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course MTH 173
10/27 23:40
Text Section 2.3Problem 2.3.3
The function is a straight line passing through (1,0) and (0,2). Thus the slope of the line is 2/-1 = -2.
The derivative function is the slope function and since the slope of the straight line is constant, thus the derivative function is a constant function of constant value of -2. Thus the derivative function is y = -2.
Problem 2.3.6
The value of the derivative at x = -3 using straightedge is about -3
The value of the derivative at x = -2 using straightedge is about -1
The value of the derivative at x = -1 using straightedge is about 0.5
The value of the derivative at x = 0 using straightedge is about 1.3
The value of the derivative at x = 1 using straightedge is about 1.3
The value of the derivative at x = 2 using straightedge is about 0.5
The value of the derivative at x = 3 using straightedge is about -1
Thus from the approximate values of the derivative it can be seen that the derivative function is like a concave down parabola with x = 0.5 the axis of symmetry and passing through points (-3,-3) , (-2,-1) , (-1,0.5) , (0,1.3) , (1,1.3) , (2,0.5) and (3,-1)
Problem 2.3.9
The value of the derivative at x = -3 using straightedge is about 3
The value of the derivative at x = -2 using straightedge is about 0.5
The value of the derivative at x = -1 using straightedge is about -1
The value of the derivative at x = 0 using straightedge is about -2
The value of the derivative at x = 1 using straightedge is about -1
The value of the derivative at x = 2 using straightedge is about 0.5
The value of the derivative at x = 3 using straightedge is about 3
Since we see that derivative values are same for both negative and positive values of a particular x value, it can be assumed that the derivative function must be symmetric about y axis. Also it can be seen that as we move away from 0 on either sides the slope changes more rapidly thus the graph of the differential function must be like a parabola passing through points ( +-3,3) , (+-2,0.5) , (+-1,-1) and (0,-2)
Problem 2.3.12
The value of the derivative at x = -2 using straightedge is about 5
The value of the derivative at x = -1 using straightedge is about 0.5
The value of the derivative at x = 0 using straightedge is about 0
The value of the derivative at x = 1 using straightedge is about 1
The value of the derivative at x = 2 using straightedge is about 1.5
The value of the derivative at x = 3 using straightedge is about 0
The graph of the derivative is like a cubic function which touches the x axis at points x = 0 and x = 3, the graph is concave up from x = -2 to x = c and concave down from x = c to x = 3 where c is a value between 0 and 2 around x = 1.
Problem 2.3.16
Given function f(x) = e ^ x. the graph of e ^ x is a like a graph which tends to 0 as x tends to - infinity and tends to + infinity as x tends to + infinity. The function slowly increases from 0 to 1 and then rapidly increases to + infinity. The derivative function when plotted using the straight line edge also turns out to be the same f(x) function. Thus the derivative function is also e ^ x.
Problem 2.3.19
Given function k(x) = 1/x = x^-1
According to the power rule for a function x ^ n its derivative it n*(x^n-1)
Thus the derivative of the above function using power rule will be k’(x) = -1*(x^-2) = -1/(x^2)
Problem 2.3.22
Given function m(x) = 1/(x+1)
According to the difference quotient method the derivative of the function is calculated as
m’(x) = lim h tending to 0 ( m(x+h) - m(x) ) / h = lim h tending to 0 ( 1/( x + h + 1 ) - 1 / ( x + 1) ) / h
m’(x) = lim h tending to 0 ( x + 1 - 1 - x - h ) / (h * (x+h+1)(x+1))
m’(x) = lim h tending to 0 -1 / (x + h + 1)*(x + 1) = -1 / (x+1)^2
Thus the derivative of the function m(x) is - 1 / (x+1)^2
Problem 2.3.25
Given f(x) = lnx
To calculate the derivative of the function at x = 1 we use the values of the function at x = 0.9999 and x = 1.0001
The value of the function at x = 1 is f(1) = 0
The value of the function at x = 0.9999 is f(0.9999) = -0.000100005
The value of the function at x = 1.0001 is f(1.0001) = 0.000099995
The average rate of change of function between the interval of x = 0.9999 to x = 1 will be
= ( 0 + 0.000100005 ) / 0.0001 = 1.00005
The average rate of change of function between the interval of x = 1 to x = 1.0001 will be
= ( 0.000099995 - 0 ) / 0.0001 = 0.99995
The approximate value of f’(1) will be the average of the average rate of change of function on the left and the right side of x = 1.
