Math Help- Application of Derivatives +REP

Tally · **Posted:** Wed Mar 10, 2010 2:12 am

Ok, so I have a test tomorrow and I was wondering if someone could help me out on a few problems.

1. The finction T(x)= 200/1+x^2 represents the temerature in degrees Celcius percieved by a person standing x meters away from a fire.

a) If the person moves away from the fire at 2m/s, how fast is the temperature changing when the person is 5m away?

If someone can help me with this that would be greatly appreciated and +rep

Ohh and we're supose to use Leibniz notation.. dy/dx

Drumstick · **Posted:** Wed Mar 10, 2010 2:25 am

Lol Wow I failed

Tally · **Posted:** Wed Mar 10, 2010 2:31 am

Syphon_02 wrote:

Clarification: It's 200/ (1+x²) right? Cause you have 200/1 (which is 200) + x².

I'm pretty sure you don't need the speed that's given, if you are, I apologize for doing this wrong

Anyway,
t(x) = 200/ (x²+1)
t'(x) = 200/ (2x)
t'(5) = 200/(2(5))
t'(5) = 20

Do you know what the answer is supposed to be? Like, do you have answers at the back of the book?

Yup it's 200/ (1+x²)
But I believe in this case the quotient rule needs to be used f'(x)g(x)-f(x)g'(x) /[g(x)]^2
And the answer is suppose to be 5.9

hszmv · **Posted:** Wed Mar 10, 2010 2:45 am

Grant me that I haven't had a need for calculus in a 1 1/2.

You need speed in so much as to get where he was at a seconds interval. So, he moves 2 m/s. That means we need to know T(x) = 5 and T(x) = 3.

The correct derivation of T(x) = 200(x^2 + 1) is T'(x) = -400*x/(x^4+2*x^2+1) dx/dy.

Answer = T'(5) - T'(3)

Answer = -400*5/(5^4+2*5^2+1) - (-400*3/(3^4+2*3^2+1))

I'm to lazy to figure it out, but there is the final break down. Good luck.

Tally · **Posted:** Wed Mar 10, 2010 2:52 am

hszmv wrote:

Grant me that I haven't had a need for calculus in a 1 1/2.

You need speed in so much as to get where he was at a seconds interval. So, he moves 2 m/s. That means we need to know T(x) = 5 and T(x) = 3.

The correct derivation of T(x) = 200(x^2 + 1) is T'(x) = -400*x/(x^4+2*x^2+1) dx/dy.

Answer = T'(5) - T'(3)

Answer = -400*5/(5^4+2*5^2+1) - (-400*3/(3^4+2*3^2+1))

I'm to lazy to figure it out, but there is the final break down. Good luck.

I think that's right, thx for the help +rep

hszmv · **Posted:** Wed Mar 10, 2010 2:58 am

Used this website to get the correct derivation. Like I said, it's been years, but if your trouble is how to derive the function, then this sight will help with the rules. Just plug in simple versions to figure it out.

Tally · **Posted:** Wed Mar 10, 2010 3:01 am

hszmv wrote:

Used this website to get the correct derivation. Like I said, it's been years, but if your trouble is how to derive the function, then this sight will help with the rules. Just plug in simple versions to figure it out.

What site? I dont see it.

hszmv · **Posted:** Wed Mar 10, 2010 3:11 am

Forgot the link: http://www.numberempire.com/derivatives.php. Sorry.

Tally · **Posted:** Wed Mar 10, 2010 3:14 am

Ok, thank you:)

Math Help- Application of Derivatives +REP

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