I am not sure if I put this in the right spot. I probably put it in the wrong spot, but I do not know where it should go.
On Friday February 6, 2009 the B period worked on a worksheet. It was an AP problem from 2005. It involved indefinite integrals also known as “accumulators.” We worked on this work sheet all class. Question A equaled 31.81593137. We had the integral from 0 to 6 of 2 +5sin(4piX/25)dt. On question B. we had to write an expression for y(t). Y(t) equaled 2500 (the amount of sand at time 0) plus the integral from o to t of s(t) – r(t). On question C, we found the rate of which the total amount of sand on the beach was changing at time 4. We first found y’(t). y’(t) = s(t)-r(t). Then we plugged in 4 and got -1.908750647. Then we solved for question D. We made y’ = 0 and then solved for it and got r(t)=s(t). We then graphed r(t) and s(t) on our calculators and got 5.12. We plugged 5.12 into y(t) and got the minimum value which was 2492.37 cubic yards.
Today, February 9th, we worked on finding the volume of washers (ring like objects, or objects that have hallowed out centers). In order to do this, we found the volume of the entire object including the center of the washer, even though there was nothing there. Then, we subtracted out this smaller center volume. So, we found an integral for the higher equation (top of the object) and subtracted the integral for the lower equation. For example, we did a problem that had a curve and a line (connected to look like the side of a bowl) and wanted to flip it over the x axis to create a 3D shape. The equation of the curve of this shape was y=4-x^2 and the equation of the line was y=3. So the Volume of the washer was (pi *Integral from -1 to 1 of (4-x^2)^2 *dx)-(pi*Integral from -1 to 1 of (3)^2*dx). The original equations for this process were (pi* Larger Radius^2*height)-(pi*Smaller Radius^2 *height), so the equations of the line and curve were just plugged in and the boundaries of the integral were the maximum and minimim points on the x axis of the shape.
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I am not sure if I put this in the right spot. I probably put it in the wrong spot, but I do not know where it should go.
On Friday February 6, 2009 the B period worked on a worksheet. It was an AP problem from 2005. It involved indefinite integrals also known as “accumulators.” We worked on this work sheet all class. Question A equaled 31.81593137. We had the integral from 0 to 6 of 2 +5sin(4piX/25)dt. On question B. we had to write an expression for y(t). Y(t) equaled 2500 (the amount of sand at time 0) plus the integral from o to t of s(t) – r(t). On question C, we found the rate of which the total amount of sand on the beach was changing at time 4. We first found y’(t). y’(t) = s(t)-r(t). Then we plugged in 4 and got -1.908750647. Then we solved for question D. We made y’ = 0 and then solved for it and got r(t)=s(t). We then graphed r(t) and s(t) on our calculators and got 5.12. We plugged 5.12 into y(t) and got the minimum value which was 2492.37 cubic yards.
Today, February 9th, we worked on finding the volume of washers (ring like objects, or objects that have hallowed out centers). In order to do this, we found the volume of the entire object including the center of the washer, even though there was nothing there. Then, we subtracted out this smaller center volume. So, we found an integral for the higher equation (top of the object) and subtracted the integral for the lower equation. For example, we did a problem that had a curve and a line (connected to look like the side of a bowl) and wanted to flip it over the x axis to create a 3D shape. The equation of the curve of this shape was y=4-x^2 and the equation of the line was y=3. So the Volume of the washer was (pi *Integral from -1 to 1 of (4-x^2)^2 *dx)-(pi*Integral from -1 to 1 of (3)^2*dx). The original equations for this process were (pi* Larger Radius^2*height)-(pi*Smaller Radius^2 *height), so the equations of the line and curve were just plugged in and the boundaries of the integral were the maximum and minimim points on the x axis of the shape.
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