QuestionIncorrect .forest fire covers 2004 acres at time0. The fire growing at rate of 8V acres per hour; where is in hours. How many acres are covered 24 hours lat...

A forest fire covers 2000 acres at time $t=0 .$ The fire is growing at a rate of $8 \sqrt{t}$ acres per hour, where $t$ is in hours. How many acres are covered 24 hours later?

So in this problem, we're told that a fire doubles inside every eight outwards. Now we're told out of fire initial size was one acre, so I don't know how many acres will be covered in three days. Well, if it's initially one acre, how much is it going to get? If what member is going to double every eight hours? So every eight hours, we're going to double it. So after the 1st 8 hours, it would then be two acres, so that would be after eight hours. And now we can continue this pattern. So after 16 hours, it would double again. It would be four acres and then look sorry, my cursor one a little crazy there. So we couldn't do this pattern again. So essentially we could just keep doubling it. Remember, in for three days, that would be 72 hours. So if I double it would get eight and this would be after 24 hours. We doubles again to be 16 after added again 32 hours. If we double 16 we get 32 this would be after for the hours. If we double this again, we're going to get 64 which is going to be after 48 hours, which is two days. Okay, we can keep doubling. So we have 128 and this would be after 56 hours. If we double 128 we're going to get 256 which would be after 64 hours. And if we double 256 we're gonna get 512 which is after 72 hours, which is three days. So our final answer would be 512 acres would be covered after three days.

Okay, so we're given a function of of access stands for an actual forested area. So um also within F of X. We have a start value plus an integral. And remember an integral is all about a change. So, um we'll be considering kind of geometric, we'll we'll sketch this out and we'll consider that it has a region underneath the curb. So notice it says that R. M. T. Has only 10 and it's at T equals 40. So let's kind of start to sketch this will put the 40 right there. And then it also says that the derivative of our at 40 is less than zero. So that means that we are a decreasing function. Now, if it only goes through 40, 1 time there's a few different ways that we can draw this, but we know that we can't go across the X axis again. Um so this is just one kind of sketch of what it could be. There are multiples out there, but this gives us a good representation because that we know that we have a positive area underneath our curve before 40 and then a negative area um after 40. So when it asks us for the maximum value of the forest, what we can see that we were accumulating all the way until we got to 40 and then we start decreasing from there. So we definitely have to be our max at 40. But let's go ahead and just kind of write out why that is true. So normally if you have to maximize something or minimize it, you take the derivative and set it equal to zero. So consider for a second that we do take that derivative of F. Of X. And we look at when it goes from positive to negative negative. Okay, So if we're taking the derivative of F of X, okay, if we're taking the derivative of F. Of X, that 8 20 would go away. And then we would use our second fundamental theorem of calculus to take the derivative of our next piece. And remember that just places the T. Straight in and you multiply by the derivative what you put in, which is one if you put zero in, you'd be multiplying by its derivative which is zero. And so that's not needed. So really um are derivative is all about our functions are in terms of like what the slopes doing so are was positive before 40. So it always has. So we can say that F was increasing before 40 And then it's negative after 40. And so f would be decreasing after 40. And so our maximum is actually at 40. And they give us that value in the problem as 1170. Now. Um the next part asked how large is the forest area at time equals 30. So we know that it has to be less and they already told us that the integral from 30 to 40 would be um 82. So let's go ahead and look at setting this up more formally. So if we need to find a value at 30 we can consider that we started at a value of 40 like a known 40 and then we would integrate from 40 to 30 of our function are empty. So what that means is that we're starting at that value of 1170 then notice we're integrating from 40 to 30. So you can do the negative and flip it and integrate from 30 to 40. So that's why we're subtracting the 82 which gives us a value of 1000 and 88 of our unit here.

The equation tells us that the number of acres left to mow is 1350 minus 1.2 times the number of hours worked. Well, if 20 hours were worked, we just put a 20 in here for X. I'm gonna put that in a calculator. So it's just 13. 50 minus 1.2 halves. 20. And that would mean that there are 1326 acres. Bring me.

Same question is last time. Only this time we're just gonna put in 100 for X. So I'll put that in a calculator. 13 15 minus 12 times. 100. Of course. We could just do that. Um, Or 1.2 times. 100. Sorry. 13 50 Minus one point. Two times 100. And that's going to give me 1230 acres.