# Calculus Question

• Mar 2nd 2010, 03:05 PM
skeltonjoe
Calculus Question
I do not know how to find part b of this problem can anyone help me.

When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (Round the answers to one decimal place.)
(a) What is the temperature of the drink after 35 minutes?
http://www.webassign.net/wastatic/common/img/tick.png°C

(b) When will its temperature be 16°C?
http://www.webassign.net/wastatic/common/img/cross.png minutes
• Mar 2nd 2010, 03:10 PM
skeeter
Quote:

Originally Posted by skeltonjoe
I do not know how to find part b of this problem can anyone help me.

When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (Round the answers to one decimal place.)
(a) What is the temperature of the drink after 35 minutes?
http://www.webassign.net/wastatic/common/img/tick.png°C

(b) When will its temperature be 16°C?
http://www.webassign.net/wastatic/common/img/cross.png minutes

I assume the green check means you got part (a) correct ... how did you figure that value?
• Mar 2nd 2010, 03:13 PM
skeltonjoe
i did the problem out, then for part b I know your suppose to take like 16=20 - 16 * 2/3 raised to the t/25 and solve it for t which I get 85.5 but apparently its wrong.
• Mar 2nd 2010, 03:28 PM
skeeter
Quote:

Originally Posted by skeltonjoe
i did the problem out, then for part b I know your suppose to take like 16=20 - 16 * 2/3 raised to the t/25 and solve it for t which I get 85.5 but apparently its wrong.

perhaps I wasn't clear in my request ...

what equation did you use to solve part (a) , and how did you go about getting that equation?
• Mar 2nd 2010, 03:47 PM
skeltonjoe
y(t) = -15 * 2/3 ^ t/25

I keep getting 85.5 I just redid it 3 times (Headbang) (Angry)
• Mar 2nd 2010, 04:04 PM
skeeter
Quote:

Originally Posted by skeltonjoe
y(t) = -15 * 2/3 ^ t/25

I keep getting 85.5 I just redid it 3 times (Headbang) (Angry)

your equation is incorrect. if y is the temperature, then ...

y(0) should = 5

y(25) should = 10

your equation does not yield those values for the given times.

you should be deriving the temperature function from Newton's law of cooling/heating ...

$\frac{dy}{dt} = k(20 - y)$ , using the initial conditions y(0) = 5 and y(25) = 10
• Mar 2nd 2010, 04:27 PM
ione
Quote:

Originally Posted by skeltonjoe
y(t) = -15 * 2/3 ^ t/25

I keep getting 85.5 I just redid it 3 times (Headbang) (Angry)

Did you mean $y(t)=20-15(\frac{2}{3})^{\frac{t}{25}}$ ?
• Mar 2nd 2010, 05:22 PM
skeltonjoe
yes thats what I meant, can anyone please answer this question its due in 2 hours
• Mar 2nd 2010, 05:39 PM
skeltonjoe
i got it the answer was 81.5