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1.1 Homework Solutions 1. f (x)= x
2 +3x− 4 → f ′(x)= 2x+3
3. y = x−
23 → dy
dx = − 23x−
53
5.
v(r)= 4
3πr3 → v′(r)= 4πr2
7. y = x
2 + 4x+3x
= x3
2 + 4x1
2 +3x−1
2
dydx =
32x
12 + 2x
−12 − 3
2x−3
2 = 3 x2
+ 2x− 3
2 x3
9. z = Ay10 + Be
y = Ay−10 + Bey
′z = −10Ay−11 + Bey 11. If f (x) = 3x5 − 5x3 + 3 , find f '(x). ′f x( ) = 45x14 −15x2 You can see that when f is decreasing the graph of ′f is below the x - axis, i.e. ′f is negative - if f is increasing the graph of ′f is above the x - axis.
13. y = e x = ex12 ⇒ dy
dx = ex12 ⋅12 x
−12 = ex12
2x12= e x
2 x y = e x
2
15. Given the following table of values, find the indicated derivatives.
x f (x) ′f (x)2 1 78 5 −3
a. g x( ) = f x( )⎡⎣ ⎤⎦
3
′g x( ) = 3 f x( )⎡⎣ ⎤⎦2⋅ ′f x( )⇒ ′g 2( ) = 3 f 2( )⎡⎣ ⎤⎦
2⋅ ′f 2( ) = 3⋅12 ⋅7 = 21
b. h x( ) = f x3( ) ′h x( ) = ′f x3( )⋅3x2 ⇒ ′h 2( ) = ′f 23( )⋅3⋅22 =12 ′f 8( ) =12 ⋅−3= −36
17. If
f (x) = x3 + 2x( )37
, find f '(x) .
⇒ ′f x( ) = 37 x3 + 2x( )36 ⋅ 3x2 + 2( ) = 37 3x2 + 2( )⋅ x3 + 2x( )36
19. f x( ) = 4 − 49 x
2 ; find f ' 5( )
f x( ) = 4 − 49 x2 = 4 − 49 x
2⎛⎝⎜
⎞⎠⎟
12⇒ ′f x( ) = 12 4 − 49 x
2⎛⎝⎜
⎞⎠⎟
−12⋅ − 89 x⎛⎝⎜
⎞⎠⎟= −4x
9 4 − 49 x2
21. y = x3 − 2x7 ; find dy
dx
y = x3 − 2x7 = x3 − 2x( )17 ⇒ ′f x( ) = 17 x3 − 2x( )−67 ⋅ 3x2 − 2( ) = 3x2 − 2
7 x3 − 2x( )67
3
1.2 Homework Solutions 1. sin 4y x=
′y = 4cos4x 3. y = a3 + cos3 x
′y = 3cos2 x ⋅ −sin x( ) = −3cos2 xsin x 5. f (t) = 1+ tant3
′f t( ) = 13 1+ tant( )−23 ⋅sec2 t = sec2 t1+ tant( )23
7.
y = cos a3+x3( )
′y = −sin a3 + x3( )⋅3x2 = −3x2 sin a3 + x3( ) 9.
f (x) = cos ln x( )
′f x( ) = −sin ln x( )⋅1x = −sin ln x( )
x
11.
f (x) = log10 2+ sin x( ) ′f x( ) = 1
2+ sin x ⋅1ln10 ⋅cosx =
cosxln10 2+ sin x( )
13.
y = sin−1 ex( )
′y = 11− e2x
⋅ex = ex
1− e2x
15.
y = sin−1 2x +1( )
′y = 11− 2x+1( )2
⋅2 = 21− 4x2 + 4x+1( )
= 2 −4x2 − 4x( )−12 = −1
−x2 − x( )12
4
17. y = csc−1 x2 +1( ) ′y = −1
x2 +1( ) x2 +1( )2 −1⋅2x = −2x
x2 +1( ) x2 +1( )2 −1
5
1.3 Homework Solutions 1. Find the equation of the tangent line to f (x) = x5 − 5x +1 at x = −2 and use it to get an approximate value of f (-1.9).
f −2( ) = −21 ′f x( ) = 5x4 − 5⇒ ′f −2( ) = 75 y+ 21= 75 x+ 2( )y = −21+ 75 x+ 2( )⇒ y x=−1.9 = −13.5
f −1.9( ) ≈ −13.5 3. Find an equation of the line tangent to the curve y = x
4 + 2ex at the point
(0, 2) .
