dude:

Butter is taken out of the fridge and left to warm up. The temperature of the butter, T (in Celsius degrees), after m minutes is given by the equation T=c-k*1.2^{-m}. The initial temperature of the butter is 5º, and after 5 minutes its temperature has increased to 15º, as shown on the graph a) Find the values of k and c

4 months ago
dude:
4 months ago

Allison:

Yeaa...nope. @Vocaloid

4 months ago
sillybilly123:

\(T(m) = c- k \cdot 1.2^{-m}\) \( T(0) = 5 \implies c = k + 5\) \(T(5) = 15 \implies c - k ~ 1.2^{-5} = 5 \) solve from there

4 months ago
sillybilly123:

\(T(5) = c - k ~1.2^{-15} = 15\)

4 months ago
sillybilly123:

oh balls :( you get the picture. just plug in the numbers and solve :)

4 months ago
dude:

For T(5), I got \[c=15+\frac{1}{2.48832}k\]

4 months ago
dude:

So then I would do \[k+5=15+\frac{1}{2.48832}k?\]

4 months ago
Vocaloid:

the key is to write both equations and solve by elimination 5 = c - k(1.2)^(0) 15 = c - k(1.2)^(-5) you can subtract these two equations to eliminate c 10 = c - k(1.2)^(-5) - [c - k(1.2)^(0)] after you eliminate c you should be able to solve for k, then back-substitute to get c

4 months ago
Vocaloid:

ok I got it now 10 = c - k(1.2)^(-5) - [c - k(1.2)^(0)] 10 = -k(1.2)^(-5) + k k is about 16.719, then you can select one of the initial points, plug in and solve for c

4 months ago
dude:

\[ 5=c-16.719*1.2^{-0}\\ c=21.719\]

4 months ago
Vocaloid:

good just to be on the safe side I would double-check those k and c values by plugging them back into the points to see if you still get 5 and 15 ;;

4 months ago
dude:

\[ 5=21.719-16.719*1.2^{0}\\ 5=5\]

4 months ago
dude:

Thank you Vocaloid :P

4 months ago
sillybilly123:

So: here, Completing the Square is a real timesaver. Not too shabby 😀

4 months ago