Question

derivative of (2x(x)^1/2 +x)/(x^1/2)

Answer 1

\[\Large\bf\sf \left(\frac{2x\cdot x^{1/2}+x}{x^{1/2}}\right)'\]Is that the problem?

Answer 2

yes

Answer 3

Hmm it would probably be good to simplify before taking the derivative..

Answer 4

If you factor an x^(1/2) out of each term in the numerator,\[\Large\bf\sf\left[\frac{x^{1/2}(2x+x^{1/2})}{x^{1/2}}\right]'\]Then you have a nice simplification from there,\[\Large\bf\sf\left[\frac{\cancel{x^{1/2}}(2x+x^{1/2})}{\cancel{x^{1/2}}}\right]'\]And you're left to differentiate this:\[\Large\bf\sf (2x+x^{1/2})'\]

Answer 5

Lemme know if you're confused on that first step.

Answer 6

i got it thanks

Answer 7

ok good.

Answer 8

yea i got one more problem . can u help me with that ?

Answer 9

sure

Answer 10

At a given time the height of a person is 5 feet and it increases at a rate of 1 inch per year. The person's weight is 100 lb and it increases at a rate of 5 lb/year. (a) What is the body fattness (BMI) of the person? (b) Is the person getting slimmer? (c) Assuming that the instantaneous rate of change of BMI does not change over a year, predict the person's BMI one year from now. Recall that the BMI index is the fraction of the body weight measured in kilograms divided by the square of the body height measure in meters.

Answer 11

can you help me with this ? i didnt get this one at all

Answer 12

Let's try to interpret the stuff before part a.
Height is changing over time.
So height is a function of time. As time changes, the height changes accordingly.
We'll write a function for the height, our units of t will be years.\[\Large\bf\sf h(t)\]The starting point is 5 feet (that value is not changing, is not directly associated with t in any way).
And then the person grows 1 inch (1/12 ft) per year (per t).
So the height is the 5 feet + 1 inch per t.

Answer 13

\[\Large\bf\sf h(t)=5+\frac{1}{12}t\]

Answer 14

So after 12 years, the person will have grown to 6 feet.

Answer 15

Does that make sense? The way I modeled the data?

Answer 16

yea i get the function u come up with h(t)=5+1/12t

Answer 17

Ok good. Using the rest of the data, do you see how we can write a function for the person's weight?

\(\Large\bf\sf w(t)=?\)

Answer 18

is it 100+ 5/12 sorry for late reply

Answer 19

100+5/12(t)

Answer 20

why 5/12? :)

Answer 21

Remember, for height we were converting inches to ft.
So 1in = (1/12)ft.

But here they gave us 100 in pounds, and the 5 is also in pounds, yes?

Answer 22

oh ok so it is 100 +5(t) ?

Answer 23

Ok good so we've got our functions for height and weight.
\[\Large\bf\sf h(t)=5+\frac{1}{12}t\]\[\Large\bf\sf w(t)=100+5t\]

Answer 24

yea but im still not getting how we calculate body fatnes.

Answer 25

In part A, they tell us the relationship for BMI.\[\Large\bf\sf BMI\quad=\quad \frac{w(t)}{h(t)^2}\]

Answer 26

But they made it a little trickier.
They're changing the units on us.. grrr.

Answer 27

Our weight is currently measured in lbs.
We need to convert that to kilograms.

Answer 28

oh so we divide wt by 2.2?

Answer 29

yes.

Answer 30

Part a is probably the hardest part.
We're looking for a general formula for calculating body fatness.

So not at any particular t value.

Answer 31

then how we do it ?

Answer 32

\[\Large\bf\sf w(t)=(100+5t)\;lbs\cdot \left(\frac{1kg}{2.2lbs}\right)\]Gives us,\[\Large\bf\sf w(t)=(45.5+2.3t)\;kg\]

Answer 33

I divided each component by 2.2.

Answer 34

and height should be in meters

Answer 35

\[\Large\bf\sf h(t)=\left(5+.083t\right)\;ft\cdot \left(\frac{?m}{?ft}\right)\]What's the conversion for ft to meters? I can't remember..

Answer 36

1m=3.28 ft

Answer 37

so ht is 1.52+.0253(t)

Answer 38

\[\Large\bf\sf h(t)=(1.5+0.025t)\;m\]Mmm ok good!

Answer 39

ok so for a part what value of t we put or we take derivative of both

Answer 40

and then punt in that w/h^2 formula

Answer 41

Ya, I guess the BMI would just be calculated by...\[\Large\bf\sf BMI(t)\quad=\quad \frac{(45.5+2.3t)\;kg}{\left[(1.5+0.025t)\;m\right]^2}\]

Answer 42

ok ill try

Answer 43

hmm

Answer 44

Oh you have to input it into an online thing?
Are you allowed multiple guesses?

Answer 45

i dont think so . and for part a i guess we just divide 45.5 by 1.5^2

Answer 46

no i have to write and solve everything

Answer 47

Oh ok, they want us to start at time t=0?
Ok ok fair enough :)

Answer 48

i saw an example and thats what they have done

Answer 49

What value did you get for part a?

Answer 50

but how we figure if persin is slimmer or not ? they got 22.11 as their weight was 50 and we 20.22 as our weight was 45.5 and height being same for both

Answer 51

For part b?
So as t increases, is the BMI output getting larger or smaller?

It might make sense to just solve part c and then answer part b afterwards.

Answer 52

yea ill try

Answer 53

so part c i got 20.33 and part a it is 20.22 and when i calculate the rate of bmi it is 0.348
i dont what i just said , does this make any sense ?