3.9 RELATED RATES dv
∧dx
In order to connect functions dt , we first
dt
1. Problem Statement: express V as a function of x .
If V is the volume of a cube with edge length x and the V (x )=x
3
dv
cube expands as times passes, find . For us to use the given information, we differentiate
dt
each side of the equation with respect to t. To
Required: differentiate the right side, we use Chain Rule.
The unknown in the problem is the rate of the volume We have,
of the cube with respect to time which is expressed as,
dv dv dx
dv = ∙
=? dt dx dt
dt
c. Calculations:
Solution:
We then use the derivative function of V with
a. Figures respect tot in relation to x .
dv dv dx
= ∙
b. Formulas used: dt dx dt
In order to express these quantities We use the value ofV ( x )=x 3to solve for the
mathematically, we introduce some suggestive derivative ofV as a function of x .
notation:
We have,
Let V be the volume of a cube and; 3
dv d (x ) dx
x be the length of a cube. = ∙
dt dx dt
Both V ∧x are functions of time. The rate increase of By applying Power Rule to x 3to get its derivative,
the volume of a cube with respect to time is we get
dv
expressed as . The rate of increase of the length
dt dv 2 dx
=3 x ∙
dx dt dt
of a cube with respect to time is expressed as .
dt
d. Summary of Answers:
Thus, the rate increase of the volume of the
dv dv 2 dx
cube with respect to time is =3 x ∙ .
dt dt dt
, dA
expressed as . The rate of increase of the side of
dt
dx
the square with respect to time is expressed as .
dt
dA
∧dx
In order to connect functions dt , we first
dt
express A as a function of x .
2. Problem Statement: Area of a square=x
2
Each side of the square is increasing at a rate of 6
2
cm/s. At what rate is the area of the square increasing A(x )=16 cm
when the area of the square is 16cm 2 .
For us to use the given information, we differentiate
Required: each side of the equation with respect to t. To
differentiate the right side, we use Chain Rule.
The unknown in the problem is the rate of the area of
the square with respect to time which is expressed as, We have,
dA dA dA dx
=? = ∙
dt dt dx dt
Solution: c. Calculations:
a. Figures We then use the derivative function of A with
respect tot in relation to x .
dA dA dx
b. Formulas used: = ∙
dt dx dt
In order to express these quantities
mathematically, we introduce some suggestive First, we use the given A( x )=16 cm 2 to solve for the
notation: side x .
Let A be the area of the square and; Since the Area of a square=x 2, we use x 2as to
substitute of A ( x )in the equation A ( x )=16 cm 2 .
x be the side of the square.
So we have,
Both A∧x are functions of time. The rate increase of
the area of the square with respect to time is 2
x =16 cm
2
We take the square root of both sides, giving us
∧dx
In order to connect functions dt , we first
dt
1. Problem Statement: express V as a function of x .
If V is the volume of a cube with edge length x and the V (x )=x
3
dv
cube expands as times passes, find . For us to use the given information, we differentiate
dt
each side of the equation with respect to t. To
Required: differentiate the right side, we use Chain Rule.
The unknown in the problem is the rate of the volume We have,
of the cube with respect to time which is expressed as,
dv dv dx
dv = ∙
=? dt dx dt
dt
c. Calculations:
Solution:
We then use the derivative function of V with
a. Figures respect tot in relation to x .
dv dv dx
= ∙
b. Formulas used: dt dx dt
In order to express these quantities We use the value ofV ( x )=x 3to solve for the
mathematically, we introduce some suggestive derivative ofV as a function of x .
notation:
We have,
Let V be the volume of a cube and; 3
dv d (x ) dx
x be the length of a cube. = ∙
dt dx dt
Both V ∧x are functions of time. The rate increase of By applying Power Rule to x 3to get its derivative,
the volume of a cube with respect to time is we get
dv
expressed as . The rate of increase of the length
dt dv 2 dx
=3 x ∙
dx dt dt
of a cube with respect to time is expressed as .
dt
d. Summary of Answers:
Thus, the rate increase of the volume of the
dv dv 2 dx
cube with respect to time is =3 x ∙ .
dt dt dt
, dA
expressed as . The rate of increase of the side of
dt
dx
the square with respect to time is expressed as .
dt
dA
∧dx
In order to connect functions dt , we first
dt
express A as a function of x .
2. Problem Statement: Area of a square=x
2
Each side of the square is increasing at a rate of 6
2
cm/s. At what rate is the area of the square increasing A(x )=16 cm
when the area of the square is 16cm 2 .
For us to use the given information, we differentiate
Required: each side of the equation with respect to t. To
differentiate the right side, we use Chain Rule.
The unknown in the problem is the rate of the area of
the square with respect to time which is expressed as, We have,
dA dA dA dx
=? = ∙
dt dt dx dt
Solution: c. Calculations:
a. Figures We then use the derivative function of A with
respect tot in relation to x .
dA dA dx
b. Formulas used: = ∙
dt dx dt
In order to express these quantities
mathematically, we introduce some suggestive First, we use the given A( x )=16 cm 2 to solve for the
notation: side x .
Let A be the area of the square and; Since the Area of a square=x 2, we use x 2as to
substitute of A ( x )in the equation A ( x )=16 cm 2 .
x be the side of the square.
So we have,
Both A∧x are functions of time. The rate increase of
the area of the square with respect to time is 2
x =16 cm
2
We take the square root of both sides, giving us
