Here we can find problems that how we are finding composition of two functions from word problems.
To find composition of two functions f and g, we have to follow the procedure given below.
Step 1 :
In (f∘g) (x),
Write f and remove the composition sign. Inside the bracket put the function g(x). So, we will get
(f∘g) (x) = f[g(x)]
Step 2 :
In the place of g(x), put the respective function.
Step 3 :
Now the function g(x) is like a input for the function f(x). So, apply the function g(x) in the place of x in the function f(x).
Problem 1 :
The surface area S (in square meters) of a hot - air balloon is given by
S(r) = 4πr^{2}
Where r is the radius of the balloon (in meters). If the radius r is increasing with time t (in seconds) according to the formula
r(t) = 2/3t^{3}, t ≥ 0
find the surface area S of the balloon as a function of the time t.
Solution :
S = surface area
Surface area is S(r) = 4πr^{2} --- (1)
Where r is the radius of the balloon.
If the radius r is increasing with time t (in seconds).
r(t) = 2/3t^{3}
r(t) = 2/3t^{3} substitute in equation (1).
S(r(t)) = 4π(r(t))^{2}
Problem 2 :
The volume V (in cubic meters) of the hot – air balloon described in Problem 65 is given by
V(r) = 4/3 πr^{3}
If the radius r is the same function of t as in Problem 65, find the volume V as a function of the time t.
Solution :
Problem 3 :
The number N of cars produced at a certain factory in one day after t hours of operation is given by
N(t) = 100t – 5t^{2}, 0 ≤ t ≤ 10
If the cost C (in dollars) of producing N cars is
C(N) = 15,000 + 8000N
find the cost C as a function of the time t of operation of the factory.
Solution :
N(t) = 100t – 5t^{2}
C(N) = 15,000 + 8000N
C(N(t)) = 15,000 + 8000N(t)
= 15,000 + 8000(100t - 5t^{2})
C(N(t)) = 15,000 + 800, 000t - 40,000t^{2}
Problem 4 :
The spread of oil leaking from a tanker is in the shape of a circle. If the radius r (in feet) of the spread after t hours is
r(t) = 200 √t
find the area A of the oil slick as a function of the time t.
Solution :
area A = πr^{2}
Given, r(t) = 200 √t
A(t) = π(200√t)^{2}
= π(200)^{2}(t^{1/2})^{2}
A(t) = 40000πt
Problem 5 :
The price p, in dollars, of a certain product and the quantity x sold obey the demand equation
p = -1/4 x + 100 0 ≤ x ≤ 400
Suppose that the cost C, in dollars, of producing x units is
C = √x/25 + 600
Assuming that all items produced are sold, find the cost C as a function of the price p.
[Hint : Solve for x in the demand equation and then form the composite]
Solution :
First we need to solve for x in the demand equation :
p = -1/4 x + 100
Add 1/4 x on both sodes.
p + 1/4 x = 100
Subtract p on each sides.
1/4 x = 100 - p
Multiply 4 on each sides.
x = 4(100 - p)
x = 400 - 4p
Problem 6 :
Cost of a Commodity : The price p, in dollars, of a certain commodity and the quantity x sold obey the demand equation
p = -1/5 x + 200 0 ≤ x ≤ 1000
Suppose that the cost C, in dollars, of producing x units is
C = √x/10 + 400
Assuming that all items produced are sold, find the cost C as a function of the price p.
Solution :
Given, p = -1/5 x + 200
Add 1/5 x on both sodes.
p + 1/5 x = 200
Subtract p on each sides.
1/5 x = 200 - p
Multiply 5 on each sides.
x = 5(200 - p)
x = 1000 - 5p
Problem 7 :
The volume V of a right circular cylinder of height h and radius r is V = πr^{2}h. If the height is twice the radius, express the volume V as a function of r.
Solution :
V = πr^{2}h
The height is twice the radius.
So, V(r) = 2r(πr^{2})
V(r) = 2πr^{3}
Problem 8 :
The volume V of a right circular cone is
V = 1/3 πr^{2}h
If If the height is twice the radius, express the volume V as a function of r.
Solution :
V = 1/3 πr^{2}h
The height is twice the radius.
So, V(r) = 2r(1/3 πr^{2})
V(r) = 2/3 πr^{3 }
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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