′y = 4x3 + 2ex ⇒ ′y x=0 = 2 Tangent Line: y− 2 = 2 x− 0( )
5. Find the equation of the tangent line to cosy x x= + at the point
0,1( ) .
′y =1− sin x⇒ ′y x=0 =1 Tangent Line: y−1=1 x− 0( )
7. Find the equation of the line tangent to y = 2
π x + cos 4x( ) when x = π
2.
′y = 2π − 4sin 4x( )⇒ ′yx=π 2
= 2π − 4sin 2π( ) = 2π
yx=π 2
= 2π ⋅π2 + cos 2π( ) = 2
6
9. Use Euler’s Method with 4 equal step sizes to find an approximation for
f 1.4( ) , given that
f x( ) = ln 2x −1( ) and
f 1( ) = 0 .
f 1.4( )≈ 0.635
11. Given the differential equation,
dydx
= y8
6− y( ) where y = f t( ) and
f 0( ) = 8 ,
use Euler’s Method with two steps of equal size to approximate f 1( ) .
7
13. Let y = f x( ) be the solution to the differential equation dy x y
dx= + with the
initial condition f 1( ) = 2 . What is the approximation for
f 2( ) if Euler’s method is
used, starting at x = 1 with a step size of 0.5? (A) 3 (B) 5 (C) 6 (D) 10 (E) 12
8
1.4 Homework Solutions 1. 3 cosy t t=
′y = t 3 ⋅ −sint( )+ cost ⋅3t2 = t2 3cost − t sint( )
3. y = tan x −1
sec x
′y =secx ⋅sec2 x− tan x−1( )⋅secx tan x
sec2 x =sec2 x− tan x−1( )tan x
secx = sec2 x− tan2 x+ tan x
secx Recall 1= sec2 x− tan2 x , so substitute into numerator:
′y = 1+ tan xsecx
5. y = xe−x2 ′y = x ⋅e−x2 ⋅ −2x( )+ e−x2 = e−x2 1− 2x2⎡⎣ ⎤⎦ 7. y = excosx ′y = excosx x ⋅−sin x+ cosx⎡⎣ ⎤⎦ = e
xcosx cosx− xsin x⎡⎣ ⎤⎦
9. y = xsin 1
x
′y = x ⋅cos 1x ⋅−1x2
+ sin 1x = − 1x cos1x + sin
1x
11. f (x) = x ln x
′f x( ) = x ⋅12 ln x( )−12 ⋅1x + ln x( )12 = 1+ 2ln x2 ln x( )12
9
13. Find the equation of the line tangent to y = x2e−x at the point 1, 1
e( ) .
′y = x2 ⋅e−x ⋅ −1( )+ e−x ⋅2x m = ′y x=1 = −1e +2e =
1e
y− 1e =1e x−1( )
15. h(t) = 1+ x2
1− x2
⎛
⎝⎜⎞
⎠⎟
17
, find h '(t)
′h t( ) =17 1+ t2
1− t2⎛
⎝⎜⎞
⎠⎟
16
⋅1− t2( )⋅2t − 1+ t2( )⋅−2t
1− t2( )2=17 1+ t2( )16 2t − 2t 3 + 2t + 2t 3⎡⎣ ⎤⎦
1− t2( )18
′h t( ) = 68t 1+ t22( )16
1− t 2( )18
17. f (x) = xsin 2x( ) + tan4 x7( )⎡
⎣⎤⎦
5, find f ' x( ) .
′f x( ) = 5 xsin 2x( )+ tan4 x7( )⎡⎣⎢
⎤⎦⎥4⋅ xcos2x ⋅2+ sin2x+ 4 tan3 x7( )⋅sec2 x7( )⋅7x6⎡⎣⎢
⎤⎦⎥
= 5 xsin 2x( )+ tan4 x7( )⎡⎣⎢
⎤⎦⎥4⋅ 2xcos2x+ sin2x+ 28x6 tan3 x7( )sec2 x7( )⎡⎣⎢
⎤⎦⎥
19. Find the equation of the line tangent to y = ex sin 4x( ) + 2 when x = 0
y x=0 = e0sin0 + 2 = 3⇒ 0,3( )
′y = exsin4x x ⋅cos4x ⋅4 + sin4x⎡⎣ ⎤⎦
′y x=0 = e0 0 ⋅cos0 ⋅4 + sin0⎡⎣ ⎤⎦ = 0
y− 3= 0 x− 3( )⇒ y = 3
20. Find the equation of the lines tangent and normal to y = xsin 1
x⎛
⎝⎜⎞
⎠⎟ when
21.
