f−1 isdefinedinanopenintervalcontaining b = f(a), and
(f−1)′(b) =1
f′(f−1(b))
“Proof”.If y = f−1(x), then
f(y) = x,
Sobyimplicitdifferentiation
f′(y)dydx
= 1 =⇒ dydx
=1
f′(y)=
1
f′(f−1(x))
. . . . . .
Thederivativeofarcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
Tosimplify, lookatarighttriangle:
cos(arcsin x) =√1− x2
So
ddx
arcsin(x) =1√
1− x2 .
.1 .x
..y = arcsin x
.√1− x2
. . . . . .
Thederivativeofarcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
Tosimplify, lookatarighttriangle:
cos(arcsin x) =√1− x2
So
ddx
arcsin(x) =1√
1− x2
.
.1 .x
..y = arcsin x
.√1− x2
. . . . . .
Thederivativeofarcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
Tosimplify, lookatarighttriangle:
cos(arcsin x) =√1− x2
So
ddx
arcsin(x) =1√
1− x2
.
.1 .x
..y = arcsin x
.√1− x2
. . . . . .
Thederivativeofarcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
Tosimplify, lookatarighttriangle:
cos(arcsin x) =√1− x2
So
ddx
arcsin(x) =1√
1− x2
.
.1 .x
..y = arcsin x
.√1− x2
. . . . . .
Thederivativeofarcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
Tosimplify, lookatarighttriangle:
cos(arcsin x) =√1− x2
So
ddx
arcsin(x) =1√
1− x2
.
.1 .x
..y = arcsin x
.√1− x2
. . . . . .
Thederivativeofarcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
Tosimplify, lookatarighttriangle:
cos(arcsin x) =√1− x2
So
ddx
arcsin(x) =1√
1− x2
.
.1 .x
..y = arcsin x
.√1− x2
. . . . . .
Thederivativeofarcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
Tosimplify, lookatarighttriangle:
cos(arcsin x) =√1− x2
So
ddx
arcsin(x) =1√
1− x2 .
.1 .x
..y = arcsin x
.√1− x2
. . . . . .
Graphingarcsinanditsderivative
I Thedomainof f is[−1, 1], butthedomainof f′ is (−1, 1)
I limx→1−
f′(x) = +∞
I limx→−1+
f′(x) = +∞ ..|.−1
.|.1
.
. .arcsin
.1√
1− x2
. . . . . .
Thederivativeofarccos
Let y = arccos x, so x = cos y. Then
− sin ydydx
= 1 =⇒ dydx
=1
− sin y=
1− sin(arccos x)
Tosimplify, lookatarighttriangle:
sin(arccos x) =√1− x2
So
ddx
arccos(x) = − 1√1− x2 .
.1.√1− x2
.x..y = arccos x
. . . . . .
Thederivativeofarccos
Let y = arccos x, so x = cos y. Then
− sin ydydx
= 1 =⇒ dydx
=1
− sin y=
1− sin(arccos x)
Tosimplify, lookatarighttriangle:
sin(arccos x) =√1− x2
So
ddx
arccos(x) = − 1√1− x2 .
.1.√1− x2
.x..y = arccos x
. . . . . .
Graphingarcsinandarccos
..|.−1
.|.1
.
. .arcsin
.
. .arccos
Note
cos θ = sin(π2− θ)
=⇒ arccos x =π
2− arcsin x
Soit’snotasurprisethattheirderivativesareopposites.
. . . . . .
Graphingarcsinandarccos
..|.−1
.|.1
.
. .arcsin
.
. .arccosNote
cos θ = sin(π2− θ)
=⇒ arccos x =π
2− arcsin x
Soit’snotasurprisethattheirderivativesareopposites.
. . . . . .
Thederivativeofarctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
Tosimplify, lookatarighttriangle:
cos(arctan x) =1√
1+ x2
So
ddx
arctan(x) =1
1+ x2 .
.x
.1..y = arctan x
.√1+ x2
. . . . . .
Thederivativeofarctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
Tosimplify, lookatarighttriangle:
cos(arctan x) =1√
1+ x2
So
ddx
arctan(x) =1
1+ x2
.
.x
.1..y = arctan x
.√1+ x2
. . . . . .
Thederivativeofarctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
Tosimplify, lookatarighttriangle:
cos(arctan x) =1√
1+ x2
So
ddx
arctan(x) =1
1+ x2
.
.x
.1
..y = arctan x
.√1+ x2
. . . . . .
Thederivativeofarctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
Tosimplify, lookatarighttriangle:
cos(arctan x) =1√
1+ x2
So
ddx
arctan(x) =1
1+ x2
.
.x
.1..y = arctan x
.√1+ x2
. . . . . .
Thederivativeofarctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
Tosimplify, lookatarighttriangle:
cos(arctan x) =1√
1+ x2
So
ddx
arctan(x) =1
1+ x2
.
.x
.1..y = arctan x
.√1+ x2
. . . . . .
Thederivativeofarctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
Tosimplify, lookatarighttriangle:
cos(arctan x) =1√
1+ x2
So
ddx
arctan(x) =1
1+ x2
.
.x
.1..y = arctan x
.√1+ x2
. . . . . .
Thederivativeofarctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
Tosimplify, lookatarighttriangle:
cos(arctan x) =1√
1+ x2
So
ddx
arctan(x) =1
1+ x2 .
.x
.1..y = arctan x
.√1+ x2
. . . . . .
Graphingarctananditsderivative
. .x
.y
.arctan
.1
1+ x2
.π/2
.−π/2
I Thedomainof f and f′ areboth (−∞,∞)
I Becauseofthehorizontalasymptotes, limx→±∞
f′(x) = 0
. . . . . .
ExampleLet f(x) = arctan
√x. Find f′(x).
Solution
ddx
arctan√x =
1
1+(√
x)2 d
dx√x =
11+ x
· 12√x
=1
2√x+ 2x
√x
. . . . . .
ExampleLet f(x) = arctan
√x. Find f′(x).
Solution
ddx
arctan√x =
1
1+(√
x)2 d
dx√x =
11+ x
· 12√x
=1
2√x+ 2x
√x
. . . . . .
Thederivativeofarcsec
Trythisfirst.
Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
Tosimplify, lookatarighttriangle:
tan(arcsec x) =
√x2 − 11
So
ddx
arcsec(x) =1
x√x2 − 1 .
.x
.1..y = arcsec x
.√
x2 − 1
. . . . . .
Thederivativeofarcsec
Trythisfirst. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
Tosimplify, lookatarighttriangle:
tan(arcsec x) =
√x2 − 11
So
ddx
arcsec(x) =1
x√x2 − 1 .
.x
.1..y = arcsec x
.√
x2 − 1
. . . . . .
Thederivativeofarcsec
Trythisfirst. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
Tosimplify, lookatarighttriangle:
tan(arcsec x) =
√x2 − 11
So
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1..y = arcsec x
.√
x2 − 1
. . . . . .
Thederivativeofarcsec
Trythisfirst. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
Tosimplify, lookatarighttriangle:
tan(arcsec x) =
√x2 − 11
So
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1..y = arcsec x
.√
x2 − 1
. . . . . .
Thederivativeofarcsec
Trythisfirst. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
Tosimplify, lookatarighttriangle:
tan(arcsec x) =
√x2 − 11
So
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1
..y = arcsec x
.√
x2 − 1
. . . . . .
Thederivativeofarcsec
Trythisfirst. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
Tosimplify, lookatarighttriangle:
tan(arcsec x) =
√x2 − 11
So
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1..y = arcsec x
.√
x2 − 1
. . . . . .
Thederivativeofarcsec
Trythisfirst. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
Tosimplify, lookatarighttriangle:
tan(arcsec x) =
√x2 − 11
So
ddx
arcsec(x) =1
x√x2 − 1
.
.x
.1..y = arcsec x
.√
x2 − 1
. . . . . .
