math 163
OK I started working on a pratice test for precal and heres all the problems I'm having with it.
Number 1Interest on principle is $7000 accumulates annually at the rate of 3.7percent per yearHow much money will there be in 1, 2, 3, and 10 years after the inital investment?Ok on that one I know that I would take 7000x.037= $259 interest for the first year. So you'd have 7259Ok but on the second year would you take 7000 + (259x2) or would you take 7259x.037= 268.58 and then add that.
You would do the latter. Figure the next year's interest on the principle at the end of the previous year, which would include the previous year's interest.
When you say principle would I only get the interest on that inital 7000 dollars or the interest and the principle?
What function P(t) gives principle as a function of the number t of years?P(t)=P + interest(t) Is that correct?
We haven't covered this yet, but the function would be 7000 * 1.037^t. You'll see why by the end of the week.
What are the growth rate and growth factor of this function? The growth rate is .037 percent and I don't know about the growth factor. Is this correct?
That is the growth rate. The growth factor is 1 + growth rate = 1.037. Again we haven't covered this, so you're doing really well.
Number 2
Solve for x, using the laws of exponenets and showing each step:
5(5x/3)^(-3/2)=32
Would I start by dividing by 5 which would give me
(5x/3)^(-3/2)=6.4
Then I have no idea how to deal with that exponent.
Could you help me?
Yup. you're off to a great start.
We'll be talking about this in class over the next day or two also.
(a^b)^c = a^(bc), so to get rid of that -3/2 power you have to take the -2/3 power of both sides.
You would get
( (5x/3)^(-3/2) ) ^(-2/3) = 6.4^(-2/3) or
(5x / 3)^-( (3/2) * (-2/3) ) = 6/4^(-2/3) or
(5x / 3)^ 1 = 6.4^(-2/3) so that
5 x / 3 = 6.4^(-2/3).
Use your calculator to evaluate the right-hand side.
Then proceed to solve for x (multiply both sides by 3/5).
The next problem is
x^5/5=32
So is that x up to the fifth divided by five?
exponentiation precedes division so it's the same as (x^5) / 5.
I don't know whether I am supposed to multiply everything through by five or not. Then would I have the square root five of 32?
You would multiply by 5.
I understand what you are saying with 'square root five' but 'square' is the inverse of the second power, and you wouldn't use that here. You would read that 'fifth root of ...'. You take the fifth root of both sides.
The way I'll be explaining it is that you take the 1/5 power of both sides, which is the same thing as the fifth root, and you use multiplication of exponents as in the preceding problem.
number 3
What equation would you have to solve to find the doubling time, starting at t=3 of a population that starts at 700 organisms and grows at a annual rate of 7.1%?
I am completely unsure of how to do this problem!
Would it be 700 + (3x2)(7.1)?
We haven't covered this yet. But you already understand a good bit so I'll explain:
At clock time t the number of organisms is P(t) = 700 * 1.071^t. Note that .071 is growth rate and 1.071 is growth factor.
The doubling time is the time required for the population to double. Initially the population is 700, so you would solve the equation
700 * 1.071^t = 1400
for t. The solution would be the doubling time.
Right now (and on Test 1) you could find this doubling time by trial and error. Soon you'll be able to solve it by logs.
The thing about 'doubling time starting at t = 3' shouldn't be on this test. Just writing down the doubling time equation would be enough on Test 1.
number 4
At clock times 13.4, 20.1, 26.8, 33.5 seconds, we observe water depths of 28, 20.9, 16, and 13.4cm.
At what average rate does the depth change during each time interval?
Would I just subtract (13.4-16)/(33.5-26.8) and do this between each set?
Right.
Sketch a graph of depth vs. clock time, which i can do.
Then it says use a sketch to explain what the slope of this graph between 13.4 and 20.1 seconds represents the average rate at which depth changes during this time interval. Why would it be between 13.4 and 20.1 seconds instead of 20.1 and 26.8 which is the middle set of times?
The slope corresponding to a given time interval is rise / run. Since rise is change in depth and run is change in clock time, the slope is equal to (change in depth) / (change in clock time), and this is the definition of the average rate of change of depth with respect to clock time.
The same question could have been asked for the middle interval, or it could have been asked for the last interval. This question could be asked of any interval.
If f(x)=x^2, what are the vertex and the three basic points of the graphs of f(x-.75), f(x)-1.35, 5f(x), and 5f(x-.75)+1.35
The three basic points for y=x^2
are (-1,1),(0,0), and (1,1)
So for y=(x-.75) the vertex is (0,.75) and the three basic points are
(-1,1.75)
(0,.75)
(1,1.75)
No, the function f(x-.75) shifts the f(x) function .75 units in the x direction.
If you shift the three points .75 units in the x direction they become
(-1 + .75, 1), (0 + .75, 0) and (1 + .75, 1), which is the same as
(-.25, 1), (.75, 0) and (1.75, 1).
ok for y=x^2-1.35
the vertex is (-1.35,0)
and the three basic points are
(-2.35, 1)
(-1.35,0)
(-.35,1)
This function shifts the graph -1.35 units in the y direction, shifting the basic points to
(-1, 1 - 1.35), (0, 0-1.35) and (1, 1-1.35), which is the same as
(-1, -.35), (0, -1.35) and (1, -.35).
and for 5f(x)=x^2
the vertex is (0,0)
and the three basic points are
(-5,1)
(0,0)
(5,1)
This function stretches the graph vertically be factor 5, taking the basic points to
(-1, 1 * 5), (0, 0 * 5) and (1, 1 * 5), which is the same as
(-1, 5), (0, 0) and (1, 5).
and for the 5f(x-.75)^2+1.35
the vertex is (0,-2.4)
and the three basic points are
(-1,-7.4)
(0,-2.4)
(1,2.6)
Is this all correct?
