#$&*
course MTH 173
11/5 19:45
Text Section 2.5
Problem 2.5.2
(a) f(2) is a value below x axis and thus corresponds to a negative y value and thus is negative.
(b) f’(2) is the slope of the graph at x = 2 which is the slope of the tangent line at x = 2 which is also negative, thus f’(2) is negative.
(c) f’’(2) signs depend upon the concavity of the graph at x = 2. The graph shows that at x = 2 the graph is concave up and thus the value of f”(2) is positive.
Problem 2.5.4
At point A
The value of the function is almost zero but below x axis thus actual value is negative. Thus f(A) is negative.
The value of the slope of the function is 0, thus f’(A) = 0
The graph is concave up at point A thus f”(A) is positive.
At point B
The point is above x axis thus the value of the function is positive thus f(B) is positive.
The value of the slope at B is 0 thus f’(B) = 0
The graph is concave down at B thus f”(B) is negative.
At point C
The point is above x axis thus the value of the function is positive thus f(C) is positive.
The slope of the graph at point C has a negative value and thus the derivative will also be negative, thus f’(C) is negative.
The graph is concave down at C thus f”(C) is negative.
At point D
The point is below x axis thus the value of the function is negative thus f(D) is negative.
The slope of the graph at point D has a positive value and thus the derivative will also be positive, thus f’(D) is positive.
The graph is concave up at D thus f”(D) is positive.
Problem 2.5.7
For the graph of height of a particle against time f(t), the velocity is the slope of the graph that is f’(t) and acceleration is the second derivative that is f”(t). Given that velocity is positive, this indicates f’(t) > 0 and acceleration is negative indicates that f”(t) < 0.
The graph of this function will be like an concave down parabola starting from (0,0) and extending up to the highest point of the parabola. The situation shown above is similar to an object thrown towards the sky from the ground.
Problem 2.5.10
The given graph of the function in concave up thus the sign of the second derivative f”(x) will be positive. The slope of the function is always positive thus the sign of f’(x) is also always positive.
Problem 2.5.11
For the given graph of the function the slope is negative for all values of x, thus the sign of f’(x) is always negative. The graph is concave up for all x values thus the sign of f”(x) will be always positive.
Problem 2.5.13
For the given graph of the function the slope is negative for all values of x, thus the sign of f’(x) is always negative also the function is concave down thus the sign of f”(x) will also be always negative.
Problem 2.5.16
Average acceleration for the first 2 seconds is calculated using the velocity at t = 0 and t = 2 sec.
Thus average acceleration = (v(2) - v(0))/2 = (52 - 0)/2 ft/sec/sec = 26 ft/sec^2
In mph average acceleration = 26 * (15/22) mph/sec = 17.727272 mph/sec approx.
Problem 2.5.17
(a) The graph of the function will be a graph which has a concave up curve for negative values of x almost like a y = x^1/3 curve with increasing slope and at x = 0 changes its curve and then is a concave down graph also resembling something like y = x^(1/3) where the slope is positive and decreasing
(b) The graph of the derivative function will be a concave up graph for both negative and positive values of x. It’s like you have a parabola y = x^2, cut the parabola about its axis flip it and at the sharp point, smoothen it. The graph will have positive increasing slopes for negative x values, negative increasing slopes for positive values of x.
(c) the graph of the second derivative function of (a) that is the first derivative of (b). The graph of the second derivative function is almost like the graph of (b) for extreme values of x on either sides of x = 0. For value of x near x = 0, we have a concave down hill for negative values of x and a concave up hill for positive values of x. The slope for the entire graph will be a graph with positive increasing slopes for extreme values of negative x away from x = 0, negative slope decreasing for remaining negative values, negative increasing slopes for positive x values near x = 0, positive decreasing slopes for remaining positive x values. The function has positive values for negative x values and negative values for positive values of x.
Problem 2.5.20
The given function is like a cubic function and thus the first derivative of this function will be like a parabola whose axis will be x = positive constant. Thus the second derivative of the function is a straight line function with positive x intercept and negative y intercept.