H(x) = 1+ x2( ) tan−1(x)
10
′H x( ) = 1+ x2( )⋅ 11+ x2 + tan
−1 x ⋅2x
23. If f (x) = ex − x2 arctan x , find
f ' x( ) .
′f x( ) = ex − x2 ⋅ 1
1+ x2 + arctan x ⋅2x⎡⎣⎢
⎤⎦⎥
= ex − x21+ x2 − 2xarctan x
25. y = sec−1 xx
′y =x ⋅ 1x x2 −1
− sec−1 x
x2=
1x2 −1
− sec−1 x
x2
11
1.5 Homework Solutions 1. f (x) = x5 + 6x2 − 7x
′f x( ) = 5x4 +12x− 7′′f x( ) = 20x3 +12
3. y = x3 +1( )23
′y = 23 x3 +1( )−13 ⋅3x2 = 2x2
x3 +1( )13
′′y =x3 +1( )13 ⋅4x− 2x2 ⋅13 x3 +1( )−23 ⋅3x2
x3 +1( )23=x3 +1( )−23 x3 +1( )⋅4x− 2x4⎡
⎣⎢⎤⎦⎥
x3 +1( )23
= 4x4 + 4x− 2x4
x3 +1( )4 3=2x x3 + 2( )x3 +1( )4 3
5. g(t) = t3e5t
′g t( ) = t 3 ⋅e5t ⋅5+ e5t ⋅3t2 = t2e5t 5t + 3⎡⎣ ⎤⎦
′′g t( ) = t2e5t ⋅5+ t2 5t + 3( )⋅e5t ⋅5+ e5t ⋅ 5t + 3( )⋅2t= te5t 5t + 5t 5t + 3( )+ 2 5t + 3( )⎡
⎣⎤⎦
= te5t 25t2 + 30t + 6⎡⎣ ⎤⎦
12
7. y = sin 3x
′y = 3sin2 xcosx
′′y = 3 sin2 x ⋅−sin x+ cosx ⋅2sin xcosx⎡⎣ ⎤⎦ = 3sin x 2cos2 x− sin2 x⎡⎣ ⎤⎦
9. d2 dx2
5x4 + 9x3 − 4x2 + x − 8⎡⎣ ⎤⎦
= ddx −20x3 + 27x2 − 8x+1⎡⎣ ⎤⎦
= −60x2 + 54x− 8
11. y = cosx2 , find ′′y
dydx = −sin x
2 2x( ) = −2xsin x2
d2ydx2
= −2x cosx2 2x( )⎡⎣ ⎤⎦ + sin x2 −2( ) = −2 2x2 cosx2 + sin x2⎡⎣ ⎤⎦
13. y = sec3x , find d2ydx2
dydx = sec3x tan3x 3( ) = 3sec3x tan3xd2ydx2
= 3sec3x sec2 3x 3( )( ) + 3tan3x 3sec3x tan3x( ) = 3sec3x 3sec2 3x+ tan2 3x( )
15. f x( ) = ln x2 + 3( ) , find
′′f x( )
f ′ x( ) = 2xx2 + 3
′′f x( ) =x2 + 3( ) 2( )− 2x( ) 2x( )
x2 + 3( )2=
2x2 + 6( )− 4x2( )x2 + 3( )2
=−2 x2 − 3( ) x2 + 3( )2
13
17. h x( ) = x2 + 5 , find ′′h x( )
h x( ) = x2 + 5 = x2 + 5( )12 h′ x( ) = x2 + 5 = 12 x2 + 5( )−12 2x( ) = x
x2 + 5( )12
′′h x( ) =x2 + 5( )12 1( )− x ⋅ x
x2 + 5( )12
x2 + 5( )1
=x2 + 5( )1 − x2
x2 + 5( )32
= 5
x2 + 5( )32
19. y = x2 − 3
x2 −10, find d
2ydx2
dydx =
x2 −10( ) 2x( )− x2 − 3( ) 2x( )x2 −10( )2
=2x( ) x2 −10( )− x2 − 3( )⎡
⎣⎢⎤⎦⎥
x2 −10( )2= −14xx2 −10( )2
d2ydx2 =
x2 −10( )2 −14( )− −14x( )2 x2 −10( )1 2x( )x2 −10( )4
=−14 x2 −10( ) x2 −10( )− 4x2( )⎡
⎣⎢⎤⎦⎥
x2 −10( )4
=−14 −3x2 −10( )
x2 −10( )3=
14 3x2 +10( )x2 −10( )3
21. y = x3 + x2 − 7x −15
y′ = 3x2 + 2x− 7y″ = 6x+ 2
14
23. y = −4xx2 + 4
dydx =
x2 + 4( ) −4( )− −4x( ) 2x( )x2 + 4( )2
=−4x2 −16( )− −8x2( )
x2 + 4( )2= 4x
2 −16x2 + 4( )2
d2ydx2
=x2 + 4( )2 8x( )− 4x2 −16( )2 x2 + 4( )1 2x( )
