This week we started Chapter 5. We learned how to estimate the area under a curve using LRAM, MRAM, and RRAM. For LRAM, you would break up the x-axis into 4-5 intervals. Then you would draw 2 vertical lines up from that interval, making sure that the left corner of the block touches the curve. This would work the same for MRAM except the middle of the block would touch the curve, and RRAM except the right corner would touch the curve. Then you would add up the heights of each block, and multiply that by the interval you used.
An example of this would be:
Use RRAM to estimate the area under the curve and above the x-axis for a function between 0 and 2 when f(x)=(1/(x+1))+1. Since the closed interval is between 0 and 2, I would go by halves. I'd graph the curve, then find what y equalled when x=1/2. Then I'd find what y equalled when x=1, 3/2, and 2. The points would be (1/2,5/3) (1,3/2) (3/2,7/5) and (2,4/3). Since I went by halves, I'd multiple the sum of the y values by 1/2 so it would be 1/2(5/3+3/2+7/5+4/3). After doing that calculation, I would find that my answer is 2.95.
At the beginning of the week we took a quiz over optimization. I thought it was pretty easy, I just had trouble with the problem about Mr. Cresswell and his dog. One of the problems said Find the area of the largest rectangle that can be inscribed inside a semi circle with radius 5. (Given by the equation y=((25-x^2)^1/2) Since the y coordinate was given by the equation of the semi circle, I knew I was looking for the x coordinate. I used the area formula to find the area of the inscribed rectangle (A=l*w). l=2x because from 0 to some point is unknown so it'd be x, and if you reflect it on the other side of the y-axis it would be 2x. w is given by the function, so w=((25-x^2)^1/2). Then I multiplied them together A=2x((25-x^2)^1/2). Next, I took the derivative and got 2x*1/2(25-x^2)^(-1/2)*(-2x)+(25-x^2)^(1/2)*2. I'd solve for x and get that x equals about 3.536. I'd plug that back into the length equation of 2x to get 2(3.536)=7.072. I'd also plug that x value into the width equation of (25-x^2)^1/2) to get (25-(3.536)^2)^1/2)=3.535. I'd multiply the length and width together to get the area to get 7.072*3.535=24.9 so about 25.
An example of this would be:
Use RRAM to estimate the area under the curve and above the x-axis for a function between 0 and 2 when f(x)=(1/(x+1))+1. Since the closed interval is between 0 and 2, I would go by halves. I'd graph the curve, then find what y equalled when x=1/2. Then I'd find what y equalled when x=1, 3/2, and 2. The points would be (1/2,5/3) (1,3/2) (3/2,7/5) and (2,4/3). Since I went by halves, I'd multiple the sum of the y values by 1/2 so it would be 1/2(5/3+3/2+7/5+4/3). After doing that calculation, I would find that my answer is 2.95.
At the beginning of the week we took a quiz over optimization. I thought it was pretty easy, I just had trouble with the problem about Mr. Cresswell and his dog. One of the problems said Find the area of the largest rectangle that can be inscribed inside a semi circle with radius 5. (Given by the equation y=((25-x^2)^1/2) Since the y coordinate was given by the equation of the semi circle, I knew I was looking for the x coordinate. I used the area formula to find the area of the inscribed rectangle (A=l*w). l=2x because from 0 to some point is unknown so it'd be x, and if you reflect it on the other side of the y-axis it would be 2x. w is given by the function, so w=((25-x^2)^1/2). Then I multiplied them together A=2x((25-x^2)^1/2). Next, I took the derivative and got 2x*1/2(25-x^2)^(-1/2)*(-2x)+(25-x^2)^(1/2)*2. I'd solve for x and get that x equals about 3.536. I'd plug that back into the length equation of 2x to get 2(3.536)=7.072. I'd also plug that x value into the width equation of (25-x^2)^1/2) to get (25-(3.536)^2)^1/2)=3.535. I'd multiply the length and width together to get the area to get 7.072*3.535=24.9 so about 25.