I have been set a probelm to do from my High school calculus class, I have gone through it and don't think I had any major probelms but I would love if someone on this forum could go through the question, so I can get a level of my understanding. Its 1 question with 7 parts (a-f):

The concentration of alcohol in a persons bloodstreams 't' hours after the person stopped drinking is given by c(t)= 0.8t / t^2+1. when C(t) is less than or equal to 0.08

A) interpret meaning of C(1/2)
B) find experssion for C'(t), which is the rate at which the concentrations of alcohol is changing.
C) write equation for the tangent line to C(t) when t=1
D) Interpret the meaning of the equation of the tangent line found in part C
E) Find the time it will take for the concentration level to get up to and then back below the legal .08 level.
F)theoretically, is there always a trace amount of alcohol in your bloodstream after some has been consumed? use limit theory to explain your answer.

desperately looking for help from anyone, & any help is much appreiciated!

2. Originally Posted by Ashish Patel
I have been set a probelm to do from my High school calculus class, I have gone through it and don't think I had any major probelms but I would love if someone on this forum could go through the question, so I can get a level of my understanding. Its 1 question with 7 parts (a-f):

The concentration of alcohol in a persons bloodstreams 't' hours after the person stopped drinking is given by c(t)= 0.8t / t^2+1. when C(t) is less than or equal to 0.08

A) interpret meaning of C(1/2)
B) find experssion for C'(t), which is the rate at which the concentrations of alcohol is changing.
C) write equation for the tangent line to C(t) when t=1
D) Interpret the meaning of the equation of the tangent line found in part C
E) Find the time it will take for the concentration level to get up to and then back below the legal .08 level.
F)theoretically, is there always a trace amount of alcohol in your bloodstream after some has been consumed? use limit theory to explain your answer.

desperately looking for help from anyone, & any help is much appreiciated!
show your solutions, then someone may have a look and provide any necessary correction(s).

3. A) i belive the 1/2 just stands for 't' hours. so 1/2 would stand for half an hour meaning the level of concentration after 30 mins. or would it mean half of the concentration levl as C(t) can be less than or equal to 0.08..then would the mean 0.04

b) C'(t) to find the rate of change would be the derivative, so you woud use the quotient rule, and the answer would be the rate at which the concentration is changing in the blood stream.

c and D) is where im quite stuck on and would like help

f) i believe there is always a trace of alcocol in your bloodstream, looking at the limit theory, it says as H->0, but never is actually 0. Therefore there is always a slight trace.

4. btw for part b)
c't is the derivative of c(t)= 0.8t/t^2+1
which i worked out to be C'(t)= 0.8t^2+0.8 /(t^2+1)^2)

part C is where i am stuck on.

5. Originally Posted by Ashish Patel
A) i belive the 1/2 just stands for 't' hours. so 1/2 would stand for half an hour meaning the level of concentration after 30 mins. <<< correct

b) C'(t) to find the rate of change would be the derivative, so you woud use the quotient rule, and the answer would be the rate at which the concentration is changing in the blood stream. <<<< your considerations are OK - but the result is wrong

....
$c(t)=\frac{0.8t}{t^2+1}~\implies~c'(t)=\frac{(t^2+ 1) \cdot 0.8 - 0.8t \cdot 2t}{(t^2+1)^2} = \frac{-0.8(t^2-1)}{(t^2+1)^2}$

6. Originally Posted by Ashish Patel
...
C) write equation for the tangent line to C(t) when t=1
...
1. Calculate the coordinates of the point on the graph of c if t = 1. That means determine P(1, c(1)).

2. Calculate the slope of the tangent which is m = c'(1).

3. Now use the point-slope-formula to get the equation of the tangent.

7. ok thanks mate, I'll give it a go.

8. I got the equation of the tangent line Y= .4 , would that be correct

I got 0 as the slope and -0.4 as the y- int

so the point slope formula would looked like this:

y - 0.4 = 0 (x-1)
y - 0.4 = 0
y= 0.4

what does y= 0.4 stand for?

and now how would I go about part E.

once again all your help is much apprieciated.

9. Originally Posted by Ashish Patel
I got the equation of the tangent line Y= .4 , would that be correct <<<<<<< OK

I got 0 as the slope and +0.4 as the y- int

so the point slope formula would looked like this:

y - 0.4 = 0 (x-1)
y - 0.4 = 0
y= 0.4

what does y= 0.4 stand for? <<<<<<<< that's a constant function. The graph of this function is a parallel to the x-axis.

...
Since the slope of the tangent is zero the point P(1, 0.4) must be the peak of the concentration. (Remember: In general a function has an extreme value - maximum or minimum - if the first derivation equals zero).

Draw the graph of c.