Question

Need help with calculus ??!!
I'll post the picture up in a bit

Answer 1

need help with # 2&3

Answer 2

slope is change in heart rate over change in time

Answer 3

which on do you want to do?

Answer 4

2

Answer 5

ok from \(t=36\) to \(t=42\) is \(42-36=6\) seconds

Answer 6

and
\[2948-2530=418\]

Answer 7

so slope is
\[\frac{418}{6}=69.666...\]

Answer 8

it is all the same
change in \(y\) over change in \(x\) but in this case it is
change in heartbeat over change in time

Answer 9

why is it 42-36

Answer 10

6 seconds elapsed

Answer 11

minutes rather

Answer 12

ohh

Answer 13

just like computing the slope of a line
change in heart rate over change in time
it is boring and tedious, but that is what you are asked to do
i would whip out a calculator and knock these out

Answer 14

Do I do that too every problem?

Answer 15

yes, unfortunately
should keep you busy for a few minutes

Answer 16

i mean that is what you do for all parts of question 2, yes

Answer 17

Can 6ou help me on #3 when I finisg #2?

Answer 18

@satellite73
Did I get them right??
A) 69.67
B) 71.75
C) 71
D) 66

Answer 19

i can help you with #2 but cannot check these answers, unless you really really want me to

Answer 20

i know the first one is right

Answer 21

I think they are correct. What would I write for the conclusion?

Answer 22

here is b
http://www.wolframalpha.com/input/?i=%282948-2661%29%2F4

Answer 23

c
http://www.wolframalpha.com/input/?i=%282948-2806%29%2F2

Answer 24

conclusion? i guess it is between 66 and 71 beats per minute
i guess

Answer 25

Ohhhh I see (:

Answer 26

yeah your answers are right

Answer 27

Yayyy. #3 kindalooks like #2

Answer 28

yeah and it is another pain
get out a spread sheet

Answer 29

you got
\[\frac{1}{1-x}\] and you want the slope of the secant lines at \((2,-1)\) for various values of \(x\)

Answer 30

at \(x=1.9\) you get
\[\frac{1}{1-1.9}=\frac{1}{-.9}=-\frac{10}{9}\] so the slopes is
\[\frac{-\frac{10}{9}+1}{2-1.9}=\frac{-\frac{1}{9}}{.1}=-\frac{10}{9}\]

Answer 31

Where do you get -2+1?

Answer 32

for the first one?

Answer 33

Yes

Answer 34

ah damn i meant
\[\frac{y_2-y_1}{x_2-x_1}\] for each one
i really screwed that up sorry

Answer 35

Ohhhhh

Answer 36

but \(y_1=-1\) and \(x_1=2\) for each pair

Answer 37

For all equations?

Answer 38

so each pair looks like
\[\frac{y_2+1}{x_2-2}\]
you are given the \(x_2\) but you have to find the \(y_2\)

Answer 39

so to recap, if \(x_2=1.5\) then you need \(y_2=\frac{1}{1-1.5}\)

Answer 40

actually maybe i screwed this up so lets be careful here
\[x_2=1.5, y_2=\frac{1}{1-1.5}=\frac{1}{-.5}=-2\]

Answer 41

the slope will be
\[\frac{-2+1}{1.5-2}=\frac{-1}{-.5}=2\] yeah i messed up the sign sorry

Answer 42

next one is similar, but this time \(x_2=1.9\)

Answer 43

Its fine. Im learning (:

Answer 44

so first you need \(y_2=\frac{1}{1-1.9}=\frac{1}{.9}=\frac{10}{9}\)

Answer 45

man you might be learning but i cannot keep my signs straight

Answer 46

Its fine

Answer 47

\[y_2=\frac{1}{1-1.9}=\frac{1}{-.9}=-\frac{10}{9}\]

Answer 48

slope is
\[\frac{-\frac{10}{9}+1}{1.9-2}=\frac{-\frac{1}{9}}{-.1}=\frac{10}{9}\]

Answer 49

So for problem iii) its -1.01

Answer 50

and so on and so one ad infinitum

Answer 51

i think they should all be positive

Answer 52

they are going to be getting closer and closer to 1 i believe
you are being set up to take the derivative, which is change in y over the change is x but at one point instead of 2
i.e. you are going to compute a limit
and this tediousness will be over

Answer 53

here is iii
http://www.wolframalpha.com/input/?i=%281%2F%281-1.99%29%2B1%29%2F%281.99-2%29

Answer 54

wanna do some algebra? i mean with this problem?

Answer 55

I gotit right (:

Answer 56

Sure

Answer 57

yay

Answer 58

this is what you are computing each time
the change in y over the change in x at the point \((2,-1)\) so all the slopes look like this:
\[\large \frac{\frac{1}{1-x}+1}{x-2}\]

Answer 59

Shouldn't it be 2-x?

Answer 60

now lets simplify this expression a little, by multiplying top and bottom by \(1-x\)
you get
\[\large \frac{\frac{1}{1-x}+1}{x-2}\times \frac{1-x}{1-x}\]
\[=\frac{1+(1-x)}{(x-2)(1-x)}\]
\[=\frac{2-x}{(x-2)(1-x)}\]
\[=\frac{-1}{1-x}\]

Answer 61

to answer your question, the denominator could be \(2-x\) but then the numerator would have to be \(1-\frac{1}{1-x}\)

Answer 62

Okay

Answer 63

lol screwed up the signs again
the denominator could be \(2-x\) then the numerator would have to be
\[-1-\frac{1}{1-x}\]

Answer 64

in any case when we get done with the algebra, we end up with a rather simple expression
\[-\frac{1}{1-x}\]

Answer 65

and now if you want you can go ahead an replace \(x\) by \(1.9\) right here and get the same answer as we got before

Answer 66

probably a lot easier actually, you can compute them all this way

Answer 67

-1/1-1.9

Answer 68

right which is the same as
\[\frac{1}{.9}=\frac{10}{9}\] as before

Answer 69

you can use this to finish up all the rest in this section
but that is not why we did it
now we can do something we could not do before
before we were computing the slope of the secant lines at numbers close to 2, like 1.9, 1.99 etc

Answer 70

but unlike the previous expression, in this one you can actually replace \(x\) by \(2\) directly!

Answer 71

Do I also have to do the slop thin

Answer 72

\[-\frac{1}{1-x}\] is a formula for the slope at the point \((2,-1)\) so you can use it for any \(x\) you like
for example you can use it for \(2.5\) or \(2.1\) or \(2.01\) just plug it in and you will get the right answer

Answer 73

but the important part is that you can also plug in \(x=2\) and get
\[-\frac{1}{1-2}=1\] which is not the slope of the secant line, it is actually the slope of the tangent line
that is the start of taking derivatives, what makes calculus different from algebra

Answer 74

Ohhhhh!!! Makes life easier

Answer 75

you will get to this in the next section
good luck

Answer 76

Thanks!!!! (: