math 163
On the question where you're asking us to show the two ways to find the slope, on the second way to find the slope you said to graph the two points given and the slope line through those two points and beyond, and then I labeled the rise and the run on a triangle below the slope line.
The slope was -1/2
so I got -1 rise and run as 2
Would this be the correct way to graph this problem in order to recieve credit?
I need to see the full statement of that question. Please resubmit this question and include the full statement.
number 7
Sketch a graph of basic exponential function y=2^x. Sketch the graph of this function stretched vertically by factor -1.58 then shifted -2.68units vertically.
Show that the graph is different than that obtained if the vertical shift precedes the vertical stretch. Give the algebraic form of the resulting function.
Ok the three basic points to that equation is (-1, .5) (0,1) (1,2)
So to vertically stretch the graph I would multiply each of the Y points by -1.58 Then I would add a -2.68 to each of those. Is this correct?
Right. Your points would become
(-1, -.79), (0, -1.58) and (1, -3.16) on the vertical stretch, then (-1, -3.47), (0, -4.26) and (1, -5.84) on the vertical shift, assuming my mental arithmetic is correct.
To show that the graphs are different I would just do this in opposite order and graoh each
If you vertically shift first then the points become
(-1, -2.18), (0, -1.68) and (1, -.68) after the vertical shift, then after the vertical stretch you will get (-1, -2.18 * (-1.58) ), (0, -1.68 * (-1.58) ) and (1, -.68 * (-1.58) ) (you can do the multiplication; all y coordinates will end up positive).
The resulting algebraic form would be y=-1.58(x)-2.68
The form if vertical stretch goes first would be
y = -1.58 * 2^x after the vertical stretch then y = -1.58 * 2^x - 2.68 after the subsequent vertical shift.
If the shift is first then you get
y = 2^x - 2.68 after the shift and y = -1.58 * ( 2^x - 2.68) after the vertical stretch (you should then distribute the multiplication to get y = -1.58 * 2^x -1.58 * (-2.68), where again you can do that last multiplication on your calculator).
Correct?
Then the problem asks at clock times t= 15, 30, 45, and 60 seconds the depth of water in a uniform cylinder was observed to be 69.25, 58, 51.25, and 49cm. At what average rate was the depth changing during each of the three time intervals? Look at the rates you have calculated and predict what the next average rate would be. If your prediction is correct, then what will be the depth at t=75 seconds. ok so i would just take (15-30)/(69.25-58) and that would give me -1.3333
then I would repeat this for each interval. (30-45)/(58-51.25)=-2.222222222 (45-60)/51.25-49)=-6.6666666666667
You divided change in t by change in depth, which is average rate of change of clock time with respect to depth, not ave rate of change of depth with respect to clock time.
You should reverse your division.
If you do you will find that the rates are
-.75, -.45 and -.15.
Do you see why the next rate would be +.15?
Ok but now I don't know how to calculate what my next average rate would be.
** How much change is there from ave rate -.75 to ave rate -.45?
How much from -.45 to -.15?
How much change do you think there will be between -.15 and the next ave rate?
So what will be the next ave rate?
If the average rate of change of depth with respect to clock time is .15, then how much depth change will there be over the next 15-second interval?**
And I don't know how to find the depth at time 75 seconds
Once you see why the next average rate is .15, meaning .15 cm / sec, then how much change would there be in depth between t = 60 sec and t = 75 sec? Answer this before you look at the rest of the solution.
Between t = 60 sec and t = 75 sec the change in clock time is 15 sec. At an average rate of .15 cm/sec, the change in depth would therefore be
change in depth = ave rate of change of depth with respect to clock time * change in clock time = .15 cm/s * 15 sec = 2.25 cm.
So the depth would change by 2.25 cm.
The t = 60 depth is 49 cm, so the t = 75 depth would be 49 cm + 2.25 cm = 51.25 cm.
On this question I understand everything until you come to the last part
I understand why you multiply .15*15 and get 2.25
But I don't understand if time is increasing while the depth decreases why you added 2.25 to 49 instead of subtracting it
Is this correct?
** The average rate of depth change was originally negative, meaning that depth was decreasing.
However if your follow the pattern, you find that the average rate is now positive, and that will result in an increase in depth.
That doesn't mean that the actual container will increase in depth. Only that if the pattern continues, it will. If the container is leaking without being refilled, then the model will cease being valid when the last of the water has leaked out somewhere between t = 60 sec and t = 75 sec.