Thederivativeofarcsec
Trythisfirst. Let y = arcsec x, so x = sec y. Then
sec y tan ydydx
= 1 =⇒ dydx
=1
sec y tan y=
1x tan(arcsec(x))
Tosimplify, lookatarighttriangle:
tan(arcsec x) =
√x2 − 11
So
ddx
arcsec(x) =1
x√x2 − 1 .
.x
.1..y = arcsec x
.√
x2 − 1
. . . . . .
AnotherExample
ExampleLet f(x) = earcsec x. Find f′(x).
Solution
f′(x) = earcsec x · 1
x√x2 − 1
. . . . . .
AnotherExample
ExampleLet f(x) = earcsec x. Find f′(x).
Solution
f′(x) = earcsec x · 1
x√x2 − 1
. . . . . .
Outline
InverseTrigonometricFunctions
DerivativesofInverseTrigonometricFunctionsArcsineArccosineArctangentArcsecant
Applications
. . . . . .
Application
ExampleOneoftheguidingprinciplesofmostsportsisto“keepyoureyeontheball.” Inbaseball, abatterstands 2 ftawayfromhomeplateasapitchisthrownwithavelocityof 130 ft/sec (about90mph). Atwhatratedoesthebatter’sangleofgazeneedtochangetofollowtheballasitcrosseshomeplate?
. . . . . .
Let y(t) bethedistancefromtheballtohomeplate, and θ theanglethebatter’seyesmakewithhomeplatewhilefollowingtheball. Weknow y′ = −130 andwewant θ′ atthemomentthaty = 0.
Wehave θ = arctan(y/2).Thus
dθdt
=1
1+ (y/2)2· 12dydt
When y = 0 and y′ = −130,then
dθdt
∣∣∣∣y=0
=1
1+ 0·12(−130) = −65 rad/sec
Thehumaneyecanonlytrackat 3 rad/sec!
..2 ft
.y
.130 ft/sec
.
.θ
. . . . . .
Let y(t) bethedistancefromtheballtohomeplate, and θ theanglethebatter’seyesmakewithhomeplatewhilefollowingtheball. Weknow y′ = −130 andwewant θ′ atthemomentthaty = 0.
Wehave θ = arctan(y/2).Thus
dθdt
=1
1+ (y/2)2· 12dydt
When y = 0 and y′ = −130,then
dθdt
∣∣∣∣y=0
=1
1+ 0·12(−130) = −65 rad/sec
Thehumaneyecanonlytrackat 3 rad/sec!
..2 ft
.y
.130 ft/sec
.
.θ
. . . . . .
Let y(t) bethedistancefromtheballtohomeplate, and θ theanglethebatter’seyesmakewithhomeplatewhilefollowingtheball. Weknow y′ = −130 andwewant θ′ atthemomentthaty = 0.
Wehave θ = arctan(y/2).Thus
dθdt
=1
1+ (y/2)2· 12dydt
When y = 0 and y′ = −130,then
dθdt
∣∣∣∣y=0
=1
1+ 0·12(−130) = −65 rad/sec
Thehumaneyecanonlytrackat 3 rad/sec!
..2 ft
.y
.130 ft/sec
.
.θ
. . . . . .
Let y(t) bethedistancefromtheballtohomeplate, and θ theanglethebatter’seyesmakewithhomeplatewhilefollowingtheball. Weknow y′ = −130 andwewant θ′ atthemomentthaty = 0.
Wehave θ = arctan(y/2).Thus
dθdt
=1
1+ (y/2)2· 12dydt
When y = 0 and y′ = −130,then
dθdt
∣∣∣∣y=0
=1
1+ 0·12(−130) = −65 rad/sec
Thehumaneyecanonlytrackat 3 rad/sec!
..2 ft
.y
.130 ft/sec
.
.θ
. . . . . .
Recap
y y′
arcsin x1√
1− x2
arccos x − 1√1− x2
arctan x1
1+ x2
arccot x − 11+ x2
arcsec x1
x√x2 − 1
arccsc x − 1
x√x2 − 1
I Remarkablethatthederivativesofthesetranscendental functionsarealgebraic(orevenrational!)