Thus f’(1) = ( 1.00005 + 0.99995 ) / 2 = 1.0000
To calculate the derivative of the function at x = 2 we use the values of the function at x = 1.9999 and x = 2.0001
The value of the function at x = 2 is f(2) = 0.693147
The value of the function at x = 1.9999 is f(1.9999) = 0.693097
The value of the function at x = 2.0001 is f(2.0001) = 0.693197
The average rate of change of function between the interval of x = 1.9999 to x = 2 will be
= ( 0.693147 - 0.693097 ) / 0.0001 = 0.500
The average rate of change of function between the interval of x = 2 to x = 2.0001 will be
= ( 0.693197 - 0.693147 ) / 0.0001 = 0.500
Thus the approximate value of f’(2) will be 0.500
To calculate the derivative of the function at x = 5 we use the values of the function at x = 4.9999 and x = 5.0001
The value of the function at x = 5 is f(5) = 1.609437
The value of the function at x = 4.9999 is f(4.9999) = 1.609417
The value of the function at x = 5.0001 is f(5.0001) = 1.609457
The average rate of change of function between the interval of x = 4.9999 to x = 5 will be
= ( 1.609437 - 1.609417 ) / 0.0001 = 0.20
The average rate of change of function between the interval of x = 1 to x = 1.0001 will be
= ( 1.609457 - 1.609437 ) / 0.0001 = 0.20
Thus the approximate value of f’(5) will be 0.20
To calculate the derivative of the function at x = 10 we use the values of the function at x = 9.9999 and x = 10.0001
The value of the function at x = 10 is f(10) = 2.302585
The value of the function at x = 9.9999 is f(9.9999) = 2.302575
The value of the function at x = 10.0001 is f(10.0001) = 2.302595
The average rate of change of function between the interval of x = 9.9999 to x = 10 will be
= ( 2.302585 - 2.302575 ) / 0.0001 = 0.1
The average rate of change of function between the interval of x = 10 to x = 10.0001 will be
= (2.302595 - 2.302585) / 0.0001 = 0.1
Thus the approximate value of f’(10) will be 0.1
Above we estimated f’(1) = 1.0000, f’(2) = 0.500, f’(5) = 0.20 and f’(10) = 0.1, thus we can estimate the function of f’(x) using the above trends in the value which is f’(x) = 1/x.
Problem 2.3.32
The given graph of f(x) is like that of a concave down parabola. Thus the slope of such a parabola is initially positive and then decreases to reach a negative value. Since it is a parabola the slope changes at a constant rate. Since the parabola has a point of maxima at x = 1, thus the value of the slope at x = 1 will become 0. Thus the graph of f’(x) will be a straight line with a negative slope passing through (1,0) and the slope depending upon what the coefficient of x^2 is in the f(x) parabola.
Problem 2.3.35
The given f(x) function is a quartic function where the roots of the function are x = 0, x = 0, x = +-2. We consider 2 roots for x = 0 because the function touches the point (0,0) rather than cutting it. Thus the derivative f’(x) will be a cubic function whose roots are the points of f(x) at which the slope is 0. The slope of f(x) is 0 at x = 0 and an approximate x = +-1.5. Thus the roots of f’(x) are x = 0 and x = +- 1.5.
Problem 2.3.37
The given f(x) function is a straight line function from x = -1 to x = 1, a curve that is basically change of slope from x = 1 to x = 2, a parabola from x = 2 to x = 4, a curve that is slope change from x = 4 to x = 5 and then a straight line beyond x = 5.
The value of the derivative is a constant value from x = -1 to x = 1 and another constant beyond x = 5. The value of the derivative is 0 for an x value almost greater than x = 1. The slope of the function in negative decreases and then again increases to a 0 at an x value almost near to 3. For further x values the slope increases to a maximum value and then decreases to a value at which it becomes a constant value for further x values.
Problem 2.3.40
(a) Given that the bus on a popular route, with no traffic. The function f(t) is the function of the distance starting from the initial point at time t. thus f’(t) will be the graph of the speed of the bus. Since the bus moves on a popular route the bus stops at regular intervals to pick up passengers. Thus the bus takes up speed, reaches maximum speed and then decreases to 0, waits for some time at the stop then again speeds up to the maximum speed. Thus this is represented by the graph (II)
(b) A car with no traffic and all green lights, the car will usually move with an increasing speed to reach a maximum speed. For the speed the car will increase its speed at a decreasing rate, this is because it is difficult for the engine to increase speed as speed increases, thus the graph (I) represent such graph of speed.