x2 + 4( )4
=x2 + 4( ) 8x( )− 4x2 −16( )2 2x( )
x2 + 4( )3
=8x3 + 32x( )− 16x3 − 64( )
x2 + 4( )3
=−8x3 + 96x( )x2 + 4( )3
=−8x x2 −12( )x2 + 4( )3
25. y = x 8− x2 = x 8− x2( )12 dydx = x ⋅
12 ⋅ 8− x
2( )−12 −2x( )+ 8− x2( )12 1( )
= −x2
8− x2( )12+ 8− x2( )12
=−x2 + 8− x2( )18− x2( )12
= 8− 2x2
8− x2( )12
15
d2ydx2
=
8− x2( )12 −4x( )− 8− 2x2( ) −x2
8− x2( )128− x2( )1
=8− x2( )1 −4x( )− −x2( ) 8− 2x2( )
8− x2( )32= −2x4 + 4x3 + 8x2 − 32x
8− x2( )32
27. y = xe−x
dydx = xe
−x −1( )+ e−x 1( ) = e−x x+1( )d2ydx2
= e−x 1( )+ x+1( )e−x −1( ) = e−x x− 2( )
29. y = x
x2 − 9
dydx =
x2 − 9( ) 1( )− x( ) 2x( )x2 − 9( )2
= −x2 − 9x2 − 9( )2
d2ydx2
=x2 − 9( )2 −2x( )− −x2 − 9( )2 x2 − 9( )1 2x( )
x2 − 9( )4
=x2 − 9( ) −2x( )− −x2 − 9( )2 2x( )
x2 − 9( )3
=−2x3 +18x( )− −4x3 − 36x( )
x2 − 9( )3
=2x3 + 54x( )x2 − 9( )3
=2x x2 + 27( )x2 − 9( )3
16
1.6 Homework Solutions 1. xy + 2x + 3x2 = 4 Implicit Explicit
x dydx + y ⋅1+ 2+ 6x = 0 xy = 4 − 2x− 3x2
dydx =
−6x− y− 2x y = 4x − 2− 3x = −4x
−2 − 3= − 4x2
− 3
dydx =
−4 − 3x2x2
dydx =
−6x− y− 2x =
−6x− 4x − 2− 3x
⎛⎝⎜
⎞⎠⎟− 2
x = −6− 4x2
+ 2x + 3−2x = −3− 4
x2= −4 − 3x2
x2
3. x + y = 4 Implicit Explicit 12 x
−12 + 12 y−12 dy
dx = 0 y = 4 − x⇒ y =16− 8 x + x
dydx =
− 12 x−12
12 y
−12= − y
12
x12
dydx = −4x
−12 +1= − 4x+1= −4 + x
x
dydx = − y
12
x12= − 4 − x
x= −4 + x
x
5. x
3 +10x2 y + 7 y2 = 60
3x2 +10 x2 dydx + y ⋅2x⎛⎝⎜
⎞⎠⎟+14y dydx = 0
3x2 +10x2 dydx + 20xy+14ydydx = 0
dydx =
−3x2 − 20xy10x2 +14y
17
7. 4cos x( )sin y( ) =1
4 cosx ⋅cosy dydx + sin y ⋅−sin x⎡
⎣⎢
⎤
⎦⎥ = 0
4cosxcosy dydx = 4sin xsin y
dydx = tan x tan y
9. tan x - y( ) = y
1+ x2
sec2 x− y( ) 1− dydx⎛⎝⎜
⎞⎠⎟=1+ x2( )− y ⋅2x1+ x2( )2
sec2 x− y( )− sec2 x− y( )dydx⎡
⎣⎢
⎤
⎦⎥ ⋅ 1+ x2( )2 = dydx + x2
dydx − 2xy
sec2 x− y( ) 1+ x2( )2 − sec2 x− y( ) 1+ x2( )2 dydx =dydx + x
2 dydx − 2xy
sec2 x− y( ) 1+ x2( )2 + 2xy = dydx 1+ x2 + sec2 x− y( ) 1+ x2( )2⎡⎣⎢
⎤⎦⎥
dydx =
sec2 x− y( ) 1+ x2( )2 + 2xy1+ x2 + sec2 x− y( ) 1+ x2( )2
11. 9x2 + 4y2 + 36x −8y + 4 = 0 at
0, − 2( )
18x+ 8y dydx + 36− 8dydx = 0
−16 dydx + 36− 8dydx = 0
dydx =
−36−24 =
32
y+ 2 = 32 x
18
13. Find the equation of the lines tangent and normal to
y − 4
π 2 x2 = 2eysin x + y3 − 3 through the point
π2
,0⎛
⎝⎜⎞
⎠⎟.
