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Calculus/higher derivatives with ordinary language

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Question
Hi Scotto, Sorry I'm sending another message, The previous one I sent I ended up getting both answers..so you don't need to reply if you haven't already..
So for another one, I have to find all answers in order to receive full credit, so I have no idea which ones are right and wrong.

The problem is:

Consider these statements written in ordinary language:
A The speed of the car is proportional to the distance it has traveled.
B The car is speeding up.
C The car is slowing down.
D The car always travels the same distance in the same time interval.
E We are driving backwards.
F Our acceleration is decreasing.

Denoting by s(t) the distance covered by the car at time t, and letting k denote a constant, match these statements with the following mathematical statements by entering the letters A through E on the appropriate boxes:

1)s''<0
2)s' is a constant
3)s'<0
4)s'''<0
5)s''>0
6)s'=ks

I have to match the letter of the written statement with the derivative part. I really don't understand how to know which phrase goes with the derivative. Any help??


Answer
Don't worry about it - some people have sent well over 30 messages.
For some, this is their third year.  I have responded to ever question, but around 2% of them I have told how I didn’t understand what they were asking.  Of those, there were very few I didn’t understand.
I really enjoy the questions and it is very refreshing for me.

A) For the speed to be proportional to the distance traveled, this means that the car is accelerating at even more all the time, which means s" is positive.  That would be 5.  The speed being proportional to the distance traveled means s’ = ks.

B) The car is speeding up means the velocity is increasing.  
Velocity is given as the derivative.  For this to be increasing, it’s derivative must be positive.  This means the f”(x) > 0.

C) The car is slowing down means the change in speed is less than 0.  The change in speed is the second derivative.

D) For the car to be travelling at a constant speed means that s’ is a constant.

E) To be driving backwards mean the speed is negative.  The speed is given by s’.

F) For acceleration to be decreasing, that means the derivative of acceleration is negative.  Since acceleration is the 2second the derivative of distance, this would refer to the derivative of that, which is the third derivatve.

Note that answer (4) has s’”.

Calculus

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Scotto

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Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

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Experience in the area: I have tutored students in all areas of mathematics for over 25 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

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Maybe not a publication, but I have respond to well oveer 7,500 questions on the PC. Well over 2,000 of them have been in calculus.

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I have been nominated as the expert of the month several times. All of my scores right now are at least a 9.8 average (out of 10).

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My past clients have been students at OSU, students at the college in South Seattle, referals from a company, friends and aquantenances, people from my church, and people like you from all over the world.

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