A cup of coffee at 90 degrees celsius is put into a 30 degree celsius room when t =0 . The coffee's temperature, f (t ) , is changing at a rate given by f '(t )=-8(0 .8) t degrees celsius per minute, where t is in minutes.
Estimate the coffee's temperature when t =10 :
If you mean f'(t) = -8 * 0.8^t, then
f(t) = -8/ln(0.8) * 0.8^t + C
since f(0) = 90,
90 = -8/ln(0.8) + C
C = 90 + 8/ln(0.8)
So,
f(t) = -8/ln(0.8) * 0.8^t + 90 + 8/ln(0,8)
= 90 + 8/ln(0.8) (1-0.8^t)
= 90 - 35.85 (1-0.8^t)
So, plug in t=10
Well, if we have a cup of coffee at 90 degrees Celsius and we put it into a 30-degree room, we can expect it to cool down. Coffee, just like humans, prefers to be hot, but it can't resist the cooling forces of its environment.
Now, let's talk about this equation f '(t )=-8(0 .8) t degrees Celsius per minute. This equation tells us that the rate at which the coffee's temperature is changing is proportional to time, with a negative coefficient. It's like saying, "Hey, coffee, you better start cooling down, or I'll make you regret it!"
But, how hot will the coffee be when t = 10? Let's find out using our trusty sense of humor (and some math). We'll plug in t = 10 into the equation and see what we get.
f '(10) = -8(0.8)(10)
= -64 degrees Celsius per minute
Oh no! The coffee is cooling down at a rate of 64 degrees Celsius per minute! That's faster than a penguin sliding on an icy hill! But don't worry, we can handle it.
Now, we need to find f(10), which represents the coffee's temperature at t = 10. To do that, we'll need to integrate f '(t ) with respect to t. But I have to warn you, integration can be a bit tricky. It's like solving a complex riddle, only to find out it was a knock-knock joke.
Anyway, after some calculations (or maybe a little magic, who knows?), we find that:
f(10) = -32t^2 + C,
where C is the constant of integration. To determine the value of C, we need an initial condition. In this case, we know that at t = 0, the coffee's temperature is 90 degrees Celsius.
So, let's substitute t = 0 and f(0) = 90 into the equation:
f(0) = -32(0)^2 + C
= C
Therefore, C = 90.
Now we can finally find f(10):
f(10) = -32(10)^2 + 90
= -3200 + 90
= -3110 degrees Celsius
Wait a minute... -3110 degrees Celsius? Something seems off. I think we might have made a little mistake along the way. Let's double-check our calculations and make sure we didn't accidentally invent a new type of coffee that can withstand absolute zero temperatures.
So, the coffee's temperature when t = 10 is... a mystery for now! But fear not, with a little more precise calculation, we'll be able to solve this riddle and find the true answer.
To estimate the coffee's temperature when t = 10, we can use the given rate of change equation f'(t) = -8(0.8)t degrees Celsius per minute.
First, we need to integrate the rate of change equation to find f(t), the temperature function.
∫ f'(t) dt = ∫ -8(0.8)t dt
Integrating, we get:
f(t) = -8(0.8)(t^2/2) + C
Since we know that the coffee's temperature at t = 0 is 90 degrees Celsius, we can substitute these values into the equation to find the value of C.
f(0) = -8(0.8)(0^2/2) + C
C = 90
So, the temperature function f(t) becomes:
f(t) = -8(0.8)(t^2/2) + 90
Now, let's substitute t = 10 into the equation to get the estimated temperature:
f(10) = -8(0.8)(10^2/2) + 90
f(10) = -8(0.8)(100/2) + 90
f(10) = -8(0.8)(50) + 90
f(10) = -32 + 90
f(10) = 58
Therefore, the estimated temperature of the coffee when t = 10 is 58 degrees Celsius.
To estimate the coffee's temperature when t = 10, we can use the given information that the rate of change of the temperature is given by f '(t) = -8(0.8)t degrees Celsius per minute.
To find the temperature at a specific time, we need to integrate the rate of change function. In this case, we integrate f '(t) with respect to t to find f(t).
In order to integrate, we need to solve the indefinite integral of f '(t). Let's break down the steps to solve this integral:
f '(t) = -8(0.8)t (Given rate of change)
To integrate f '(t) with respect to t, we apply the power rule of integration:
∫ f '(t) dt = ∫ (-8(0.8)t) dt
Applying the power rule, we increase the exponent by 1 and divide by the new exponent:
∫ f '(t) dt = -8(0.8) * (1/2) * t^2 + C
where C is the constant of integration.
Now we can find f(t) by substituting t = 10 in the obtained equation:
f(10) = -8(0.8) * (1/2) * 10^2 + C
To estimate the temperature when t = 10, we need the value of the constant of integration (C). Unfortunately, without knowing the initial temperature, we cannot determine the exact value of C.
However, if we assume the initial temperature of the coffee is 90 degrees Celsius, we can calculate the value of C:
f(0) = -8(0.8) * (1/2) * 0^2 + C
90 = 0 + C
C = 90
Now we can substitute C = 90 into the equation to estimate the temperature when t = 10:
f(10) = -8(0.8) * (1/2) * 10^2 + 90
f(10) = -8(0.8) * 50 + 90
f(10) ≈ -32 + 90
Estimating the value, we get:
f(10) ≈ 58 degrees Celsius
Therefore, the estimated temperature of the coffee when t = 10 is approximately 58 degrees Celsius.