The simple scribe post

 

Last Thursday, we were given this sheet.

 

 

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With the answers on the other side:

 

 

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Its due Monday (tomorrow).

 

I’ll try and post solutions to the best of my ability. Didn’t get far with this one…

 

I have questions 1 and 2… sad. I’ll explain how I solved these and I sort of grasp the concept to solving 4 and 5 so I’ll try and describe those.

 

 

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Solution 1, is pretty straightforward. Since the variables are already on their respective sides, just antidifferentiate. Since we’re solving for C, we’ll put that on the left and move the rest to the right. Simplify. As Benchmen pointed out, because C is a constant, even if you multiply it it will remain C, hence the part that says “let (2) C = C”

 

Solution 2 is a little longer, but also pretty straight forward. Organize your variables to their respective sides, integrate/antidifferentiate both sides. Take the ln of both sides, and solve for A which is e^c. The solving for A is a bit of simple algebra.

 

Solution 3 I tried twice but kept getting stuck. I figure the way to solve it is to follow the same steps as in 2, IE organize and integrate, and once you have an equation (this is where I got stuck) solve for t using the given values.

 

Solution 4 I have no clue, sorry guys. Feel free to post your solutions in the comments.

 

Solution 5 I know HOW to solve, but cant actually do it because I get stuck (as you can see) at the point after integrating. However, this is very much like question 3, where you are given inital values. In this question, you would organize, integrate, and then use the inital values to pinpoint the graph since when you integrate/antidifferentiate you are finding the family of functions and you want the specific function that passes through a given point.

 

Solution 6 - 10 use Newtons cooling law.

 

 

As usual, this has been a late scribe post, although from what I remember on Thursday you guys are probably ahead of me on this stuff anyway. See you all tomorrow.

 

Oh and the next scribe is Francis.