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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 9

Lesson 3: AP Calculus AB 2011 free response- 2011 Calculus AB free response #1a
- 2011 Calculus AB Free Response #1 (b, c, & d)
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c

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# 2011 Calculus AB free response #2 (c & d)

Second fundamental theorem of calculus application. Created by Sal Khan.

## Want to join the conversation?

- In this video Khan rounds at the hundreths place but I heard that for the AP Exam you need to round at the thousandths place. Is this true?(6 votes)
- From College Board:
*"As on previous AP Calculus Exams, a decimal answer must be correct to three decimal places unless otherwise indicated. Students should be cautioned against rounding values in intermediate steps before a final calculation is made. "*

So I guess the answer is yes!(7 votes)

- In a question like this, do I need to worry about the number of significant digits? The table of the temperatures of the tea gives us data with two significant digits. When we find the temperature of the biscuits to four significant digits and compare them to the temperature of the tea (which has two), aren't the two digits to the right of the decimal point meaningless?(4 votes)
- When you integrate speed you get position, and when you differentiate speed you get acceleration. Is there a correlation with temperature? I can see when you differentiate it you get change in temperature over time. Is there any way to interpret the integral of a temperature function?(3 votes)
- @Cody - if you're reading this I want to say thank you for your very clear explanation. What you said makes perfect sense.(3 votes)

- at3:50when it's the integral of the derivative why isn't it the same as in part c? I thought it was the same thing where it's H'(10)'H'(0) in c so that part d is just B'(10) -B'(0) ? Why does he actually integrate it?(3 votes)
- From my understanding of the information given this is because in the table they give you values for H(t) because they don't give you the equation for h'(t), so you can't integrate it yourself. However they do give you the equation for b'(t) and you have to integrate it to find the B(t) function to find out the values at the given times(3 votes)

- At around6:39: Wherefore do you subtract 65.817247 from 100?(2 votes)
- B(t) = integral B'(t) from 0 to t + y-intercept

It's said that when t = 0,

B(0) = 100

You can think of it as.. if x = 0, then y = 100

y-intercept is 100

B(0) = integral B'(t) from 0 to 0 + y-intercept

B(0) = 0 + y-intercept

B(0) = y-intercept

B(0) = 100

so when actually solving the problem where t = 10...

B(10) = integral B'(t) from 0 to 10 + 100

B(10) = -65.8172 + 100

does that help clarify?(3 votes)

- At3:48, why do you divide by -.173?(2 votes)
- Why didn't he just integrate b'(t) to get b(t) and use the initial condition of b(0)=100 to solve for the constant of integration and then just plug in t=10?(2 votes)
- Is it okay if in my answer to part d, I actually divided -13.84 by -0.173 to get 80 before proceeding with the question. After dividing the two and evaluating the integral, I got 80(e^-0.173t)+C. Knowing the temperature is 100 when t=0, I set t=0 and evaluated. That then helped me figure out that C must be 20. Then, using my equation B(t)= 80(e^-0.173t)+20, I figured out that when t=10, the temperature of the biscuits is 34.18. 43-34.18 gave me 8.82.(2 votes)
- For problem D, the question states that the equation represents the temperature of the biscuits at time t. Can't you just substitute 10 for t in the equation, solve, then subtract the answer from 100 to get the temperature difference?

I know I must be missing something here, because it doesn't seem necessary to integrate. There's no rate; only a value exists.(2 votes)

