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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?
• 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!
• 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?
• 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?
• @Cody - if you're reading this I want to say thank you for your very clear explanation. What you said makes perfect sense.
• at when 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?
• 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
• At around : Wherefore do you subtract 65.817247 from 100?
(1 vote)
• 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?
• At , why do you divide by -.173?
(1 vote)
• 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?
(1 vote)
• 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.
(1 vote)
• 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.
(1 vote)