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# Maths help c1 Watch

1. Can anyone help me with these questions?

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8a, 8b and 10b

Ermm well I don't really know where to start! help
for 8b i worked out the area under the curve between the two points, but idk if that was right or what to do next
2. (Original post by emma363)
Can anyone help me with these questions?

8a, 8b and 10b

Ermm well I don't really know where to start! help
for 8b i worked out the area under the curve between the two people, but idk if that was right or what to do next
I dunno what you're on about by saying 'people', but i'll assume you meant lines/curves.

8a) You've gotta integrate. Find the equation of the straight line that passes through P and Q. You've got the two coordinates, so just find the gradient, and use y=mx+c. You've now got 2 equations, one for the straight line, and one for the curve. If you look at the diagram, you should see the the shaded area is the area underneath the straight line with the area underneath the curved line subtracted from it. If you let the straight line = y1 and the curved line = y2, then you do:
where y1 and y2 are their respective equations.

8b) A perpendicular bisector cuts the line between two points, in this case P & Q, at 90° to the line into two equal sections. From part a), you should have found the gradient of the straight line passing through P & Q. The gradient of a line perpendicular to that line is the negative reciprocal, or . Find the midpoint of the straight line PQ and with this coordinate and the gradient, you can find the equation of the perpendicular bisector using y=mx+c.

10b) The area underneath the curve between 0 and 2 can be worked out using normal integration, as you did in part a), but the equation is slightly different. The area bound by 2 and k is under the x-axis, so in order to work out and expression for this area, you need to take the negative of the function given (i.e flipping the whole graph vertically), so you get for this new area. Integrate this as normal with the limits 2 and k, k > 2, and you should get an expression in terms of k. Equate this to the area underneath the curve between 0 and 2 to get a value of k.

Hope i made sense.
3. (Original post by F1Addict)
I dunno what you're on about by saying 'people', but i'll assume you meant lines/curves.

8a) You've gotta integrate. Find the equation of the straight line that passes through P and Q. You've got the two coordinates, so just find the gradient, and use y=mx+c. You've now got 2 equations, one for the straight line, and one for the curve. If you look at the diagram, you should see the the shaded area is the area underneath the straight line with the area underneath the curved line subtracted from it. If you let the straight line = y1 and the curved line = y2, then you do:
where y1 and y2 are their respective equations.

8b) A perpendicular bisector cuts the line between two points, in this case P &amp; Q, at 90° to the line into two equal sections. From part a), you should have found the gradient of the straight line passing through P &amp; Q. The gradient of a line perpendicular to that line is the negative reciprocal, or . Find the midpoint of the straight line PQ and with this coordinate and the gradient, you can find the equation of the perpendicular bisector using y=mx+c.

10b) The area underneath the curve between 0 and 2 can be worked out using normal integration, as you did in part a), but the equation is slightly different. The area bound by 2 and k is under the x-axis, so in order to work out and expression for this area, you need to take the negative of the function given (i.e flipping the whole graph vertically), so you get for this new area. Integrate this as normal with the limits 2 and k, k &gt; 2, and you should get an expression in terms of k. Equate this to the area underneath the curve between 0 and 2 to get a value of k.

Hope i made sense.
I have NO idea why I put people O_O
anyways I get it now thankyou
4. (Original post by emma363)
I have NO idea why I put people O_O
anyways I get it now thankyou
no problem. glad to help.

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