[related rates][section 12]
(1.) xy = 6 and dx/dt = 5 given
Find dy/dt when x = 3. here is the problem
x(dy/dt) + y(dx/dt) = 0 take the derivative implicitly
3y = 6 replace x with 3
__ ___
3 3
divide each side by 3
y = 2 divide and cancel
3(dy/dt) + 2(5) = 0 make
substitutions
3(dy/dt) + 10 = 0 multiply
-10 -10
subtract 10 from each side
_______________________
3(dy/dt) = -10
subtract
_________ ____
3 3
divide each side by 3
dy/dt = -10/3 cancel
(2.) Given x/y = 2 and dx/dt = 4; Find dy/dt when
x = 2.
2/y = 2 replace x with 2
2 = 2y multiply each side by y, cancel
___
____
2
2 divide each side by
2
1 = y divide and cancel
y(dx/dt) - x(dy/dt)
______________________ = 0 take the derivative
y2 implicitly
1(4) - 2(dy/dt)
____________________ = 0 make substitutions
12
4 - 2(dy/dt) = 0 multiply and divide
-4 + 2(dy/dt) = 0 multiply thru by -1
+
4 + 4
add 4 to each side
_______________________
2(dy/dt) = 4 add
________ ___
2 2
divide each side by 2
dy/dt = 2 divide and cancel
(3.) A 10-foot ladder is leaning against
the side of a
house. As the foot of the ladder is pulled away
from
the house, the top of the ladder
slides down along the
side of the house. Suppose the foot of the ladder is
pulled away at a rate of 2
ft/sec. How fast is the top
of the ladder sliding down when the
foot is 8 ft from
the house?
[use a 6 - 8 - 10 pythagorean
triangle]
a2 + b2 = c2 use the pythagorean theorem
2a(da/dt) + 2b(db/dt) =
2c(dc/dt) take the derivative
implicitly
a(da/dt) + b(db/dt) =
c(dc/dt) divide thru by 2, cancel
6(da/dt) + 8(2) = 10(0) make substitutions
6(da/dt) + 16 = 0 multiply
- 16
-16 subtract 16 from each side
_____________________________
6(da/dt) =
-16 subtract
________ _____
6 6 divide each side by 6
da/dt = -8/3 reduce and cancel
(4.) Two roads intersect at right
angles. A car travelling
80 km/hr reaches the intersection
half an hour before a
bus that is travelling on the other
road at 60 km/hr. How
fast is the distance between the car
and the bus increasing
1 hour after the bus reaches the
intersection?
a2 + b2 = c2 use the pythagorean theorem
1202 + 602 = c2 replace a with 120 and b with 60
14,400 + 3600 = c2 square
18,000 = c2 add like terms
(3600)(5) = c2 factor like this
_
c = 60√5 take
the square root of each side
2a(da/dt) + 2b(db/dt) = 2c(dc/dt)
[take the derivative implicitly]
a(da/dt) + b(db/dt) =
c(dc/dt) divide thru by 2, cancel
_
(120)(80) + (60)(60) = (60√5)(dc/dt)
make substitutions
_
7200 + 3600 = (60√5)(dc/dt)
multiply
_
10,800 = (60√5)(dc/dt)
add like terms
________ _____________
_ _
60√5
60√5 divide each side by this
_
180/√5
= dc/dt reduce and cancel
_ _
dc/dt = 180√5/5
multiply top and bottom by √5
_
dc/dt = 36√5
divide
(5.) A sailor standing on the edge of a
dock 15 ft above the
lake surface is pulling in his boat
by means of a line
attached to the boat's bow. He pulls in the line at
a rate of 5 ft/min. How fast is the boat approaching the
foot of the dock when the boat is 20
ft away?
Use a 15 - 20 - 25 right triangle.
a2 + b2 = c2 use the pythagorean theorem
2a(da/dt) + 2b(db/dt) =
2c(dc/dt) take the derivative
implicitly
a(da/dt) + b(db/dt) = c(dc/dt) divide thru by 2, cancel
15(0) + 20(db/dt) = 25(5) make substitutions
20(db/dt) = 125 multiply add
_________ _____
20 20 divide each side by 20
db/dt = 6.25 divide and cancel
(6.) An airplane at a height of 1000 m
is flying horizontally
at a velocity of 500 km/hr and
passes directly over an
observer. How fast is the plane receding from the
observer when it is 1500 m away,
(along a direct line
of sight), from the observer?
a2 + b2 = c2 use the pythagorean theorem
10002 + b2 =
15002 replace a with 1000
& c with 1500
1,000,000 + b2 =
2,250,000 square 1000 and 1500
-1,000,000 -1,000,000 subtract this fr ea side
__________________________________
b2 =
1,250,000 subtract
b2 = (5)(250,000) factor like this
_
b = 500√5 take
square roots
a2 + b2 = c2 use the pythagorean theorem again
2a(da/dt) + 2b(db/dt) = 2c(dc/dt) take the derivative
implicitly
a(da/dt) + b(db/dt) = c(dc/dt) divide thru by 2
1000(0) + (500√5)(500,000) = 1500(dc/dt)
[make
substitutions]
_
(500√5)(500,000) = 1500(dc/dt) multiply, add
_
(5√5)(5000)
= 15(dc/dt) divide thru by 100
_
25,000√5
= 15(dc/dt) multiply
__________ _________
15 15 divide each side by 15
_
(5000/3)√5
= dc/dt reduce and cancel
(7.) Sand is dropped onto a conical pile
at a rate of 15 m3/min.
Suppose the height of the pile is
always equal to its
diameter. How fast is the height increasing when the
pile is 7 m high?
V = (1/3)(pi)r2h use this formula
V = (1/3)(pi)(h/2)2h replace r with h/2
V = (h3/12)(pi) multiply
dV/dt = (h2/4)(pi)(dh/dt) take the derivative implicitly
15 = (72/4)(pi)(dh/dt) replace dV/dt with 15 & h with 7
15 = (49/4)(pi)(dh/dt) square 7
60 = 49(pi)(dh/dt) multiply each side by 4, cancel
___
______________
49pi 49pi divide each side by 49pi
60
_________ = dh/dt cancel
49pi
(8.) A rock is thrown into a pool of
water. A circular wave
leaves the point of impact and
travels so that its radius
increases at a rate of 25
cm/sec. How fast is the
circumference of the wave increasing
when the radius is
1 m?
C = 2∏r use the circumference formula
dC/dt = 2∏(dr/dt) take the
derivative implicitly
dC/dt = 2∏(25) make substitutions
dC/dt = 50∏
multiply
(9.) Bacteria grow in circular
colonies. The radius of one
colony is increasing at the rate of
4 mm/day. On
Wednesday, the radius of the colony
is 1 mm. How fast
is the area of the colony changing
one week later?
A =
∏r2 use the area
formula for circle
dA/dt = 2∏r(dr/dt) take the derivative implicitly
dA/dt = 2∏(7)(4) replace r with 7, replace dr/dt with 4
dA/dt = 56∏ multiply
(10.) A spherical mothball is dissolving
at a rate of 8pi
cm3/hr. How fast is the radius of the mothball de-
creasing when the radius is 3 cm.
V = (4/3)∏r3 use the volume formula
dV/dt = 4∏r2(dr/dt)
take the derivative implicitly
8∏ = 4∏(3)2(dr/dt) replace dV/dt with 8∏
replace r
with 3
8∏ = 36∏(dr/dt) multiply
____
__________
36∏ 36∏ divide each side by 36∏
2/9 = dr/dt reduce, cancel