dydx −
8π 2 x = 2e
ysinx ycosx+ sin x dydx⎡
⎣⎢
⎤
⎦⎥+ 3y2
dydx
dydx −
8π 2
π2
⎛⎝⎜
⎞⎠⎟= 2e0 0+ sinπ2
dydx
⎡
⎣⎢
⎤
⎦⎥+ 0
dydx −
4π = 2 dydx⇒
dydx = −
4π
Tangent line: y− 0 = −4π x−π 2( ) , Normal line: y− 0 =π 4 x−π 2( )
15. Find the equation of the line tangent to x
2 + 3xy + y2 =11, through the point (1,2).
2x+ 3 x dydx + y⎡
⎣⎢
⎤
⎦⎥+ 2y
dydx = 0
2+ 3 dydx + 2
⎡
⎣⎢
⎤
⎦⎥+ 4
dydx = 0⇒ 2+ 3dydx + 6+ 4
dydx = 0⇒ 7 dydx = −8⇒
dydx = −
8 7
y− 2 = −8 7 x−1( )
17. Find d2ydx2
if 4x2 + 9y2 = 36
8x+18y dydx = 0⇒dydx =
−8x18y =
−4x9y
d2ydx2
=9y ⋅−4 − −4x( )⋅9 dydx
81y2 =−36y+ 36x ⋅ −4x9y
81y2 =−36y−144x
2
9y81y2
= −324y2 −144x2729y3 = −36y2 −16x2
81y2
19
18. y = 2x +1( )4 x3 − 3( )5
19.
z = y3 − 3( )e 2 y+1( )
lnz = ln y3 − 3( )+ 2y+11zdzdy =
3y2y3 − 3+ 2
dzdy = z
3y2y3 − 3+ 2
⎡
⎣⎢
⎤
⎦⎥
dzdy = y3 − 3( )e2y+1 3y2
y3 − 3+ 2⎡
⎣⎢
⎤
⎦⎥
20.
y = sin2 x tan4 x
x2 +5( )2
ln y = 2lnsin x+ 4ln tan x− 2ln x2 + 5( )1ydydx = 2 ⋅
1sin x ⋅cosx+ 4
1tan x ⋅sec
2 x− 4xx2 + 5
dydx = y 2cot x+
4sin xcosx −
4xx2 + 5
⎡⎣⎢
⎤⎦⎥
dydx =
sin2 x+ tan4 xx2 + 5( )2
2cot x+ 4sin xcosx −
4xx2 + 5
⎡⎣⎢
⎤⎦⎥
20
21.
g t( ) = t ln t( )
′g t( ) = t ⋅1t + lnt =1+ lnt 22.
y = lnx x( )
ln y = x ln ln x( )1ydydx = x ⋅
1ln x ⋅
1x + ln ln x( )⋅1
dydx = y
1ln x + ln ln x( )⎡
⎣⎢⎤⎦⎥
dydx = ln
x x 1ln x + ln ln x( )⎡
⎣⎢⎤⎦⎥
23.
