Pre-Calculus Assignment: Analyzing Demand, Rate of Change and More

This pre-calculus assignment explores key concepts including demand functions, rate of change, market equilibrium, and tangent lines. It demonstrates how to calculate demand at a given price, determine the rate of change of demand with respect to price, and find the market equilibrium price by equating demand and supply functions. The assignment also covers finding the rate of change of drug concentration over time using derivatives and determining the equation of a tangent line to a given function at a specific point. The solutions are detailed and step-by-step.

Pre-Calculus
Pre-Calculus Assignment
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Pre-Calculus
Questions 14-16
Data: Demand function,
x=1000−60 √ p+25 , where x=number of bicycle helmets customers are willing ¿ buy
20 ≤ p ≤ 100
Question 14: Demand when price is $ 45
This will be determined by substituting p in the demand function with 45
x=1000−60 √45+25
¿ 1000−60 √70
¿ 1000−501.996=498.00 4
≈ 498 helmet s
Therefore, the demand at p=$ 45 is 498 helmets
Question 15: Rate of change of demand with respect to price, when p=$ 45
To obtain be the rate of change of demand, the first derivative of demand function will
be determined first, then p in the derivative will be substituted with 45 to obtain the real
rate of change.
Derivative

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The terms of the demand function will be differentiated separately
x=1000−60 √ p+25 can also be written as x=1000−60 ( p+25 )
1
2
Derivative of 1000 is 0
Derivative of y=60 ( p+25 )
1
2
Chain rule will be applied; dy
dx [g ( x ) ]n=n [ g(x )]n−1∗dg
dx
Let g ( p )= p+25, then y=( g ( p ) )
1
2
Differentiate the two functions
dg( p)
dp =g ( p ) ' =1
dy
dg( p)= 1
2 (g ( p ) )
−1
2
The derivative of y will be given by
y' =
60∗dg ( p )
dp ∗dy
dg ( p ) =60∗1∗1
2 ( g ( p ) )
−1
2
y' =30 ( g ( p ) )
−1
2 , substituting g ( p ) with( p+25)
y' =30 ( 1+25 )
−1
2
Finally the derivative of demand function(x' ) will be given by

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Derivative of 1000 minus y '
x' =0−30 ( p+25 )
−1
2
x ' =−30 ( p+25 )
−1
2 … …… i
Substitution of p=45 in i above
x' =−30 ( 45+25 )
−1
2
x'=−3.5857
x' ≈−3.586
Therefore, rate of change of demand at p=45 is -3.586
Interpretation
The rate at which demand is changing is negative(-3.586), implying that any change in price
in will influence demand in the opposite direction, for instance if the price increase the
demand of bicycle helmets will decrease as most customers will be willing to buy at
relatively low prices.

Pre-Calculus
Question 16: Equilibrium market price and rate of Demand at market price
Data:
Demand function
x=1000−60 √ p+25
Supply function
x=80 √ p+25−400
Equilibrium in the market is attained when Demand is equal to Supply; therefore, to obtain the
Equilibrium market price, Demand function and Supply function will be equated as shown below
1000−60 √ p+25=80 √ p+ 25−400 … … … . putting like terms together
1000+ 400=80 √ p+25+60 √ p +25
1400=140 √ p+25 , … … … . divide both sides by 140
1400
140 =140
140 √ p+25
10= √ p+25 , … … … . squaring both sides
102= [ √ p+25 ]
2

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100= p+ 25 … … … putting like terms together
p=100−25
p=75
Therefore, the market equilibrium price is $ 75
The rate of rate of change of demand at market equilibrium price will be obtained by
substituting p=75 in equation i in question 15 above, x' =−30 ( p+25 )
−1
2
x' =−30 ( 75+25 )
−1
2
x' =−30 ( 100 )
−1
2
¿−30∗1
√ 100
¿−30∗1
10 =−3
Therefore, rate at which demand is changing at p=75 is -3.00
Question 17: Rate at which Concentration of drug is changing after t=3 hours
Data: Concentration function
C ( t )=9.5 e−0.75 t ,t=time∈hours

Pre-Calculus
To obtain the rate of change of concentration, the first derivative of the C ( t ) will be
determined first, then t in the derivative will be substituted with 3 to obtain the real rate.
Derivative
Differentiation of exponential function
giventhat y=ef (x ) , then dy
dx =f ( x )'∗ef ( x )
Let f ( t ) =−0.75t then C ( t )=9.5 e−f (t ) ,
Differentiate the f ( t )
df
dt =−0.75
Therefore, the derivative of C ( t ) will be given by,
dC
dt ¿ C(t )' = 9.5∗df
dt ∗e−f ( t )
C (t)'=9.5 (−0.75 ) e−0.75 t
C (t)' =−7.125 e−0.75 t … … . i
Substitution of t=3 in equation i to obtain the rate at which concentration of drug is changing
after three
C ( t ) '=−7.125 e−0.75 ( 3 ) =−7.125 e−2.25
¿−7.125∗0.1054=−0.751

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C ( t ) ' ≈−0.75
Therefore, the rate at which concentration of drug is changing after 3 hours is -0.75
Question 18: Equation of a line tangent to the graph f ( x ) = √ 10 x +6 at x=4
First, the value of f ( x ) at x=4 will be determined to obtain the point where tangent and the
curve intersect in terms of ( x , y ) coordinates.
f ( x ) = √ 10( 4)+6= √ 46=6.7823
Thus, the tangent line intersects with the graph at ( x , y ) =( 4 , 6.7823)
Second, the gradient function will be determined by differentiating the equation of the graph,
f ( x ), which will be used to determine the gradient of the tangent line.
Writing the f ( x ) to set it into a differentiable form;
f ( x )= (10 x +6 )
1
2
To differentiate this function chain rule will be applied, which states given f (x) as a function of
g ( x ) to the power of n then derivative of y is given by f ( x ) ' = df
dx [g ( x ) ]n=n [ g(x )]n−1∗dg
dx
Let g ( x )=10 x+ 6, then f ( g ( x ) )= [ g ( x ) ] 1
2
Differentiate the two functions
dg
dx =10

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df
dg = 1
2 [g ( x ) ]
−1
2
Therefore, from the chain rule, the derivative y will be,
df
dx [ g ( x ) ]
n
=n [ g ( x ) ] n−1
∗dg
dx
¿ 1
2
[ g ( x ) ]−1
2 ∗dg
dx =1
2 [ g ( x ) ]−1
2 ∗10
¿ 5[ g ( x )]
−1
2 … … … .. substituting g ( x ) =10 x +6, hence, the derivative of y with respect to x will
be ;
f ( x)' = df
dx =5[10 x+6 ]
−1
2
Therefore, the gradient function of f ( x ) is
f ( x )'=5 [10 x+ 6]
−1
2
Then x=4 is substituted in the gradient function to obtain the gradient of the tangent line
f ( x )'=gradient=5[10(4 )+ 6]
−1
2 =0.7372
The equation of a tangent is given by the general equation of a straight line;
f ( x )= y=mx+ c , where mis the gradient ∧c is y−intercept of the line
The Substitution of point ( x , y ) =( 4 , 6.7823) and gradient=0.7372 in the general equation done
to obtain the unknown value, c , y−intercept

Pre-Calculus
y=mx+c
6.7823=0.73721 ( 4 ) + c
c=6.7823−0.7372 ( 4 )=3.8335
Hence, the equation of the tangent line is:
f ( x )= y=0.7372 x+3.8335