Problem 2.5.23
The first derivative function will be like a concave up graph for negative values of x and a concave down function for positive values of x. Using the first derivative function the second derivative function obtained is a concave down parabola. The parabola has y axis as its axis but unlike y = -x^2, the graph is slightly shifted towards the positive values of x.
Problem 2.5.28
(a) f(x) < 0 , the values of the function corresponding to x4 and x5 are below x axis and thus negative. Thus for x4 and x5 f(x) < 0.
(b) f’(x) < 0, the slope of the function at x3 and x4 is negative thus for x3 and x4 f’(x) will be negative.
(c) f(x) is decreasing, f(x) will be decreasing when the slope of the function at that point will be negative thus for x3 and x4 f(x) will be decreasing.
(d) f’(x) is decreasing, f’(x) will be decreasing when f”(x) will be negative, thus at points x1, x2 and x3 the graph is concave down and f”(x) will be negative and thus f’(x) will be decreasing.
(e) Slope of f(x) is positive, the value of the slope at points x1, x2 and x5 is positive and thus slope of f(x) is positive at points x1, x2 and x5.
(f) Slope of f(x) is increasing, the slope will be increasing when f”(x) will be positive. For points x4 and x5 the graph is concave up and thus the f”(x) is positive and thus the slope of f(x) is increasing.
Problem 2.5.31
Given that f(5) = 20 and the slope of the function at x = 5 that is f’(5) = 2 and f”(x) is negative throughout the graph. This indicates that the slope continuously decreases. To estimate the value of the function at x = 7, we can use the constant slope of the function at x = 5 and can predict the value of the function at x = 7 beyond which the function cannot reach that value at x = 7.
Thus for the straight line from x = 2 to x = 7
Run = 2 and slope = 2, thus the rise will be 2*2 = 4.
Thus f(7) = f(5) + rise = 20 + 4 = 24.
This indicates that the value of the function at x = 7 cannot be equal to or more than 24 since f”(x) is negative throughout.
Thus (a) 26 - impossible, (b) 24 - impossible and (c) 22 - possible.
Text Section 2.6
Problem 2.6.2
(a) The function is continuous for all values of x thus there is no value of x at which the function is not continuous.
(b) The function’s derivative can be questioned at x = 2 and x = 4. At x = 2 the function seems to have a vertical tangent and at x = 4 the graph has a corner thus the function is not differentiable at x = 2 and x = 4.
Problem 2.6.4
The given graph on the function is continuous at all values of x and all the points at which the slope changes it’s sign has a smooth curve and not a corner, thus the function is differentiable on the interval of x-values shown.
Problem 2.6.7
Given that f(x) = x^2*sin(1/x) for x not equal to 0 and f(x) = 0 for x = 0.
The function f is differentiable at x = 0 if
lim h tending to 0 (f(0+h) - f(0))/h
The value of the function f(0) = 0^2*sin(1/0)
To calculate sin(1/0), we know that sin(x) function oscillates between -1 and 1 and thus sin(infinity) must be a value within -1 and 1 and thus f(0) will be 0
Thus f’(0) = lim h tending to 0 (h^2*sin(1/h) - 0)/h = lim h tending to 0 h*sin(1/h)
The value of the limit whatever it may be tending from, right hand side of x = 0 or left hand side will be 0(as explained earlier). Thus the value of the derivative of the function at x = 0 is 0, the value of the function at x = 0 is also 0 which indicates that the function is continuous at x = 0 and thus differentiable at x = 0.
Problem 2.6.9
Given function f(x) = (x^2 + 0.0001)^1/2
To calculate whether the function has a derivative at x = 0 or not we use the derivative definition.
Value of the differential f’(x) = lim h tending to 0 (f(x+h)-f(x))/h
Thus f’(0) = lim h tending to 0 (f(h)-f(0))/h = lim h tending to 0 ((h^2+0.0001)^1/2 - 0.01)/h
Rationalizing the above function we get
= lim h tending to 0 h^2 / (h*((h^2+0.0001)^1/2+0.01))
@&
When you rationalize the numerator will become h^2 + .0001 - h^2 = .0001.
