Lesson 15: Inverse Trigonometric Functions

. . . . . .

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!)