Question

CAN SOMEONE HELP ME PLEASE!!!!
the concentration of a drug in a patient’s bloodstream h hours after it was injected is given by A(h)=(0.17h)/(h^2+2).

(a) Find and interpret A′(3).

(b) Use A′ to answer the following. For what values of h is A(t) increasing?

Answer 1

im going to poop in your mouth

Answer 2

start by finding the derivative of A(h)

Answer 3

You come up with anything for a) and b) ? :)

Answer 4

ok this what i have \[\frac{ 0.17((1/h)^2)^2 }{ 1/h^2+2}\]

Answer 5

Woahh, whut? +_+

Answer 6

That's your derivative?
Did you remember to apply quotient rule?

Answer 7

\[\Large\bf\sf A(h)\quad=\quad \frac{0.17h}{h^2+2}\]
\[\Large\bf\sf A'(h)\quad=\quad \frac{\color{royalblue}{\left[0.17h\right]'}(h^2+2)-0.17h\color{royalblue}{\left[h^2+2\right]'}}{(h^2+2)^2}\]

Answer 8

\[\frac{ h^2+2(0.17)-0.17h }{ (h^2+2) ^2}\]

Answer 9

hold on i forgot something

Answer 10

\[\Large\frac{ \color{red}{(}h^2+2\color{red}{)}(0.17)-0.17h\color{red}{(?)} }{ (h^2+2) ^2}\]

Answer 11

Don't forget brackets on that first term, the red ones. Important.
And then ya, looks like you're missing something on the second term.

Answer 12

\[\frac{ 0.17(h^2+2)-0.17h2h }{ (h^2+2)^2 }\]

Answer 13

Ok looks good!

Answer 14

No need to simplify it down too much.
Let's just plug in our h=3 to evaluate A'(3).

Answer 15

\[\frac{ (h^2+2)(0.17)-0.17h(2h) }{ (h^2+2)^2 }\]

Answer 16

ok

Answer 17

-.00983

Answer 18

Ok good.
How would you interpret that number?
What does it mean?

Answer 19

This step is a little tricky, it took me a minute to realize what it meant heh.

Answer 20

after 3hrs the drug was injected to the patients the concentration rate is changing to -.00983

Answer 21

Good, yes.
It's the `rate of change of the concentration`.

So the concentration is `decreasing` (because of the negative), but it's decreasing at a very very very slow rate.

So the concentration of drug in the bloodstream isn't changing very much at 3 hours.

Answer 22

ok so how can i find out part b when its increasing because its decreasing

Answer 23

When A' is `negative`, it tells us that A is `decreasing`.
When A' is `positive`, it tells us that A is `increasing`.

What happens when A' is `zero`?
Those are what we call critical points.
It's where the function switches from increasing to decreasing.
So we need to find those point(s).

Answer 24

So we'll let A'=0 and solve for h to find critical points of the function.\[\Large\bf\sf A'\quad=\quad\frac{(h^2+2)(0.17)-0.17h(2h) }{ (h^2+2)^2 }\]
\[\Large\bf\sf 0\quad=\quad\frac{(h^2+2)(0.17)-0.17h(2h) }{ (h^2+2)^2 }\]

Answer 25

Now solve for h.

Answer 26

Any confusion on how to solve for h? :o

Answer 27

no sorry for the delay

Answer 28

.085

Answer 29

Hmmmm....

Answer 30

You should end up withhhhh.. lemme check..\[\Large\bf\sf h=\sqrt2\]

Answer 31

so does that mean when A is increasing its at \[\sqrt{2}\]

Answer 32

Just in case there was any confusion on how you solve for h,
\[\Large\bf\sf 0\quad=\quad\frac{(h^2+2)(0.17)-0.17h(2h) }{ (h^2+2)^2 }\]Multiply both sides by the denominator:\[\Large\bf\sf 0\quad=\quad (h^2+2)(0.17)-0.17h(2h)\]Divide each side by 0.17,\[\Large\bf\sf 0\quad=\quad h^2+2-2h^2\]\[\Large\bf\sf 0\quad=\quad 2-h^2\]\[\Large\bf\sf h^2~=\quad 2\qquad\qquad\implies\qquad\qquad h=\sqrt2\]

Answer 33

That is our critical point.
I like to draw a number line personally.
Helps me understand what's going on.

Answer 34

We want to plug `test points` into our derivative function to determine regions of increasing and decreasing.

Answer 35

h=3 would be somewhere over here,

Answer 36

We already determined that to be `negative` yes?

Answer 37

So our function is decreasing over that interval.

Answer 38

Let's plug another test point into our derivative to find out what's happening in this interval.