finding the temperature after a period of time
$begingroup$
I have a homework problem that I am really struggling with. I will put the problem below and then discuss what I have tried to do.
A science geek brews tea at $195$ °F, and observes that the temperature $T(t)$ of the tea after $t$ minutes is changing at the rate of $T′(t)=−4.5e^{−0.07t}$ °F/min. What is the average temperature of the tea during the first $5$ minutes after being brewed? Round your answer to the nearest hundredth of a degree.
I tried to integrate from $0$ to $5$ minutes but I couldn't get the right answer does anyone know how to do this?
calculus
edited Jun 15 '16 at 16:06
Jennifer
8,44721837
asked Jun 15 '16 at 15:26
googlegoogle
805
$endgroup$
add a comment |
$begingroup$
I have a homework problem that I am really struggling with. I will put the problem below and then discuss what I have tried to do.
A science geek brews tea at $195$ °F, and observes that the temperature $T(t)$ of the tea after $t$ minutes is changing at the rate of $T′(t)=−4.5e^{−0.07t}$ °F/min. What is the average temperature of the tea during the first $5$ minutes after being brewed? Round your answer to the nearest hundredth of a degree.
I tried to integrate from $0$ to $5$ minutes but I couldn't get the right answer does anyone know how to do this?
calculus
edited Jun 15 '16 at 16:06
Jennifer
8,44721837
asked Jun 15 '16 at 15:26
googlegoogle
805
$endgroup$
$begingroup$
You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
$endgroup$
– Curious
Jun 15 '16 at 15:33
$begingroup$
I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
$endgroup$
– google
Jun 15 '16 at 15:38
$begingroup$
The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
$endgroup$
– Curious
Jun 15 '16 at 15:52
$begingroup$
even trying that its saying its not right
$endgroup$
– google
Jun 15 '16 at 15:53
add a comment |
$begingroup$
I have a homework problem that I am really struggling with. I will put the problem below and then discuss what I have tried to do.
A science geek brews tea at $195$ °F, and observes that the temperature $T(t)$ of the tea after $t$ minutes is changing at the rate of $T′(t)=−4.5e^{−0.07t}$ °F/min. What is the average temperature of the tea during the first $5$ minutes after being brewed? Round your answer to the nearest hundredth of a degree.
I tried to integrate from $0$ to $5$ minutes but I couldn't get the right answer does anyone know how to do this?
calculus
edited Jun 15 '16 at 16:06
Jennifer
8,44721837
asked Jun 15 '16 at 15:26
googlegoogle
805
$endgroup$
I have a homework problem that I am really struggling with. I will put the problem below and then discuss what I have tried to do.
A science geek brews tea at $195$ °F, and observes that the temperature $T(t)$ of the tea after $t$ minutes is changing at the rate of $T′(t)=−4.5e^{−0.07t}$ °F/min. What is the average temperature of the tea during the first $5$ minutes after being brewed? Round your answer to the nearest hundredth of a degree.
I tried to integrate from $0$ to $5$ minutes but I couldn't get the right answer does anyone know how to do this?
calculus
calculus
edited Jun 15 '16 at 16:06
Jennifer
8,44721837
asked Jun 15 '16 at 15:26
googlegoogle
805
edited Jun 15 '16 at 16:06
Jennifer
8,44721837
asked Jun 15 '16 at 15:26
googlegoogle
805
edited Jun 15 '16 at 16:06
Jennifer
8,44721837
edited Jun 15 '16 at 16:06
Jennifer
8,44721837
edited Jun 15 '16 at 16:06
Jennifer
8,44721837
8,44721837
asked Jun 15 '16 at 15:26
googlegoogle
805
asked Jun 15 '16 at 15:26
googlegoogle
805
asked Jun 15 '16 at 15:26
googlegoogle
805
805
$begingroup$
You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
$endgroup$
– Curious
Jun 15 '16 at 15:33
$begingroup$
I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
$endgroup$
– google
Jun 15 '16 at 15:38
$begingroup$
The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
$endgroup$
– Curious
Jun 15 '16 at 15:52
$begingroup$
even trying that its saying its not right
$endgroup$
– google
Jun 15 '16 at 15:53
add a comment |
$begingroup$
You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
$endgroup$
– Curious
Jun 15 '16 at 15:33
$begingroup$
I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
$endgroup$
– google
Jun 15 '16 at 15:38
$begingroup$
The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
$endgroup$
– Curious
Jun 15 '16 at 15:52
$begingroup$
even trying that its saying its not right
$endgroup$
– google
Jun 15 '16 at 15:53
$begingroup$
You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
$endgroup$
– Curious
Jun 15 '16 at 15:33
$begingroup$
You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
$endgroup$
– Curious
Jun 15 '16 at 15:33
$begingroup$
I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
$endgroup$
Jun 15 '16 at 15:38
$begingroup$
I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
$endgroup$
Jun 15 '16 at 15:38
$begingroup$
The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
$endgroup$
– Curious
Jun 15 '16 at 15:52
$begingroup$
The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
$endgroup$
– Curious
Jun 15 '16 at 15:52
$begingroup$
even trying that its saying its not right
$endgroup$
Jun 15 '16 at 15:53
$begingroup$
even trying that its saying its not right
$endgroup$
Jun 15 '16 at 15:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint :
The average value of a function $f(x)$ on the intervall $[a,b]$ is given by :
$$frac{1}{b-a}int_a^bf(x)dx$$
Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.
