Chain rule

[1141]

Help, please?

The question I need to do is:

A metal bar is heated to a certain temperature and then the heat source is removed. At time t, after the heat source is removed, the temperature, ? degress Celsius, of the metal bar is given by ? = 280 / 1+0.02t . At what rate is the temperature decreasing 100 minutes after the removal of the heat source?

I have to use the chain rule to work this out.
But I'm still confused. I'm not good with these kind of questions.

thanks guys!

 

[mmm4444bot]

Maybe it helps to see theta rewritten with an exponent.

\(\displaystyle \theta \;=\; 280 \cdot (1 + 0.02 t)^{-1}\)

This is a composite function (i.e., a function within a function).

The inner function is linear: 1 + 0.02t .

It's derivative is 0.02 .

The outer function is of the form Ax^(-1) . To be more general, I've used the symbol x to represent the inner function (the base), and A = 280.

By the Power Rule, it's derivative is -Ax^(-2), so we have -280(x)^(-2) .

The chain rule says that the derivative of a composite function is the product of the inner and outer derivatives.

(0.02)(-280)(1 + 0.02t)^(-2)

-5.6(1 + 0.02t)^(-2)

I'm thinking that the units on the time variable t are minutes, so the expression above gives us the rate at which the temperature is changing at any given number of minutes.

They gave us 100 minutes. Substitute that for t, and calculate the requested rate of decrease.

100 minutes after the metal began cooling, I think its temperature is dropping at a rate of roughly 62/100ths of a degree per minute.

 

[BigGlenntheHeavy]

\(\displaystyle T(t) \ = \ \frac{280}{1+.02t}\)

\(\displaystyle \frac{d[T(t)]}{dt} \ = \ \frac{-5.6}{(1+.02t)^{2}}\)

\(\displaystyle \frac{d[T(100)]}{dt} \ = \ -.6 \overslash 2^{o}/min \ (about).\)

 

[1141]

Alright, I worked it out like you guys did and got to -62 degrees per minute.

I have another question, sort of like this, which I'm struggling to get a formula/equation for.

A spherical star is collapsing in size, while remaining spherical. When its radius is one million kilometres, the radius is decreasing at the rate of 500km s[sup:1hunlmcp]-1[/sup:1hunlmcp]. Find:

a.) the rate of decrease of its volume
b.) the rate of decrease of its surface area

I know that the formula for the volume would be, V=4/3.pi.r[sup:1hunlmcp]3[/sup:1hunlmcp], and the formula for surface area = 4.pi.r[sup:1hunlmcp]2[/sup:1hunlmcp].

 

[galactus]

a. \(\displaystyle \frac{dV}{dt}=4{\pi}r^{2}\cdot \frac{dr}{dt}\)

b. \(\displaystyle \frac{dS}{dt}=8{\pi}r\cdot \frac{dr}{dt}\)

 

[1141]

a. \(\displaystyle \frac{dV}{dt}=4{\pi}r^{2}\cdot \frac{dr}{dt}\)

b. \(\displaystyle \frac{dS}{dt}=8{\pi}r\cdot \frac{dr}{dt}\)

so...if I have this correct,

a.) dv/dt = 4.pi.(1 000 000)[sup:2yfd0v81]2[/sup:2yfd0v81] x 500 km s[sup:2yfd0v81]-1[/sup:2yfd0v81]
dv/dt = 6.28 x 10 [sup:2yfd0v81]15[/sup:2yfd0v81] km[sup:2yfd0v81]3[/sup:2yfd0v81] s[sup:2yfd0v81]-1[/sup:2yfd0v81]


b.) ds/dt = 8.pi.1 000 000 x 500 km s[sup:2yfd0v81]-1[/sup:2yfd0v81]
ds/dt = 1.26 x 10[sup:2yfd0v81]15[/sup:2yfd0v81] km[sup:2yfd0v81]2[/sup:2yfd0v81] s[sup:2yfd0v81]-1[/sup:2yfd0v81]

I'm not sure about my units? the answers don't look right though...