Calculus Help?

[Yusuke Urameshi]

This problem deals with integrals and stuff. I'm not sure what to do with the temperatures.

Here's the problem:

A cup of coffee at 100 degrees celsius is put into a 15 degree celsius room when t = 0. The coffee's temperature, f(t), is changing at a rate given by f′(t) = −5(0.7)^t degrees celsius per minute, where t is in minutes.

Estimate the coffee's temperature when t=11: ________

Any legitimate attempt will be repped, and if you could explain your process, that'd be great.

 

[fiend]

In class so can't solve full

equ:-

∫(11-0) −5(0.7)^t =T(11)-100 (where 11 is the upper limit and 0 is the lower limit)

∫(11-0) −5(0.7)^t=-13.7411(approx)

Put it in the equn and solve

86.2589 approx

 

[Nebula 868]

Integral f'(t) to find f(t)

integral -5(0.7)^t = 14.0184*(.7)^t + C

we know the initial condition , so we can find C

when t=0 the coffee is 100 degrees so..


100=14.0184*(.7)^0 + C , anything to 0 power is 1

so we have 100=14.0184+C , C = 85.9816


so coffee's temperature f(t) = 14.0184*(0.7)^t + 85.9816

now put in t=11 to find coffee's temp at that time

f(11)= 14.0184*(.7)^11 + 85.9816

=86.25 degrees

Did this quickly so don't know if there are mistakes.

If you are getting problems with how I did the integral I remembered the formula

the derivative of a^x is ln(a) a^x and the anti-derivative is (1/ln(a)) a^x.

So 0.7 ^t will be ln (0.7) 0..7 ^t and the anti derivative is ( 1/ ln (0.7)) 0.7^t.

The -5 is a constant so it can be multiplied after doing thee integral.

Hope you understand it better.

 

[Dannie]

Goddamn I hate math.