Calc1131W5T4 - The chain rule

Friday, 30 March 2018

12:28 AM

Question One
Untitled picture.png Machine generated alternative text:
dQ
Informally the sensitivity of some quantity Q to changes in the independent variable t is denoted
Often this variable Q depends upon some intermediate variable y and y itself depends on t.
dQ dQ and dy
The rates of change
are related by the Chain Rule:
= 3. Then
dt ' dy
For example suppose that Q
dt
dy
2y + .5y , and that we know
dt
dt
dQ
dy
dy dt
so that
When y
1.2 we may compute that
dt
dQ
40.92
Question Two
Untitled picture.png Machine generated alternative text:
A 2 m ladder leans against a wall If the base slides out ata rate of 0.2 m/sec, then how fast is the area of the triangle underneath the ladder changing at the instant when the base of the
ladder is 1 m from the wall?
Let's start by defining variables, and translate these words into something we can work with! Let denote the distance from the base of the ladder to the wall, t denote the time in seconds,
and A denote the area of the triangle under the ladder. We are told that
and we are asked to find
O — when t
dt
O
when t
O — when
dt
O
when
O
when t
dt
when
dt
We need to find a formula for A in terms of the variable x:
A
Using our prodigious skill at differentiating
.2
dt
1/2 x sqrt(4 - x*2)
lisqä(3)
dA
So by the chain rule
dt
dc
dA dx
we deduce that at the time in question
dx dt
oasqä(3)
dt
O m}sec,
0 m/sec_
0 m/sec

Untitled picture.png Machine generated alternative text:
A 2 m ladder leans against a wall If the base slides out ata rate of 0.2 m/sec, then how fast is the area of the triangle underneath the ladder changing at the instant when the base of the
ladder is 1 m from the wall?
Let's start by defining variables, and translate these words into something we can work with! Let denote the distance from the base of the ladder to the wall, t denote the time in seconds,
and A denote the area of the triangle under the ladder. We are told that
and we are asked to find
O — when t
dt
O
when t
O — when
dt
O
when
O
when t
dt
when
dt
We need to find a formula for A in terms of the variable x:
A
Using our prodigious skill at differentiating
.2
dt
1/2 x sqrt(4 - x*2)
lisqä(3)
dA
So by the chain rule
dt
dc
dA dx
we deduce that at the time in question
dx dt
oasqä(3)
dt
O m}sec,
0 m/sec_
0 m/sec
Question Three
Untitled picture.png Machine generated alternative text:
According to Puff Daddy, '"More money more problems - it's just the way it is"
Consider the relation between money m and problems p . Say you earn
dm
23.5 dollars per hour.
The relationship between problems and money is
p = In(m)
Let's look at the rate at which money brings problems. By differentiation
dp
dm
Then by the Chain Rule, the rate at which time brings problems is
dp
23.51m
How much money do you need so that you have
dp
The answer is
1
problems per hour?
5
O dollars
Ink Drawings
Ink Drawings
Ink Drawings
Ink Drawings
Ink Drawings

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