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So stuck can someone write the solutions to these two questions (1 Viewer)
- Thread starter kpad5991
- Start date
fluffchuck
Active Member
I won't give the solutions to these just yet, however, I will give you some hints.
20. (a) An area between 2 curves is rotated about the x-axis. Try sketching the functions that you are given on a number plane (you will notice that one volume is contained within the other volume, to find the required volume, you would subtract one volume from the other V = V2 - V1).
(b) This is a regular volumes question, you should be able to do this.
21. (a) y = 6e^(-x) and y = e^x - 1
They want the point of intersection, so wouldn't you just equate these two curves?
6e^(-x) = e^x - 1
6 = (e^x)^2 - e^x
(e^x)^2 - e^x - 6 = 0
Comparing what you have and what you need, you would just use the substitution given in the question.
(b) Solve the quadratic. You should get u = 3 and u = -2, hence e^x = 3 and e^x = -2 and so on.
(c) You should be able to do this.
(d) Area between two curves, whereby the boundaries are x = 0 and x = ln(3) (from your sketch). From here, you can use
Area = integral from x = 0 to x = ln3 of (y2 - y1) dx and evaluate this.
20. (a) An area between 2 curves is rotated about the x-axis. Try sketching the functions that you are given on a number plane (you will notice that one volume is contained within the other volume, to find the required volume, you would subtract one volume from the other V = V2 - V1).
(b) This is a regular volumes question, you should be able to do this.
21. (a) y = 6e^(-x) and y = e^x - 1
They want the point of intersection, so wouldn't you just equate these two curves?
6e^(-x) = e^x - 1
6 = (e^x)^2 - e^x
(e^x)^2 - e^x - 6 = 0
Comparing what you have and what you need, you would just use the substitution given in the question.
(b) Solve the quadratic. You should get u = 3 and u = -2, hence e^x = 3 and e^x = -2 and so on.
(c) You should be able to do this.
(d) Area between two curves, whereby the boundaries are x = 0 and x = ln(3) (from your sketch). From here, you can use
Area = integral from x = 0 to x = ln3 of (y2 - y1) dx and evaluate this.
dan964
what
(Q20a)
try reading the question and drawing a diagram. You will have to done 20a first to make the comparison. Graph the region 20a and then sent it whizzing around the correct axis. Repeat by graphing the region in 20b and send it whizzing around the correct axis.
Choose your weapon:
- slices
- shells
- cross sectional area.
and go from there. If one doesn't work try another. Because it is not exam, important you learn how to do it.
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