p v( ) = vev
ln p = ev lnv1pdpdv = e
v ⋅1v + lnv ⋅ev
dpdv = p
evv + ev lnv⎡
⎣⎢
⎤
⎦⎥
dpdv = v
ev evv + ev lnv⎡
⎣⎢
⎤
⎦⎥
21
24. Use logarithmic differentiation to prove the product rule.
y = uv⇒ ln y = lnu − lnv
1ydydx =
1u ⋅dudx −
1v ⋅dvdx⇒
dydx = y
1u ⋅dudx −
1v ⋅dvdx
⎡
⎣⎢
⎤
⎦⎥
⇒ dydx =
v dudx −udvdx
v2
25. Find dtdu if tu = ut .
ddt t
u = ut( )⇒ utu−1 dudt = ut ⋅lnu ⋅1⇒ du
dt =ut lnuutu−1 ⇒ du
dt =ut−1 lnutu−1
26. For the function,
f x( ) = xln x , find an equation for a tangent line at x = e, and
use that to approximate the value for f (2.7). Find the percent difference between this and the actual value of f (2.7). ln y = ln x ⋅ln x = ln x( )21ydydx = 2ln x ⋅
1x⇒
dydx = y 2ln x ⋅
1x
⎡⎣⎢
⎤⎦⎥⇒ dy
dx = xlnx 2ln x ⋅1x
⎡⎣⎢
⎤⎦⎥
dydx x=e
= elne 2lne ⋅1e⎡⎣⎢
⎤⎦⎥= 2
Tangent Line: y− e= 2 x− e( ) f 2.7( ) ≈ e+ 2 2.7− e( ) = 2.682 f 2.7( ) = 2.7( )ln 2.7( ) = 2.682
% Difference = 0
22
1.7 Homework Solutions 1. If V is the volume of a cube with edge length, s, and the cube expands as
time passes, find dVdt in terms of
dsdt .
V =πs3 → dV
dt = 3π 2 dsdt
2. A particle is moving along the curve y = 1+ x3 . As it reaches the point (2,3), the y-coordinate is increasing at a rate of 4 cm/sec. How fast is the x-coordinate changing at that moment? dydt =
12 1+ x
3( )−12 ⋅ 3x2( )⋅dxdt ; dydt = 4
cm s
⇒ 4 = 12 1+ 23( )−12 ⋅3⋅22 ⋅dxdt
dxdt = 2
cm s
3. A plane flying horizontally at an altitude of 1 mile and a speed of 500 mph flies directly over a radar station. Find the rate at which the horizontal distance is increasing when it is 2 miles from the station. Find the rate at which the distance between the station and the plane is increasing when the plane is 2 miles from the station.
x2 +1= z2 ⇒ 2x dxdt = 2z
dzdt ; dxdt = 500
mih ; z = 2⇒ x = 3
⇒ 2 3 ⋅500 = 2 ⋅2 dzdt
Light
Plane
1 mi
z
x
23
⇒ dzdt = 250 3mih
4. If a snowball melts so that its surface area is decreasing at a rate of 1 cm2/min, find the rate at which the diameter is decreasing when it has a diameter of 10 cm.
S = 4πr2 ⇒ dSdt = 8πr
drdt ; dSdt = −1cm2
min ; D =10cm⇒ r = 5cm
⇒−1= 8π ⋅5 drdt ⇒drdt = −
140π
cmmin
D = 2r⇒ dDdt = 2 drdt ⇒
dDdt = − 1
20πcmmin
5. A street light is mounted at the top of a 15 foot tall pole. A 6 foot tall man walks away from the pole at a speed of 5 ft/sec in a straight line. How fast is the tip of his shadow moving away from the pole when the man is 40 feet from the pole?
15x+ y =
6y⇒ x+ y = 52 y⇒ x = 32 y ; dxdt = 5
fts ; x = 40 ft
⇒ dxdt =
32dydt ⇒ 5 = 32
dydt ⇒
dydt =
103ftmin
⇒d x+ y( )dt = dxdt +
dydt = 5+
103 = 253
ftmin
Dude
Light
6ft.15ft.
yx
24
6. Two cars start moving away from the same point. One travels south at 60 mph, and the other travels west at 25 mph. At what rate is the distance between the cars increasing two hours later?
x2 + y2 = z2 ; dydt = 60
mih ; dxdt = 25
mih
⇒ 2x dxdt + 2ydydt = 2z
dzdt
⇒ 2 ⋅50 ⋅25+ 2 ⋅120 ⋅60 = 2 ⋅130 ⋅dzdt
dzdt = 65
mih
7. The altitude of a triangle is increasing at a rate of 1 cm/min, while the area of the triangle is increasing at a rate of 2 cm2/min. At what rate is the base of the triangle increasing when the altitude is 10 cm and the area is 100 cm2?
A = 12bh ; dhdt =1
cmmin ; dAdt = 2
cm2
min ; h =10cm ; A =100cm2
⇒ dAdt =
12 b dhdt + h
dbdt
⎡
⎣⎢
⎤
⎦⎥⇒ 2 = 12 20 ⋅1+10 ⋅
dbdt
⎡
⎣⎢
⎤
⎦⎥⇒
dbdt = −1.6
cmmin
Car
Car
zy
x
h
b