(c) A car in heavy traffic conditions will have uneven speeds, it will have to stop at irregular intervals. It would also not have a constant speed and a high speed ever. Thus the graph (III) represents such unevenness in the speed and thus in f’(t) graph
Problem 2.3.41
(a) according to V’(t) graph the value of V’(t) is first positive after t = 3 time unit. A positive value of V’(t) indicates that the volume of the balloon is increasing, thus the boy begins to inflate that is increase the volume at t = 3 time unit
(b) The value of V’(t) is positive till t = 9 time unit, after the V’(t) becomes 0, when V’(t) becomes 0 the balloon no longer changes volume and thus stops to inflate. Thus the boy finishes to inflate the balloon at t = 9 time unit
(c) The value of V’(t) is negative the first time after t = 14 approx time units, V’(t) negative means that the volume of the balloon is decreasing and thus the boy is letting the air out of the balloon. Thus the balloon lets the air out at t = 14 time unit.
(d) If we consider the negative f’(t) between t = 14 and t = 17 to be a cup shape graph, then if the child instead of letting the entire air to come out the child starts pinching and leaving then the cup shape of the function breaks into a number of small cups similar shape but smaller magnitude lowest value as compared to the magnitude of the lowest value in the given graph.
Text Section 2.4
Problem 2.4.3
Given that the function T = f(t) is the temperature function of a cold yam placed in a hot oven. Since the yam placed in yam is cold and is placed in a hot oven, thus the temperature of the yam increases in the oven as t increases up to the temperature of the oven.
(a) thus the function f’(t) represents the rate at which the temperature of the yam increases at a particular time t. Since the temperature increases thus the sign of f’(t) will be positive.
(b) Since f’(t) is the function representing the rate at which the temperature changes at a particular time t, thus the unit of f’(t) will be temp unit / time unit which will be F / min. Thus the unit of f’(20) will be F / min.
Thus practical meaning of f’(20) = 2 will be that at t = 20 min the average rate of temperature change will be 2 F / min.
Problem 2.4.6
Given that p represents the cost of certain quantities and q represents the quantity of item sold. Thus q = f(p) will be the quantity function in terms of the cost where the cost is the independent variable and the quantity is the dependent variable.
(a) Thus f(150) = 2000 represents the quantity of item sold at a price of $150
(b) f’(150) = -25 represents the derivative of the function f(x). The derivative function f’(x) represents the rate of change of the quantity of the item sold at a particular x (price) value.
Thus f’(150) = -25 represents that the at a price of $150 the rate at which the quantity of items sold is changing is -25 items/$
Problem 2.4.10
Given that B = f(t) is the function where t represents the number of years since the initial investment and B represents the compounded balance in $ for the initial investment after t years.
Thus dB / dt represents the rate at which the compounded balance changes at a particular t. Thus the unit of dB/dt is $/year and the financial interpretation of dB / dt is that
Problem 2.4.11
Given that $1000 is invested at a rate of r% for 10 years and the balance is given as $B. Thus the function B = g(r) is a compounded investment function where r is the independent variable and B is the dependent variable.
(a) Thus g(5) = 1649 indicates that when $1000 invested at a rate of 5% for 10 years gives an approximate compound balance of $1649
(c) g’(r) thus will be the rate function which indicates the rate of change of balance with respect to the rate. Thus the unit of g’(r) will be $(compound investment balance)/ %(rate unit). Thus g’(5) = 165 indicates that rate of change of compounded investment at rate r = 5% will be approximately 165, and the unit of which will be $ / %.
Problem 2.4.13
Given that W = f(c) is a weight function where c Calories consumed per day is the independent variable and W weight in pounds is the dependent variable.
(a) Thus f(1800) = 155 indicates that the weight of the subject will be 155 pounds when the subject consumes 1800 calories per day
f’(c) is the rate of change of weight function in terms of calories consumed per day, thus f’(2000) = 0 indicates the rate of change of weight function will be 0 when the calories consumed per day is 2000.
f^-1(x) is the inverse function that is, it is the number of calories consumed per day function in terms of weight of the subject. Thus f^-1(162) = 2200 indicates that when the weight of the subject is 162 pounds the calories consumed by the subject will be 2200 calories.
(b) f’(c) = dW/dc, since f’(c) is the rate function thus the unit of f’(c) will be pounds(weight unit)/calories.
Problem 2.4.16
Given population function P = f(t) = 1.34*(1.004)^t where t is the number of years since 2011 and P is the population of China in billions.
Thus f(9) = 1.34*(1.004)^9 = 1.389 billion approximately
The f’(t) function will be f’(t) = 1.34*(1.004)^t*ln(1.004)
Thus the approximate value of the function f’(9) will be f’(9) = 1.34*(1.004)^9*ln(1.004) = 0.004 billion/year
Thus this information tells us that the population of China in 2020 is estimated to be 1.389 billion and the rate at which the population will be changing during this year will be 0.004 billion/year
Problem 2.4.18
The function should be f(r) = R, where r rating of the news which is the independent variable and R is the revenue in $ in millions which is the dependent variable.
Thus the derivative function is f’(r) whose unit is $ in millions / rating. Given that for a drop of 0.1 point rating corresponds to $5.5 million drop. Since the CBS evening news has a 4.3 point rating. Thus the derivative is estimated at point rating of 4.3, for an estimation of the change in revenue the value of the derivative at 4.3 is used as constant slope for the 0.1 drop of point rating. Thus if the slope is f’(4.3) and the run is -0.1 thus rise in the revenue will be -f’(4.3)*0.1.
Problem 2.4.21
Giving that h(t) is the depth in feet at time t in hours
(a) h(5) = 3 indicates that at time 5 hours the depth in the tank is 3 feet.
(b) h’(5) = 0.7 is a rate function which indicates that the rate of change of depth at t = 5 hours is 0.7 feet / hour
(c) h^-1(5) = 7 is the inverse function which indicates that when the depth of the function is 5 feet the time is 7 hours.
(d) (h^-1)’(5) = 1.2 indicates that rate of change of time when the depth is 5 feet is 1.2 hours / feet.
Problem 2.4.23
g(t) is the function of height in inches of Amelia Earhart. Thus g’(t) is the rate function of height change in terms of age. Thus the unit of g’(t) will be inches/year.
!!!! The data in the text says assume 0 < t < 39 the age at which Amelia Earhart’s plane disappeared. But according to the answer in the textbook the interval should be 0 < t < 29
Since g’(10) is the rate at which the height changes when age is 10 years, and the height always increases thus the sign of g’(10) will be positive. Since Amelia Earhart’s plane and she disappeared when her age was between 0 < t < 29 thus her height will no longer change after she reaches age of 29. Thus g’(30) will be 0.
Problem 2.4.26
(a) The velocity keeps increases at an increasing rate until it reaches a terminal velocity. Thus the graph of this velocity vs. time function would be a concave up function with increasing slope and when the velocity reaches the maximum velocity it becomes a horizontal line parallel to the x axis with y = terminal velocity. The horizontal line continues until the time the object reaches the ground.
(b) The concavity of the graph will be concave up from velocity 0 till the object reaches the maximum velocity, after reaching the maximum velocity the concavity becomes 0.
(c) The natural phenomena represented by the graph at t = 0 when the air resistance in negligible will be free fall. That is when air resistance is neglected at t = 0 the object moves towards the earth at an acceleration of g (an approximate value of 9.8 m / s^2). This is a case of free fall. When air resistance is considered the acceleration is less than 9.8 m/s^2.
Problem 2.4.31
Given P(x) is the number of people of height less than or equal to x inches in the US. P(66) will be the number of people having height less than 66 inches. Thus P’(66) will be the rate at which the function P(x) will change when x = 66 inches. The unit of P’(66) will be number of people / inches. In order to estimate the value of P’(66) it is given that we are supposed to use the difference quotient with h = 1. Thus for finding the right hand limit, we calculate the derivative of the function as P(67) - P(66). Thus for finding the left hand limit, we calculate the derivative of the function as P(66) - P(65). To calculate the estimate of the derivative at x = 66 we use the average of the right hand and the left hand derivative. Thus the approximate value is (P(67) - P(65))/2. P’(x) will never be negative. P(x) indicates the number of people having height less than x inches, thus P(x) as x increases always gets added up and thus always keeps increasing. Thus P’(x) will never be negative.
NOTE: I did not understand how would I estimate P’(66) by using US population of 300 million and the note that 66 inches = 5 feet 6 inches.!!!!!!!!!!!!
Problem 2.4.35
Given that the original volume of the balloon is 3 liters and it is leaking that is reducing volume at the rate of 1%. Thus the volume of the balloon is decreasing at the rate of 1% and thus the radius is decreasing too. It is given that r’(t) > 0. Since r’(t) is the rate at which the radius of the balloon is changing, and since the radius is decreasing thus r’(t) should be negative.
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