*@
= lim h tending to 0 h / ((h^2+0.0001)^1/2+0.01)
As h tends from 0+ the value of the limit is 0 and if the value tends from 0- the value of the limit still tends to 0, and since the function is continuous at x = 0, the function thus is differentiable at x = 0 even though it appears to have a sharp point in the graph.
@&
The graph would appear to have a sharp point, but in fact the graph is rounded at that point, with the curvature becoming apparent when the value of x^2 approaches .0001 (i.e., when x approaches .01).
*@
Problem 2.6.12
Given that B = r*(B0/r0) for r < r0 and B = B0*(r0/r) for r > r0
(a) The graph of the function will be a straight line with the slope of (B0/r0) for r(x values) lesser than and equal to r0. The graph of the function will be a hyperbolic curve with a constant of B0*r0 (x*y = constant), this hyperbola will be a concave up graph. The constant B0 is the value of the function at r = r0 which is the point at which the two parts of the function join.
(b) Graph will be continuous at r = r0 if the value of the function at r = r0 for both the different parts of the function will be equal. For the first part of the function that is B = r*(B0/r0), the value of the function at r = r0 will be B0. For the second part of the function that is B = B0*(r0/r), the value of the function at r = r0 will also be B0. Since the values for both the parts of the function is equal, the function will be continuous at r = r0.
(c) For the first part of the function B = r*(B0/r0), the slope of the function is B0/r0 which is the value of the derivative at r = r0 tending from left hand side. For the second part of the function B = (r0/r)*B0, the derivative of the function will be B’(r) = -(r0/r^2)*B0, the derivative of the function at r = r0 will be -B0/r0, this will be the value of the derivative of the function when the right hand derivative is considered. Since the right hand derivative and the left hand derivative of the function B(r) are unequal the function is not differentiable at r = r0.
Problem 2.6.13
Given E = kr for r < r0 and kr0^2/r for r > r0.
(a) To check whether E is continuous at r = r0, we calculate the value of the function of both the functions of E at r = r0. Thus value of E = kr at r0, E = k*r0 and the value of E = kr0^2/r at r = r0 will be E = k*r0. Since the value for both the function is same thus the function is continuous at x = r0.
(b) No E is not differentiable at r = r0, this is because the value of the slope of the function E = kr as r approaches to r0 from left hand side of r0 is k, whereas the value of the slope of the function E = kr0^2/r as r approaches to r0 from right hand side of r0 is -k. Since the value of the slopes are different as r approaches to r0 from either sides, the function is not differentiable at r = r0.
(c) The graph of the function E for r less than equal to r0 is a straight line with a slope of k. For r greater than r0, the graph starts from the same point at which it was left by the preceding function and is a hyperbolic function with negative increasing slopes, with the function tending towards 0 as r tends toward infinity.
Problem 2.6.16
(a) The graph of the function is x axis for negative x values and a concave up parabola starting from its minimum point (0,0) with positive increasing slopes for non-negative values of x. Function of x^2 has a smooth curve at x = 0 and thus the graph of this f(x) will also have a smooth curve at x = 0 and thus is differentiable at all values of x. the derivative f’(x) graph will be x axis for negative x axis and a straight line with a slope of 2 for non-negative values of x, since the derivative of the function x^2 is 2x.
(b) The function f’(x) has a sharp corner at x = 0 and thus is not differentiable at x = 0. Thus the function f’(x) is differentiable at all x values except x = 0.
Since f’(x) is not differentiable at x = 0 thus f”(x) will be discontinuous at x = 0 and will not have any value at x = 0. Thus f”(x) function graph will be x axis for negative values of x and y = 2 horizontal line for positive values of x with the function being discontinuous at x = 0 and not having any value at x = 0. The function f”(x) is also differentiable at all values of x except x = 0, the value of this differential will be 0 for all values of x except x = 0.
Problem 2.6.17
A function f when not differentiable at x = 0, not only indicates that the graph has sharp corner at x = 0. The function may also be not continuous at x = 0, in such case too the function will not be differentiable at x = 0.
