Question

Cleaning pumps remove oil at the rate modeled by the function R, given by R(t)=2+cos(pi*t/11) with t measure in hours and and R(t) measured in gallons per hour. How much oil will the pumping stations remove during the 6-hour period from t = 0 to t = 6?

Answer 1

So they want you to find the value of \(\LARGE \int_{0}^{6}R(t)dt\)

Answer 2

Wait wrong question

Answer 3

Which means you'll need to find the antiderivative of \(\Large 2+\cos\left(\pi*\frac{t}{11}\right)\)

Answer 4

you already figured this one out?

Answer 5

Yes

Answer 6

Let F(x) integral from [0, 2x] tan(t^2) dt. Use your calculator to find F″(1).

Answer 7

are you familiar with the fundamental theorem of calculus (FTC) ?

Answer 8

yes

Answer 9

If
\[\Large f(x) = \int_{a}^{x}g(t)dt\]
then
\[\Large f \ '(x) = g(x)\]

Answer 10

yes, so how does the second derivative relate?

Answer 11

A slight variation of the rule
If
\[\Large f(x) = \int_{a}^{h(x)}g(t)dt\]
then
\[\Large f \ '(x) = g(h(x))*h \ '(x)\]
where h(x) is any function of x

Answer 12

the second derivative is simply the derivative of the first derivative

Answer 13

yes, that makes sense

Answer 14

so then the derivative of tan(x^2) is 2xsec^2(x^2) right?

Answer 15

`so then the derivative of tan(x^2) is 2xsec^2(x^2) right?`
correct

Answer 16

And then just plug in 1?

Answer 17

Original
\[\Large F(x) = \int_{0}^{2x}\tan(t^2)dt\]
------------------------------------------
First Derivative
\[\Large F \ '(x) = \tan((2x)^2)*\frac{d}{dx}[2x]\]
\[\Large F \ '(x) = \tan(4x^2)*2\]
\[\Large F \ '(x) = 2\tan(4x^2)\]
------------------------------------------
Second Derivative
\[\Large F \ ''(x) = \frac{d}{dx}\left[F \ '(x)\right]\]
\[\Large F \ ''(x) = \frac{d}{dx}\left[2\tan(4x^2)\right]\]
\[\Large F \ ''(x) = 2\sec^{2}(4x^2)*\frac{d}{dx}[4x^2]\]
\[\Large F \ ''(x) = 2\sec^{2}(4x^2)*8x\]
\[\Large F \ ''(x) = 2*8x\sec^{2}(4x^2)\]
\[\Large F \ ''(x) = 16x\sec^{2}(4x^2)\]

Answer 18

what you said is correct, but there were certain pieces missing. What I just posted is how to do the full problem