- Obtain an expression for $T$.
$T(0)=195$, so $T(t)=int_0^tT'(s)ds+T(0)=int_0^t−4.5e^{−0.07s}ds+195=−4.5int_0^te^{−0.07s}ds+195=frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=frac{4,5}{0.07}(e^{-0.07t}-1)+195$ - Use the formula for the average value of the function :
$$frac{1}{5-0}int_0^5T(x)dx=frac{1}{5}int_0^5frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10int_0^5e^{-0.07t}dt-50+195=-frac{10}{0.07}(
e^{-0.35}-1)+145approx187.187$$
answered Jun 15 '16 at 15:32
JenniferJennifer
8,44721837
$endgroup$
$begingroup$
what answer did you get? I got the same answer as the guy above and its wrong
$endgroup$
– google
Jun 15 '16 at 16:02
$begingroup$
added the answer
$endgroup$
– Jennifer
Jun 15 '16 at 16:03
$begingroup$
@google Did it answer your question ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:11
$begingroup$
its still saying that its wrong. I dont know what it could be
$endgroup$
– google
Jun 15 '16 at 16:17
$begingroup$
@google I did a mistake in last line can you check the new result ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:22
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint :
The average value of a function $f(x)$ on the intervall $[a,b]$ is given by :
$$frac{1}{b-a}int_a^bf(x)dx$$
Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.
- Obtain an expression for $T$.
$T(0)=195$, so $T(t)=int_0^tT'(s)ds+T(0)=int_0^t−4.5e^{−0.07s}ds+195=−4.5int_0^te^{−0.07s}ds+195=frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=frac{4,5}{0.07}(e^{-0.07t}-1)+195$ - Use the formula for the average value of the function :
$$frac{1}{5-0}int_0^5T(x)dx=frac{1}{5}int_0^5frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10int_0^5e^{-0.07t}dt-50+195=-frac{10}{0.07}(
e^{-0.35}-1)+145approx187.187$$
answered Jun 15 '16 at 15:32
JenniferJennifer
8,44721837
$endgroup$
$begingroup$
what answer did you get? I got the same answer as the guy above and its wrong
$endgroup$
– google
Jun 15 '16 at 16:02
$begingroup$
added the answer
$endgroup$
– Jennifer
Jun 15 '16 at 16:03
$begingroup$
@google Did it answer your question ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:11
$begingroup$
its still saying that its wrong. I dont know what it could be
$endgroup$
– google
Jun 15 '16 at 16:17
$begingroup$
@google I did a mistake in last line can you check the new result ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:22
add a comment |
$begingroup$
Hint :
The average value of a function $f(x)$ on the intervall $[a,b]$ is given by :
$$frac{1}{b-a}int_a^bf(x)dx$$
Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.
- Obtain an expression for $T$.
$T(0)=195$, so $T(t)=int_0^tT'(s)ds+T(0)=int_0^t−4.5e^{−0.07s}ds+195=−4.5int_0^te^{−0.07s}ds+195=frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=frac{4,5}{0.07}(e^{-0.07t}-1)+195$ - Use the formula for the average value of the function :
$$frac{1}{5-0}int_0^5T(x)dx=frac{1}{5}int_0^5frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10int_0^5e^{-0.07t}dt-50+195=-frac{10}{0.07}(
e^{-0.35}-1)+145approx187.187$$
answered Jun 15 '16 at 15:32
JenniferJennifer
8,44721837
$endgroup$
$begingroup$
what answer did you get? I got the same answer as the guy above and its wrong
$endgroup$
– google
Jun 15 '16 at 16:02
$begingroup$
added the answer
$endgroup$
– Jennifer
Jun 15 '16 at 16:03
$begingroup$
@google Did it answer your question ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:11
$begingroup$
its still saying that its wrong. I dont know what it could be
$endgroup$
– google
Jun 15 '16 at 16:17
$begingroup$
@google I did a mistake in last line can you check the new result ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:22
add a comment |
$begingroup$
Hint :
The average value of a function $f(x)$ on the intervall $[a,b]$ is given by :
$$frac{1}{b-a}int_a^bf(x)dx$$
Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.
