to Solutions Web-Based Homework 3 This is the 'semi-detailed' solution set for the third web-based homework. Hopefully you will nd it useful and instructive. For now and in the future, these solutions will be available only after the web-based homework is due. Hopefully, I'll be sending you the solutions the day after each set is due. Please take advantage of these solutions, the recommended homework problems in your text, my oce hours, and our recitation. Based on the averages of the rst midterm, we as a class need to strengthen the basics while moving on to other topics in calculus. Problem 1: The answer to this question is simple. All that you need to do, is nd the slope (rate of change) of the line, determined by the two points specied. However they only give you thex 0 values of the two points, you must use the table to determine the corresponding y 0 values of the two points. Hence you will be able to calculate the slope of the line, and (if you wanted to) then the y 0 intercept, and write an equation of the form, y = mx + b: So, at t = 38, H (t) = 2700, and at t = 42, H (t) = 2960. So the slope will be: y 0 y2 m= 1 x 1 0 x2 2960 0 2700 260 130 = = = 65 m= 42 0 38 4 2 Thus the answer is 65 hearbeats per minute. Problem 2: This problem asks you to nd the equation of a line, which is tangent to a curve, P , at some point on P . The information given tells you that the line is tangent at the point (2; 22), which is on the graph of P. They also tell you that the slope of this tangent line is 10. So this makes life easy, all we need to do is nd the y 0 intercept of the line, and write the equation of the tangent line: y = mx + b: 1
In fact we know that the equation of the line is: y = 10x + b. Now all we need to do is 'plug in' the point, (2; 22) and we will have found b. So, 22 = 10 3 2 + b 22 0 20 = b
b=2
So
y = 10x + 2:
Problem 3: To nd the average velocity, one must use a formula like the one below: If v(t) is some velocity function, then the average velocity between some time t = and a later time t = is:
v() 0 v ( ) 0
So using the equation, one can calculuate what the average velocity is of the given function, between t = 3 and t = 3:2 and t = 3 and t = 3:02. The given velocity function is:
y = 35t 0 13t2
Now, and So,
y(3) = 35 3 3 0 13 3 (3)2 = 012ft=s y(3:2) = 35 3 (3:2) 0 13 3 (3:2)2 = 021:12ft=s
y (3:2) 0 y (3) 021:12 0 (012) 09:12 = = = 045:6 3:2 0 3 3:2 0 3 :2 And, if we replace with 3.02 while keeping = 3, then the average velocity will be: y(3:02) 0 y(3) = 3:02 0 3
012:8652 0 (012) = 0:8652 = 043:26: 3:02 0 3 :02
2
These are the respective answers for the average velocities, for 3 t 3:2, and 3 t 3:02. Problem 4: The answer to this problem is to nd the average velocity between t = 1 and t = 2 and then the average velocity between t = 2 and t = 3. These two numbers will be used to determine an approximation to the instantaneous velocity between t = 1 andt = 3. So, the average velocity between 1 and 2 is: 32 0 12 = 20 201 and the average velocity between 2 and 3 is: 78 0 32 = 46: 302 Now, the approximate instantaneous velocity between 1 and 3 is: 78 0 32 0 (32 0 12) 66 = = 33 3 0 2 0 (2 0 1) 2 And this is the answer. Problem 5: If and if lim f (x) = 4
x 50
!
x5
lim f (x) = L
i.e. if the limit exists as x approaces 5, this tells us that the limit of f (x) as x approaches 5 from the right equals the limit of f (x) as x approaches 5 from the left. So lim f (x) = 4:
x5
!
As x approaches 5 the limit must be 4. This is all based on what the denition of the "total" or "two-sided" limit is. A function has a "two-sided limit" at some point x0 if the limit from the right and left, while approaching x0 3
!
equal each other. Hence if the "two-sided" limit exists, then the right and left sided limits equal each other. Problem 6: The limit of f (x), as x approaches 1 from the right is 1:5. This is found by following the graph from the right towards x = 1, and looking at the y 0 value of the function at x = 1. By inspection, one can see that the y 0 value from the right will be 1:5. Problem 7: This problem is very similar to the previous one, problem 6. The idea is exactly the same, except for the fact that now, one must follow the graph from the left and move towards x = 1, and then look at the corresponding y 0 value of the function. If one does this, it is quite clear that the limit of f (x) as x approaches 1 from the left (x ! 10 ) is 2. By inspection, one can see that this is the y 0 value that the function approaches. Problem 8: The correct graph for the is piece-wise function is graph 3 (the last one). This is somewhat clear by inspection as well. The graph f (x) is the line y = x for 01 x < 2. However for x 2 the function looks like the parabola y = x2 , but the parabola is shifted 2 units to the right, since there is a 02 inside the parentheses. Finally, for x < 01 the function looks like the line y = 2 0 x. Clearly, the only graph that satises even the part rst (that between -1 and 2 the function looks like the line y = x) is the last one. Problem 9: The answer to this problem can be found in two ways. The rst way is simpler, but only works since we saw the graph of this function in the previous problem (the second way, involves one graphing the function himself, or analytically looking at the endpoints of the intervals of the "partial domains" of the functions which comprise the piece-wise function and seeing if the limits from the left and right mathc up: e.g. looking if f (x) = x matches f (x) = (x 0 2)2 at x = 2). The rst method only requires one to inspect the 4
graph of the function and then determine the x 0 values of the function for which left and right limits do not match. Those x 0 values are the values for which the limit of f (x) does not exist for arbitrary a. The subsequent x 0 values for which limits exist for the function f (x) are: (01; 01) [ (01; 02) [ (2; 1): The reason that these are the x 0 values at which limits exists for f (x) is simply because these are x 0 values for which the left and right sided limits match (i.e. they equal each other). If one is clever, he would notice that the end points of each of these intervals is essentially the endpoints of the intervals which specify the dierent types of functions which comprise the piece-wise function f (x). This is not the case always. It is the case here, because the piece-wise functions do not "connect" to one another at the endpoints of the intervals for which they are dened. For example, for x 2, f (x) = (x 0 2)2 ,and for 01 x < 2 f (x) = x. Hence, the limit of f (x) as x approaches 2 from the left and right do not equal each other. From the left the limit of f (x) equals 2, and from the right the limit of f (x) equals 0. Problem 10: When I went online, my computer wouldn't show me the choices for the graphs. So I can't give you the answer to this one. Hopefully, I will have covered the answer in class, if someone asks for it. Problem 11: The
x01 =1 x!1 x3 0 1 as x ! 1, f (x) is 1. The answer is found by looking at the table, which indicates the behavior of the function f (x), as x gets closer and closer to 1.
lim f (x) = 3 3
Problem 12: The method of nding the answer to this problem is exactly the same as that of the previous problem. All one needs to do is look at the behavior of the function as x ! 3. This behavior is given by the table. As x approaches 5
3, f (x) approaches .333333. That is, lim
x + e 3 0 ex = :333333 x!3 x2
Problem 13: There is one sure-re method by which to calculate limits (without the more advanced techniques which you are going to learn in the next month or so, in this course). An easy way to check if a limit exists, and if it does, what it's value is, is to appoximate the function at the limit value desired. For example in this case, the function is:
f (x) =
And we want to calclulate lim
tan(2x) : x
tan(2x) : x!0 x As one can see, we cannot directly calculate this limit, by...