Problem 2.6.20
The function y = x^(1/3) is a continuous function that too one to one function, thus the function is an invertible function and at the same time the function is not differentiable at x = 0.
Problem 2.6.22
The statement that there is a function continuous on [1,5] but not differentiable at x = 3 is definitely true. Example for this statement is the function y = |x-3| which is continuous over [1,5] but not differentiable at x = 3.
Text Section 5.1
Problem 5.1.1
(a) The rectangles represent a left sum. The rectangle for every delta t rectangle has a height associated with the value of the function related to the left x value of the delta t rectangle.
(b) Since the rectangles height is above the functions values thus the rectangles lead to an upper estimate.
(c) The value of n is the number of rectangles which is 6
(d) Delta t = 12 - 0 / n = 12 / 6 = 2 sec
(e) An approximate value of the upper estimate will be = 4*2 + 2.9*2 + 2*2 + 1.5*2 + 1*2 + 0.8*2
= 8 + 5.8 + 4 + 3 + 2 + 1.6 = 24.4
Problem 5.1.2
(a) Given that, the initial value of t = 0 time unit and final value of t = 12 time unit
(i) n = 4
Thus detla t = (12-0)/4 = 3
Thus t0 = 0, t1 = 3, t2 = 6, t3 = 9 and t4 = 12 time units.
Given f(t0) = 34 distance unit/time unit, f(t1) = 37 distance unit/time unit, f(t2) = 38 distance unit/time unit, f(t3) = 40 distance unit/time unit and f(t4) = 45 distance unit/time unit.
Since the function is increasing the upper estimate for the total distance travelled using n = 4 will be
= 37*3 + 38*3 + 40*3 + 45*3 = 480 distance unit
(ii) n = 2
Thus delta t = (12-0)/2 = 6
Thus t0 = 0, t1 = 6, t2 = 12 time units
Given f(t0) = 34, f(t1) = 38 and f(t2) = 45 distance unit/time unit
Since the function is increasing the upper estimate for the total distance travelled using n = 2 will be
= 38*6 + 45*6 = 498 distance unit
(b) The answer in part (i) is more accurate. This is because smaller the time interval used for calculating the estimates the better is the result. Also another way to look at it is that smaller the upper estimate as compared to the other more accurate the result is.
(c) For n = 4 the lower estimate of the total distance traveled will be
= 34*3 + 37*3 + 38*3 + 40*3 = 447 distance unit.
Problem 5.1.3
Initial time = 2 time unit
Final time = 12 time unit
Given n = 5
Thus delta t = (12 - 2) / 5 = 2
Thus t0 = 2, t1 = 4, t2 = 6, t3 = 8, t4 = 10 and t5 = 12
Also f(t0) = 44, f(t1) = 42, f(t2) = 41, f(t3) = 40, f(t4) = 37 and f(t5) = 35 distance unit/time unit.
(a) Since the function is decreasing the upper estimate would be
= 44*2 + 42*2 + 41*2 + 40*2 + 37*2 = 408 distance unit.
(b) Since the function is decreasing the lower estimate would be
= 42*2 + 41*2 + 40*2 + 37*2 + 35*2 = 390 distance unit.
Problem 5.1.4
(a) The lower estimate of the distance the car traveled after the brakes were applied will be = 60*1 + 40*1 + 25*1 + 10*1 + 0*1 = 60 + 40 + 25 + 10 + 0 = 135 feet
The upper estimate of the distance the car traveled after the brakes were applied will be = 88*1 + 60*1 + 40*1 + 25*1 + 10*1 = 88 + 60 + 40 + 25 + 10 = 223 feet.
(b) On the sketch of velocity vs. time graph the lower estimate will be the sum of the area of the triangles formed by using delta t of 1 and the height equal to the value of the function at the right hand x value of the delta t rectangle formed since it is a decreasing graph.
Whereas the upper estimate will be the total area of the triangles formed by using the delta t of 1 and the height equal to the value of the function at the left hand x value of the delta t rectangle formed since it is a decreasing graph.
