Question

Help with calculus 1 homework please =/

Answer 1

http://tinypic.com/view.php?pic=2q9ax08&s=8#.U6HGMPldWuo

Answer 2

Hi - Since s(t) shows the distance the particle has traveled, when t = 2, s(t) = 10 ... although it doesn't say whether that is 10 feet or 10 meters or 10 whatever-unit-of-distance they are using. So the answer to a) should be 10.

Answer 3

Okay I will just put 10 units.

Answer 4

can i give u an attachment

Answer 5

Ok. :) Then for b) if you look at the distance the particle has traveled between t=2 seconds and t= 8 seconds, what do you notice?

Answer 6

0 slope does that mean accelerating im nto sure what the right term for that is lol

Answer 7

Yep, shamim's attachment looks right. :)

Answer 8

In this case the 0 slope would refer to velocity, since it would be change in distance divided by change in time.

Answer 9

slope=0
velocity v=0

Answer 10

Yes

Answer 11

how did you get 2 for d ?

Answer 12

So from t=2 to t= 8, the particle hasn't moved, since its position stays the same at 10 units.

Answer 13

ya 2m/s

Answer 14

Particle's average velocity from 8 to 12 should be (2-10)/(12-8) = -8/4 = -2 m/s

Answer 15

Now, I am trying to think if that should be negative.

Answer 16

Okay i see, thanks guys! can you help me with 1 more lol.

Answer 17

k

Answer 18

http://i58.tinypic.com/e8suc5.jpg

Answer 19

Thanks!!

Answer 20

So you would use the power rule, then the chain rule

Answer 21

And you started with f(x) = (1-x)^1/2 so (1/2)(1-x)^(-1/2) * (-1)

Answer 22

Let me retype that\[-1/2\sqrt{1-x}\]

Answer 23

So, yes, what you found for f'(x) looks right. :)

Answer 24

cool so a and b is right ?

Answer 25

Let me look again

Answer 26

a is right, but I didn't notice that there was a b part - so just a moment, please.

Answer 27

Since f'(x) gives the slope of the tangent for a particular value of x, when x = -3, I got f'(-3) to be -1/4 . But I will double check!

Answer 28

I got -1/4 again.

Answer 29

So then you have x = -3 and you need the y that goes with it, so find f(-3), which is 2, which is what you found.

Answer 30

So now you have m = -1/4 and a point (2,-3) so we can use point-slope equation of the line y - y1 = m(x-x1)

Answer 31

So y - 2 = (-1/4)(x + 3)

Answer 32

Or y = (-1/4)x - 3/4 + 2 or y = (-1/4)x + 5/4

Answer 33

Ohh I see okay thanks!

Answer 34

So yes, your setup is right, just double check the value of the slope

Answer 35

can you help me with c and d pls ?

Answer 36

Please do that for me, since I am horrible at arithmetic!

Answer 37

Yes, I will look at c and d while you double check the calculation for the slope on part b. :)

Answer 38

Yes you're right, thank you!!

Answer 39

You're welcome :) I make arithmetic errors, though, so I always like a second pair of eyes!

Answer 40

Lol i do as well

Answer 41

Now on part d, they say to discuss the differentiability of f(x) at x = 1, and then in part e they say to discuss the continuity of f(x) at x = 1. The slope is undefined at x = 1, since you would get a division by 0 there ... and the domain of the function is [1, inf) so the graph doesn't even exist below x = 1. Do you mind if I get a book quickly to make sure we describe this correctly?

Answer 42

sure go ahead

Answer 43

I suppose you could say, for part e, that f is continuous on the interval from [1, inf) and is undefined elsewhere.

Answer 44

So then, it seems to me that you couldn't find a derivative at x = 1, since both left and right sided limits would have to exist.

Answer 45

Thanks so much for your help!!

Answer 46

You are welcome. :) I hope it was helpful!