course Mth 163
At clock times 19.8, 29.7, 39.6 and 49.5 sec, we observe water depths of 12.4, 1.9, -4.1 and -5.6 cm. •At what average rate does the depth change during each time interval?
Answer: For the first two points before the depth becomes negative there is an average change or slope of 1.06 approximately but then when the depth becomes negative the slope decreases and the average change or slope between the two points was about .15.
On the first interval the ave rate is
ave rate = change in depth / change in clock time = (1.9 cm - 12.4 cm) / (29.7 s - 19.8 s) = -1.06 cm/s.
You should show your reasoning in a manner similar to that shown above.
Your answer 1.06 is close to the correct answer but differs in two important ways:
The correct number is -1.06.
The result has units, which are cm / s.
The correct result is thus -1.06 cm/s.
Sketch a graph of depth vs. clock time.
•Use a sketch to explain why the slope of this graph between 19.8 and 29.7 sec represents the average rate at which depth changes during this time interval.
Answer: These are the more accurate points because they are not negative. They represent a much better and more precise look at what we are measuring.
To explain the meaning of a slope between two points you need to do the following:
Explain the meaning of the rise between the points.
Explain the meaning of the run between the points.
Explain what it means to divide the quantity represented by the rise by the quantity represented by the run. (This quantitiy is in general an average rate of change of one quantity with respect to another.)
Explain what the slope therefore means.
If f(x) = x^2, what are the vertex and the three basic points of the graphs of
f(x- .25)…The vertex is (-.25,0) left and right points (-1.25,1) (.75,1)
f(x) - .65 …. The vertex is (0,.65) left and right points (-1,1.65) (1,1.65)
The basic points (-1, 1), (0, 0) and (1, 1) of the graph of f(x) = x^2 are shifted .65 units to the right.
The vertex (0, 0) shifts to (.65, 0).
The point (-1, 1) shifts to (-1 + .65, 1) = (-.35, 1).
The point (1, 1) shifts to (1 + .65, 1) = (1.65, 1).
.25 f(x)…The vertex is (0,0) left and right points (-1,.25) (1,.25)
.25 f(x- .25) + -.65…. The vertex is (0,-.65) left and right points (-1,-.509) (1,-.509)
f(x - .25) shifts the basic points .25 units to the right, to (-.75, 1), (.25, 0) and (1.25, 1).
.25 f(x - .25) results in a vertical stretch by factor .25, so the points become (-.75, .25), (.25, 0) and (1.25, .25).
.25 f(x - .25) + -.65 shifts the basic points .65 units lower to (-.75, -.40), (.25, -.65), (1.25, -.40).
See my notes and be sure you ask if there's something you don't understand.
course
f(28.43875)=.02(28.43875)^2+-2.57(28.43875)+99= 42.08766253
f(12)=.02(12)^2+-2.57(12)+99= 71.04
f(81.43875)=.02(81.43875)^2+-2.57(81.43875)+99= 22.34781253
f(32)=.02(32)^2+-2.57(32)+99= 37.24
Some of these are correct and some are not. However I can't tell which without knowing specifically how the questions were stated.
Please insert these answers into a copy of the questions and resubmit.
course Mth 163
At clock time t = 10 sec the depth of water in a container is 79 cm, while at clock time t = 26 sec the depth is 56 cm. Plot the corresponding points on a graph of depth vs. clock time and determine the slope of the straight line segment connecting these points. Explain why this slope represents the average rate at which the water depth changes over this time interval.
Answer: The slope would be -1.44 or just 1.44 but the line is negatively sloped this is an accurate view of the average times because it is of two points that are spread out from each other and this number is derived from the x and y coordinates which is how it can show us the time changed through each checkpoint.
That would be -1.44 cm / sec, and would represent change in depth / change in clock time, i.e., average rate of change of depth with respect to clock time.
For the quadratic function y = f(t) = .021 t2 + -2.18 t + 99, determine the average rate of change of y with respect to t, between clock times t = 26 and t = 31.
Answer: 51.0165 is the average change in the y value for this range of functions.
This answer is not correct.
Whether correct or not, you need to show the details of how you got your solution. This is especially important when your solution is incorrect; I can't tell what you did to get the result you got so I can't help you figure out where and how you went wrong.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end). &#