Calc 1 review for Calc 2

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April 27th 2011, 11:56 AM
mathhel

Calc 1 review for Calc 2

Hello, I am just starting to take Calc 2, and I was given a Calc one review sheet. But it has been so long since I took calc one in college. So I scanned the questions that are giving me a hard time, I do not expect you to give me the answers to the questions as I would like to know how to do this myself.

ps.(for moderators) I know that I can only ask two questions per post, but since I scanned my questions I will post them as is, if that is an issue I will immediately remove them. Thank you.

I couldn't upload my scans onto this site so I had no choice but to upload them somewhere else I will provide the link below

MEGAUPLOAD - The leading online storage and file delivery service

Thank You
April 27th 2011, 01:30 PM
topsquark

Quote:

Originally Posted by mathhel

Hello, I am just starting to take Calc 2, and I was given a Calc one review sheet. But it has been so long since I took calc one in college. So I scanned the questions that are giving me a hard time, I do not expect you to give me the answers to the questions as I would like to know how to do this myself.

ps.(for moderators) I know that I can only ask two questions per post, but since I scanned my questions I will post them as is, if that is an issue I will immediately remove them. Thank you.

I couldn't upload my scans onto this site so I had no choice but to upload them somewhere else I will provide the link below

MEGAUPLOAD - The leading online storage and file delivery service

Thank You

All I got from the download page is a bunch of ads.

Let's try this. Why don't you post (as best as you can) the first problem or two that are giving you difficulties. We'll worry about the rest later.

-Dan
April 27th 2011, 01:38 PM
mathhel

Ok I will. The Position of a particle at time t is given by s, where s^3 +9st+4t^3-8t = 0. find the velocity ds/dt

A.) ds/dt = (8-9s-12t^2)/3s^2 +9t)
B). ds/dt = 8+9s-12t^2)/3s^2-9t
C). ds/dt = 8-9s - 12t^2)/ 3s^2-9t
D). ds/dt = 8+9s-12t^2)/ 3s^2 +9t
April 27th 2011, 05:31 PM
topsquark

Quote:

Originally Posted by mathhel

Ok I will. The Position of a particle at time t is given by s, where s^3 +9st+4t^3-8t = 0. find the velocity ds/dt

A.) ds/dt = (8-9s-12t^2)/3s^2 +9t)
B). ds/dt = 8+9s-12t^2)/3s^2-9t
C). ds/dt = 8-9s - 12t^2)/ 3s^2-9t
D). ds/dt = 8+9s-12t^2)/ 3s^2 +9t

You can't solve for s as a function of t (well, not easily anyway) so I'd use implicit differentiation.
$s^3 +9st+4t^3-8t = 0$

$3s^2s' + 9s't + 9s + 12t^2 - 8 = 0$

Now solve for s'.

-Dan
April 27th 2011, 08:04 PM
mathhel

Oh I see, Thank you Dan. is there anyway we could work through the worksheet its only twelve questions so eleven now. Is it possible to email you the scans ?
April 28th 2011, 03:47 AM
topsquark

Quote:

Originally Posted by mathhel

Oh I see, Thank you Dan. is there anyway we could work through the worksheet its only twelve questions so eleven now. Is it possible to email you the scans ?

I make it a policy not to answer questions by mail. For starters I get busy and can't always get to helping specific problems in a timely manner. Also it is better to post them on the forum so others might benefit as well and you can possibly get the advantage of multiple viewpoints.

I'd recommend posting them one or two problems per thread.

-Dan