(c) The difference between the upper and the lower estimate is 223 feet - 135 feet = 88 feet. On the graph this difference appears as the excess area of the rectangle formed for upper estimates as compared to the rectangle formed for lower estimates. All the rectangles are stacked aside and the height of this rectangle formed is seen to be 88 feet/sec and the width/base is 1 sec, thus the excess area which is the difference between the upper and the lower estimate is 88 feet/sec* 1 sec = 88 feet.
Problem 5.1.7
(a) Since the velocity is always positive - that is the sign of the velocity values does not change thus the particle is always moving in the same direction. The particle for t = - 3 to t = -1.7(approx.) has increasing velocity and thus is speeding up whereas for remaining t values the velocity is decreasing thus the particle is slowing down.
(b) For the under and over estimates I use intervals of x = 1. Thus from the graph we have f(-3) = 1.25, f(-2) = 2.25 f(-1) = 2.83, f(0) = 2.67, f(1) = 1.83, f(2) = 0.83 and f(3) = 0.33, these values are approximated values. Thus the underestimate for the function will be 1.25 + 2.25 + 2.67 + 1.83 + 0.83 + 0.33 = 9.16 and the overestimate for the same t interval is 2.25 + 2.83 + 2.83 + 2.67 + 1.83 + 0.83 = 13.24. Thus the underestimate is 9.16 and the overestimate is 13.24.
Problem 5.1.10
Given function is the velocity function in cm/sec in terms of time in sec. Even though the velocity is decreasing put still is positive, thus the distance is positive throughout (that is the particle moves in a straight direction). The area of the triangle will be the distance, that the particle moves(that is change in position) within this time interval = (1/2)*(5 sec)*(10 cm/sec) = 25 cm.
Problem 5.1.13
(a) Initial time t = 15 sec, final time = 23 sec and n = 4
Thus delta t = 23 sec - 15 sec / 4 = 2 sec
Thus t0 = 15 sec, t1 = 17 sec, t2 = 19 sec, t3 = 21 sec, t4 = 23 sec
Thus f(t0) = 10, f(t1) = 13, f(t2) = 18, f(t3) = 20, f(t4) = 30
(b) The left sum using n = 4 is 10*2 + 13*2 + 18*2 + 20*2 = 122
And the right sum using n = 4 is 13*2 + 18*2 + 20*2 + 30*2 = 162
(c) Here given that n = 2 gives that delta t = 23 sec - 15 sec / 2 = 4 sec
Thus t0 = 15 sec, t1 = 19 sec and t2 = 23 sec
Thus f(t0) = 10, f(t1) = 18 and f(t2) = 30
(d) The left sum using n = 2 is 10*4 + 18*4 = 112
And the right sum using n = 2 is 18*4 + 30*4 = 192
Problem 5.1.15
(a)
For the first half hour,
Time since start = 0 min = 0 hour, speed = 12 mph
Time since start = 15 min = 0.25 hour, speed = 11 mph
Time since start = 30 min = 0.5 hour, speed = 10 mph
Given that Roger’s speed is never increasing,
The lower estimates for the distance Roger ran during the first half hour = 11*0.25 + 10*0.25 = 5.25 miles
The upper estimate for the distance Roger ran during the first half hour = 12*0.25 + 11*0.25 = 5.75 miles
(b)
Time since start = 0 min = 0 hour, speed = 12 mph
Time since start = 15 min = 0.25 hour, speed = 11 mph
Time since start = 30 min = 0.5 hour, speed = 10 mph
Time since start = 45 min = 0.75 hour, speed = 10 mph
Time since start = 60 min = 1 hour, speed = 8 mph
Time since start = 75 min = 1.25 hour, speed = 7 mph
Time since start = 90 min = 1.5 hour, speed = 0 mph
The lower estimate for the distance Roger ran in total during the entire hour and a half will be
= 11*0.25 + 10*0.25 + 10*0.25 + 8*0.25 + 7*0.25 + 0*0.25 = 11.5 mile
The upper estimate for the distance Roger ran in total during the entire hour and a half will be
= 12*0.25 + 11*0.25 + 10*0.25 + 10*0.25 + 8*0.25 + 7*0.25 = 14.5 mile
(c) Delta t for the rectangles is the questions is 0.25. The difference between the velocity at the beginning and end of the observation period is 12 mph. Now if the time between successive measurements is delta t, then the difference between the upper and lower estimates is (12)*(delta t). We want 12*(delta t) < 0.1, thus delta t < 0.00833 hours.
