AP Calculus AB 1998 Multiple Choice Part VIII (86-92)

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Good day students AT part eight of the eighty Eight multiple-choice trivia questions off for nineteen ninety-eight so is going to going over questions 86 to 92 in this uhm installment. Right let's take a look at question number eighty-six so question eighty-six it says the base of a solid is the region and the first quadrant bounded by the X axes y-axis and the line X was to why equals eight is shown in the figure above. If cross-sections of the solid perpendicular to the X axes are semi circles what is the volume of of the solid okay so this is the basically assessing your understanding of solids with known cross-sections's right so. The going to make a sketch of the situation, but if I do that for me labeled this function right here so we looking for the salt the volume of the solid created by of the base down device functions this function in a access. So let's call this the top function since our segment is going to be perpendicular to the X axis and then look something like this am that the cross-section of the semi circles so we have a this is the function on the top the call this of F of X top right to going top-down and then at the bottom this is this is G of X bottom right so for one of find the length of the segment which is basically find the difference between these two functions and one thing one and notes is that this function down here is basically the function why equals zero right and in the top function is of why equals eight minus X over two ago ridiculous I got that later but just know that this is the top in this is about a function so can you visualize the solid that of is being created here to help you visualize it is go ahead and sketched of the situation on going the rotate y-axis and just lay down flat. So this is my y-axis and that's my X axis and that's the line of why equals X case but it like that okay so what we have is of so this is why this is X the have semi circle sitting right on this in such a way that you perpendicular to the X axis so this is one of the semi circles limit to one of them. So this is one semi circle perpendicular to the X axis imagine that this is ninety degrees right here okay that's ninety right so the semi circle look something like this okay so that's one of them in a was going on is that you have an infinite amount of semi circle just like that okay so the question is you have over to semi circle just like that over this entire region what is the of of volume of that solid going to be to that's what we doing in this problem okay right so of in order to do this we have to find the of area the cross-sectional area of of our semi circle in the with and evaluate an integral by the one and notes that for this problem with going from zero all the way to eight so let's take a look at this the segment for example so the semi circle right here how the damage are okay of the denominator is the and that D is basically the difference between this function on the top and the function on the bottom right so is basically F of X minus G of X where F of X is the one on top in G of X is about as the bottom function okay so F of X is it is function right here of isolate is for Y of X plus two why equals eighty by solid for why you have why equals eight minus X over two so this is my top function by top function is eight minus two X answer is minus X eight minus X over to into my bottom function is the x-axis which is why equals zero okay so that's the denominator of our of semi circle okay eight minus X over to now of let's see me write down the formula that were going to use to find the area of the resultant solid so generally speaking is one of find the area of of solids is known cross-section a be do it semi circle this is the formula right here so is putting green the formula if you going along the X axis everything that in terms of X okay so the

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For question number 91, we use the definition of a continuous function to determine that the function that we are dealing with will not have any discontinuities (corner, cusp, and vertical tangents). Using the intermediate value theorem we can use this to determine where the function as x-intercepts or zeroes. (this is states that there will be a guaranteed zero between the y interval. In this case since 0 is between -5 and 5, there are zeros). The mean value theorem states that the slope of the secant line between two point on a function that is continuous, there will be a point on the graph in which the tangent's slope will equal the secant's slope. While the video did show examples of cases for the intermediate value theorem and the mean value theorem, I still do not know the actual definitions for the theorems stated within the video. The only possible mistakes that I could think of is if you forget the definitions of the theorems mentioned. Other than that, I have no questions.

danylopez

For question number 91, we use the definition of a continuous function to determine that the function that we are dealing with will not have any discontinuities. Using the intermediate value theorem we can use this to determine where the function as x-intercepts or zeroes. The mean value theorem states that the slope of the secant line between two point on a function that is continuous, there will be a point on the graph in which the tangent's slope will equal the secant's slope. While the video did show examples of cases for the intermediate value theorem and the mean value theorem, I still do not know the actual definitions for the theorems stated within the video. The only possible mistakes that I could think of is if you forget the definitions of the theorems mentioned.

DJBJAMIN

Thanks so much
I almost fully understand most problems. You are really good

zakf

I have watched it. The review with finding lines tangent to other lines was helpful because I thought you had to derive twice, but it's only once.

liliansalino

I have watched the video, and I feel more confident in using the calculator to help me solve the problems

Tman

I have watched the video. I learned that using a Ti-89 can make solving problems much easier.

ricardodeluna

Question 90, the formulas used were the area of a triangle and the rate related formulas. A misunderstanding was that I did not know I had to use implicit differentiation for the height part of this problem. A mistake in this problem can be that when solving for da/dt you forget to use the product formula uv prime. I question I had when reading this problem was I going to determine what was true based on the options in the multiple choice. Then I figured out that you plug in the lowest number and the highest number to figure that out.

nallelylozoya