## Video transcript

We're now on part c. "Evaluate
the definite integral from 0 to 10 of
H prime of (t)dt. Using correct units,
explain the meaning of the expression in the
context of this problem." So we know that if we
just want to evaluate this definite integral from
0 to 10 of H prime of (t)dt, this is just the same thing as
evaluating the anti-derivative of this thing right over here,
which is just H of t at 10. And subtract from that the
anti-derivative of this evaluated at 0. This is the second fundamental
theorem of calculus. This is exactly how we
evaluate definite integrals. And when you look
at it from this, you just see that
when you evaluate it, it just gives us the
difference in temperature from 0 minutes to 10 minutes. We're taking the
temperature at 10 minutes and from that we're subtracting
from that the temperature at 0 minutes. So this is really our
change in temperature over those 10 minutes. And we can actually evaluate it. We know what our temperature
was after 10 minutes. H of 10 is 43 degrees Celsius. So this is 43 right over here. And from that, we're
going to subtract our initial temperature,
our temperature at 0 minutes, which
is 66 degrees Celsius. We're going to subtract
66 degrees Celsius. This gives us negative
23 degrees Celsius. So our change in temperature
is negative 23 degrees. Or our temperature has gone
down 23 degrees Celsius over the course of
the first 10 minutes. So that is part c
right over there. Now, let's do part d. "At time t equals 0, biscuits
with temperature 100 degrees were removed from an oven." So now we're talking
about biscuits. We started with
tea, now biscuits. "The temperature of
the biscuits at time t is modeled by a
differentiable function B, for which it is
known that B prime of t" is equal to this
business right over here. "Using the given
models at time t equals 10, how much cooler are
the biscuits than the tea?" Well, we know what the
temperature of the tea is. So we just have to figure out
the temperature of the biscuits to figure out how much
cooler they are than the tea. And to figure out the
temperature of the biscuits, we can essentially just
use the exact same idea. We can say, we know that
the biscuits started off at 100 degrees Celsius. And we could say, well, how
much did they cool down? Or what was their
change in temperature over the 10 minutes? If we know the change
in temperature, and we know it started
at 100 degrees, then we can use that
information to get what its temperature
is at 10 minutes. And then we can
answer their question. So the change in temperature
over those 10 minutes is just the definite
integral from 0 to 10 of this business
right over here. Let me just write it. So you see it's the
exact same pattern as what we saw over here. This was for the cookies. Now we're talking
about the biscuits. And this right
over here is going to be the definite
integral from 0 to 10. B prime of t-- they give
it to us right over here-- is negative 13.84e to
the negative 0.173t, dt. And now we just evaluate this. Well, we can evaluate this. What we'd have to
is-- because we know the derivative of e
to the a x is a e to the x. Let me write that. The derivative of e to the
ax is equal to ae to the ax, just from the chain rule. Derivative of the
inside is just a. And then we multiply
that times the derivative of the entire thing. And derivative of e to
the x is just e to the x. Or we could say that the
integral of e to the ax dx is equal to 1 over a
e to the ax plus c. And you could take
the derivative of this to see that you would
get this right over here. So using that same idea,
the anti-derivative of this over here is just
going to be the negative 13.84. And we're going to divide
by this coefficient on the t right over here. So negative 0.173. And you could use
your calculator. Calculators are allowed for
this part of the problem. But we could do
this analytically. Times e to the
negative 0.173 times t. We're going to evaluate
that from 0 to 10. So we're going to
evaluate it at 10 and subtract from that
this thing evaluated at 0. Let me just factor out
this part right over here. So it's going to be negative
13.84 over negative 0.173 times this evaluated
at when t is 10. So e to the-- if we
multiply this times 10-- this is negative 1.73. That's when I
evaluated it at 10. And from that we
want to subtract when this is evaluated at 0. If the exponent here is 0, if t
is 0, the whole exponent is 0. So e to the 0 power
is just equal to 1. And now we can get
our calculator out to evaluate this. So I'll get my TI-85 out. I'm just going to evaluate this
inner expression right over here. So e to the negative 1.73. And from that I
want to subtract 1. So that gives me negative 0.822. So that's this part
in parentheses. And then I want to multiply it
times what I have out front. So my previous
answer times-- and I won't write the negatives just
because the negative divided by negative is going
to be a positive. So it's just 13.84
divided by 0.173. And I didn't write
the negatives, because they just cancel out. And that gives me
negative 65.817. So this is our change in
temperature for the biscuits. It's our change in temperature. Let me write it down. Negative 65.82 degrees. So this is negative
65.82 degrees Celsius. This is our change in
temp for the biscuits. Now we know that they
started off at 100 degrees, and they went down by 65.82
degrees over the 10 minutes. They started at
100, and then they went down by this
previous amount. So, whoops. I want to subtract the answer. So minus the answer. Oh, sorry. I want to actually add
the change in temperature. I want to be very careful here. It went down by 65 degrees. So I really should just
say 100 minus 65.817. I could keep adding
digits if I want. 247, that's enough. So after 10 minutes the
biscuits are 34.18 degrees. And we already knew what
the temperature of the tea is after 10 minutes. We already knew
it was 43 degrees. So the biscuits
are at 34 degrees. The tea is at 43 degrees. So if we do 43 is
where the tea is, then we subtract the temperature
of the biscuits, we see that the biscuits
are 8.82 degrees cooler. 8.82 degrees Celsius cooler. So we could say the biscuits
are 8.82 degrees Celsius cooler than the tea. And we're done with
part d as well.