Every basic point is vertically stretched by factor 5, horizontally shifted .75 units and vertically shifted 1.35 units.
So the points
(-1, 1), (0, 0) and (1, 1) become
(-1, 5), (0, 0) and (1, 5) by the vertical shift,
(-.25, 5), (.75, 0) and (1.75, 5) by the horizontal shift and
(-.25, 6.35), (.75, 1.35) and (1.75, 6.35) by the vertical shift.
Just checking one more time, when its (x-1) the one is positive and a horizontal shift, when its x-1 its a negative vertical shift.
number 5
Find the equation of a line through (6,3) and (4,4) by each of the following methods:
-substitute the coordinate points into the form of a straight line and solve the equations for the appropriate parameters
-use the slope=slope form, complete with explanation and a labeled graph
Ok I'm not sure which is which here, I just solved, I hope you can clarify for me which why I did it and how to do the other way.
First I had to find the slope for I took rise over run and took the two points and (4-3)/(4-6) which gives me a slope of (-1/2).
So I took the straight line equation y=mx+b and plugged in the x and y and slope and found the b, I did this for both sets just to make sure I had done everything right.
I got 3=(-1/2)(6)+b and found b to be equal to 6
Does all of this look right to you?
This is correct, but the method is not as specified.
To substitute the coordinate points into the form of a straight line and solve the equations for the appropriate parameters:
Your form is y = m x + b.
Substituting the two points into this form you get the two equations
3 = m * 6 + b
4 = m * 4 + b.
Subtracting the second equation from the first you get
-1 = 2 m, which you solve for m to get
m = -1/2.
Substituting this into the first equation you get
3 = -1/2 * 6 + b, which you solve for b to get
b = 6.
Substituting m and b into the form you get
y = -1/2 x + 6.
To use the slope=slope form, complete with explanation and a labeled graph:
You plot the points (6, 3) and (4, 4) on a graph.
Draw a straight line through and beyond these points.
Put the point (x, y) on the line, somewhere that isn't too close to the two points and isn't between them so you can label the graph easily.
Sketch the 'slope triangle' between (6, 3) and (4, 4). Label your rise and your run. You'll probably get a rise of -1 and a run of 2; if not you'll get rise of 1 and a run of -2. Either way the slope is -1/2.
Sketch the 'slope triangle' between either of your points and (x, y). Label the rise and the run. If you use the point (6, 3) the rise will be (y - 3) and the run will be (x - 6). So the slope is (y - 3) / (x - 6).
You have two expressions for slope. One is -1/2 and the other is (y - 3) / (x - 6).
These slopes both represent the slope of the line.
So slope = slope:
(y - 3) / (x - 6) = -1/2.
This is your slope - slope equation.
You can solve this equation for y, and on some problems you are asked to do so. To do this:
Multiply both sides by (x - 6) to get
y - 3 = -1/2 ( x - 6). Add 3 to both sides:
y = -1/2 ( x - 6) + 3. Distribute and add things up:
y = -1/2 x + 3 + 3 or
y = -1/2 x + 6.
If your graph represents the range of a water stream in cm, vs. clock time in seconds:
-what does the slope of the graph represent?
slope represents a decrease of the water steam as clock time increases.
slope is change in range / change in clock time = average rate of change of range with respect to clock time.
-what is the range of clock time t=11 seconds? Would the range just be 11?
11-0=11? or would i have to take my y=(-1/2)11+6= (1/2)
Your function is y = -1/2 x + 6.
y represents range.
x represents clock time.
If clock time is 11 then
y = -1/2 * 11 + 6 = -5.5 + 6 = .5.
The range is 5.
-What is the clock time at which the range is 11cm?
would i just take 11=(-1/2)x+6 and solve to get -10.
Exactly. y represents range, so if range is 11 you let y = 11. x represents clock time so you solve for x.
Good thinking.
Number 6
Explain how to use two simultaneous linear equations, obtained from two given pointsm to obtain the equation of the line through the two points?
Ok I'm not sure how to find slope here, are you giving me two points or two sets of coordinates?
I am confused, could you explain this one to me?
My explanation on the preceding problem should clarfy this. Let me know if it doesn't.
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.
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.
number 8
Find the equation of the straight line through the t=7sec and the t=18 sec points of the quadratic fucntion depth(t)=.1t^2+-2.5+32 where depth is in centimeters when time is in seconds.
When I substitute the times I'm given into the equation to find depth I get
ok for time 7 depth is 19.4
(19.4,7)
Then to get slope I take
(19.4-19.4)/(7-18)
That gives me zero, this is line has no slope!
You get depth 19.4 when t = 7.
What depth do you get when t = 18? You don't say, but I'll bet you got 19.4.
So the change in depth is indeed 0, as you say.
You don't say that the line has no slope, you say it has a definite slope, which is 0.
If the line had no slope then there wouldn't be a number for the slope. 0 is a perfectly good number.
What does the slope of your line tell you about the depth function?
Would I just say that the line is straight, not increasing or decreasing?
A straight line can be increasing or decreasing. Or it can be horizontal, which is the case here when it has 0 slope.
Evaulate both the linear and the quadratic depth function at four equally spaced points between t=7 and t=18 seconds. HOw closely does the linear function approximate the quadratic function at each of these times.
If there is no slope for the linear model the line will not change vertically at all, it will just increase as time increases no matter what time you'd use.
that is correct but you should get the linear function, which is found by any method to be
y = 0 x + 19.4, or just plain
y = 19.4.
For the quadratic depth function the depth would remain at a constant 32 because there is no slope
The quadratic function is depth(t)=.1t^2+-2.5+32. This will not be the same at all 4 points.
Are these correct?