Question

Is there anybody on the face of this planet who is good at calculus and can help me until I am done with all my study sheets?!?!?!

Answer 1

I got sick and have finals and I need to get all of my study sheets done so I am prepared for my finals.

Answer 2

I can't spend all day here, because of my own exams. But I could help for a bit and hopefully other will continue from where I stopped.

Answer 3

use limits to determine if \[f(x)=\ln (x-1)\] is continuous at x=1.

Answer 4

i do have options.
\[A. \lim_{x \rightarrow 1+}f(x)=\lim_{x \rightarrow x-}f(x)=f(1)\] so the function is continuous at x=1
\[B. \lim_{x \rightarrow 1+}f(x)\neq \lim_{x \rightarrow 1-}f(x)\]so the function is not continuous at x=1

Answer 5

\[C. \lim_{x \rightarrow 1+}f(x)=\lim_{x \rightarrow 1-}f(x)\neq f(1)\] so the function is not continuous at x=1
\[D.\]can not determine using limits

Answer 6

when x ->1+ it means x is just a bit bigger than 1, say 0.001 as a concrete example
what is ln( 1.001 - 1) ?
now compare to x-> 1-, which means x is a bit less than 1:
what is ln( 0.999 - 1) ?

Answer 7

-6.90775528 + 3.14159265 I???

Answer 8

you can use a calculator. You will see you get -6.9 for the first case
and "math error" for the second. That means they are not the same

Answer 9

or you get a complex number. but definitely different.

Answer 10

so it would be B?

Answer 11

I think so.

Answer 12

An object in free fall has its distance from the ground measured by the function d(t)=-4.9t^2 +50, where d is in meters and t is in seconds. If gravity is the only acceleration affecting the object, what is gravity's constant value (include units in your answer)?

Answer 13

if you use x instead of d (to avoid confusion)
\[ x= -4.9t^2 +50\]
by definition
\[ \text{velocity }= \frac{dx}{dt} \\
\text{acceleration }= \frac{d^2x}{dt^2} \]

Answer 14

but what in the equation is d and what is x and t?

Answer 15

@terenzreignz

Answer 16

d is just an operator to represent derivatives (rates of change)
x is the horizontal distance
t is time

Answer 17

so it is the first derivative of x divided by the first derivative of time?

Answer 18

no, dx/dt is THE derivative of x with respect to time.

Answer 19

Sorry to interrupt guys, @kaylalynn have you learned differentiation?

Answer 20

I was sick for about 3 weeks ND my teacher just gave me these to get ready for finals. im so lost. so here I am.

Answer 21

I am so sorry. I think your teacher can do better.

Answer 22

I agree. however this is all I got to even prepare me . :(

Answer 23

Do you know the definition of differential?

Answer 24

definition? no

Answer 25

No, she missed many classes how can she know. I think your teacher should give a little more time.

Answer 26

I don't blame her completely. she needs to have final scores in by Tuesday.

Answer 27

The challenge now is how folks here will TEACH you 3 weeks work in a couple of hours. Cos we can't just solve and explain

Answer 28

yea I know cuz then I wouldn't be ready for finals:(

Answer 29

all I can think of is if somebody could answer and explain and ill try to keep up. If I get lost, I can say something??

Answer 30

When (or if) you have time, you can learn everything watching Khan's videos
This is an intro type video.
http://www.khanacademy.org/math/calculus/differential-calculus/derivative_intro/v/calculus--derivatives-1--new-hd-version

Answer 31

okay thank you. I can try that.

Answer 32

For your question
***
An object in free fall has its distance from the ground measured by the function d(t)=-4.9t^2 +50, where d is in meters and t is in seconds.
***
I renamed the distance d(t) as x(t) because calculus uses d to stand for differential

if x is distance (that changes as time time changes)
the change in distance divided by change in time is speed (or velocity)

we say
\[ V= \frac{dx}{dt} \]
where
\[ x(t) = -4.9t^2 +50 \]
take the derivative of both sides
\[ \frac{dx}{dt} = \frac{d}{dt}(-4.9t^2 + 50 ) \]

Answer 33

to do the derivative , you need to know the short-cut rules
Here is one
http://www.khanacademy.org/math/calculus/differential-calculus/power_rule_tutorial/v/power-rule

Answer 34

is it 9.8m/s^2????

Answer 35

yes

Answer 36

you should find
\[ \frac{dx}{dt} = \frac{d}{dt}(-4.9t^2 + 50 ) \\
= -4.9\frac{d}{dt}t^2 + \frac{d}{dt}50 \\
-4.9 \cdot 2 t + 0\\
= -9.8 t
\]
now take the derivative again
\[ \frac{d}{dt} \left(\frac{dx}{dt} = -9.8t\right) \\
\frac{d^2x}{dt^2} =-9.8 \frac{dx}{dt} t\\
\frac{d^2x}{dt^2} =-9.8 \frac{m}{s^2}
\]

Answer 37

Here are youtube links. Do they work?
Intro
https://www.youtube.com/watch?v=ANyVpMS3HL4

Power rule
https://www.youtube.com/watch?v=bRZmfc1YFsQ

Answer 38

how do I find the equation of a tangent line using a table?

Answer 39

Oops I've not heard of that before.

Answer 40

can you post the exact question ?

Answer 41

The table below gives specific values to a differentiable function h(t). Use this information to write an equation for the tangent line to the graph h(t) at t=9.

Answer 42

t 0 5 9 14 20
f(t) 86 77 64 59 52

Answer 43

my only thought is to find the slope through the points
(5,77) and (14,59)
and write the equation of a line
through the point (9,64)

Answer 44

slope is -2?

Answer 45

yes,
now use
y - 64 = -2(x-9)

Answer 46

or
y= -2x +82

Answer 47

But I would like to see your notes/write-up on how to do this... I am only guessing on this problem

Answer 48

I got y=-2x+192?

Answer 49

y= -2x + 18 + 64
y= -2x + 82

Answer 50

\[y-64=-2(x-64)\]
\[y-64=-2x+128\]
\[y-64+64=-2x+128+64\]
\[y=-2x+192\]

Answer 51

Oh, you are a victim of my typo
we should be using the point (9,64)
and I accidentally used 64 instead of 9
the equation of a line through the point (x0,y0) is
y - y0 = m(x - x0)
Here x0 is 9 and y0 is 64

Answer 52

Do you have *any* notes on how they would do this problem ?

Answer 53

no I don't and that equation doesn't work for the rest...

Answer 54

No the equation will only works (if it works at all) for one point on the curve. Your points, if you plot them do not line up... we have a curve that changes direction.

A tangent line at (9,64) is expected to touch the curve only at (9,64)

A tangent line at another spot on the curve will have a different slope (if the curve changes direction... and it does)

Answer 55

I have a funeral viewing I have to go to . ill be back later tonight

Answer 56

did we figure anything out?