Thus Jeff would have to measure Roger’s speed 0.00833 hours regularly to find lower and upper estimates within 0.1 mile of the actual distance he ran.
Problem 5.1.16
Given function f(t) = 6 - 2t cm/sec. We can use a graph of f(t) to find the exact change in position of the particle from t = 0 to t = 4 seconds we use the area under the graph.
The value of the function at t = 0 seconds is f(0) = 6 cm/sec, the value of the function at t = 4 seconds is f(4) = 6 - 2(4) = -2 cm/sec.
Since the sign of the velocity function changes from + to - we need to calculate the area of positive value of v and negative values separately.
Thus the value of the function becomes 0 when t = 3 sec.
Thus the change in position of the particle from t = 0 to t = 3 sec will be = (1/2)*(3)*(6) = 9 cm
Also next the change in position of the particle from t = 3 sec to t = 4 sec will be = (1/2)*(-2)*(1) = -1 cm
Thus the net change in the position of the particle from t = 0 sec to t = 4 sec is 8 cm.
Problem 5.1.19
Given f(t) = sin(t) where a = 0 and b = pi/2 and n = 100.
f(a) = sin(0) = 0 and f(b) = sin(pi/2) = 1
The value of the delta t for the function will thus be = (b - a)/n = pi/2/100 = pi/200
The difference between the highest and lowest value of the function = 1 - 0 = 1
Thus the difference between the lower and the upper estimates of the distance traveled at velocity f(t) on the interval a < t < b for n subdivisions will be
= (f(b)-f(a))*(delta t) = 1 * pi/200 = pi/200 = 0.0157 approx
Problem 5.1.22
Given that positive velocities take you away from the home and negative velocities take you towards home. Thus we calculate the distance covered by positive velocities and that covered by negative velocities separately. Thus for positive velocities the total distance travelled will be the area under the curve for the positive velocities curve. The area under curve will be area of the square under the curve. There are 6 perfect square and another 2 square formed by joining the half squares and another square of approximately 80% area as compared to the original perfect square. Thus the area under the curve will be 60 km + 20 km + 8 km = 88 km. The area under the curve for negative velocities will take you back towards home. The area of the squares under the negative velocities curve will be 10 cm + 7.5 cm + 7.5 cm + 1 cm = 26 cm. Thus the total distance travelled by the person at the end of 5 hours will be 88 km - 26 km = 62 km.
The person will be farthest from the home when the positive velocity becomes 0 and negative velocity starts returning back home. This is when t = 3 hours. The area under the curve for positive values of velocity is 88 km. Thus the person is farthest from the home at t = 3 hours and he is 88 km away from the home at that time.
Problem 5.1.25
(a) It is given that the time taken by Car 2 to start is 2 hrs after Car A. From the graph it can be seen that the intervals are regular for both car A and car B and thus approximately car A travels for 2hrs*4 = 8 hrs and car B travels for 2hrs*2 = 4 hrs.
(b) We can see that car A and car B increases its velocity at the same rate. Thus in 2 hrs if car B reaches a maximum velocity of 50 km/hr, car A will reach a maximum velocity of 100 km/hr in 4 hours.
(c) To consider how much Car B travels in its 4 hours we find the area under the curve, thus the area is (1/2)*(4 hr)*(50 km/hr) = 100 km, thus car B travels a total of 100 km
For Car A area under curve is (1/2)*(8 hrs)*(100 km/hr) = 400 km. Thus car A travels approximately 400 km in 8 hrs.
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
&#This looks good. See my notes. Let me know if you have any questions. &#