- Obtain an expression for $T$.
$T(0)=195$, so $T(t)=int_0^tT'(s)ds+T(0)=int_0^t−4.5e^{−0.07s}ds+195=−4.5int_0^te^{−0.07s}ds+195=frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=frac{4,5}{0.07}(e^{-0.07t}-1)+195$ - Use the formula for the average value of the function :
$$frac{1}{5-0}int_0^5T(x)dx=frac{1}{5}int_0^5frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10int_0^5e^{-0.07t}dt-50+195=-frac{10}{0.07}(
e^{-0.35}-1)+145approx187.187$$
answered Jun 15 '16 at 15:32
JenniferJennifer
8,44721837
$endgroup$
Hint :
The average value of a function $f(x)$ on the intervall $[a,b]$ is given by :
$$frac{1}{b-a}int_a^bf(x)dx$$
Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.
- Obtain an expression for $T$.
$T(0)=195$, so $T(t)=int_0^tT'(s)ds+T(0)=int_0^t−4.5e^{−0.07s}ds+195=−4.5int_0^te^{−0.07s}ds+195=frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=frac{4,5}{0.07}(e^{-0.07t}-1)+195$ - Use the formula for the average value of the function :
$$frac{1}{5-0}int_0^5T(x)dx=frac{1}{5}int_0^5frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10int_0^5e^{-0.07t}dt-50+195=-frac{10}{0.07}(
e^{-0.35}-1)+145approx187.187$$
answered Jun 15 '16 at 15:32
JenniferJennifer
8,44721837
edited Jun 15 '16 at 16:21
answered Jun 15 '16 at 15:32
JenniferJennifer
8,44721837
answered Jun 15 '16 at 15:32
JenniferJennifer
8,44721837
answered Jun 15 '16 at 15:32
JenniferJennifer
8,44721837
8,44721837
$begingroup$
what answer did you get? I got the same answer as the guy above and its wrong
$endgroup$
– google
Jun 15 '16 at 16:02
$begingroup$
added the answer
$endgroup$
– Jennifer
Jun 15 '16 at 16:03
$begingroup$
@google Did it answer your question ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:11
$begingroup$
its still saying that its wrong. I dont know what it could be
$endgroup$
– google
Jun 15 '16 at 16:17
$begingroup$
@google I did a mistake in last line can you check the new result ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:22
add a comment |
$begingroup$
what answer did you get? I got the same answer as the guy above and its wrong
$endgroup$
– google
Jun 15 '16 at 16:02
$begingroup$
added the answer
$endgroup$
– Jennifer
Jun 15 '16 at 16:03
$begingroup$
@google Did it answer your question ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:11
$begingroup$
its still saying that its wrong. I dont know what it could be
$endgroup$
– google
Jun 15 '16 at 16:17
$begingroup$
@google I did a mistake in last line can you check the new result ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:22
$begingroup$
what answer did you get? I got the same answer as the guy above and its wrong
$endgroup$
Jun 15 '16 at 16:02
$begingroup$
what answer did you get? I got the same answer as the guy above and its wrong
$endgroup$
Jun 15 '16 at 16:02
$begingroup$
added the answer
$endgroup$
– Jennifer
Jun 15 '16 at 16:03
$begingroup$
added the answer
$endgroup$
– Jennifer
Jun 15 '16 at 16:03
$begingroup$
@google Did it answer your question ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:11
$begingroup$
@google Did it answer your question ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:11
$begingroup$
its still saying that its wrong. I dont know what it could be
$endgroup$
Jun 15 '16 at 16:17
$begingroup$
its still saying that its wrong. I dont know what it could be
$endgroup$
Jun 15 '16 at 16:17
$begingroup$
@google I did a mistake in last line can you check the new result ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:22
$begingroup$
@google I did a mistake in last line can you check the new result ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:22
add a comment |
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$begingroup$
You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
$endgroup$
– Curious
Jun 15 '16 at 15:33
$begingroup$
I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
$endgroup$
– google
Jun 15 '16 at 15:38
$begingroup$
The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
$endgroup$
– Curious
Jun 15 '16 at 15:52
$begingroup$
even trying that its saying its not right
$endgroup$
– google
Jun 15 '16 at 15:53