anonymous
  • anonymous
The diagram shows the points A(1,2) and B(4,4) on the curve y=2 sqrt x. The line BC is the normal to the curve at B, and C lies on the x-axis. Lines AD and BE are perpendicular to the x-axis. (i) find the equation of the normal BC (ii) FInd the area of the shaded region.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1332855027947:dw|
anonymous
  • anonymous
What are the boundaries for part (ii) You have to integrate y=2 sqrt x, right? But what are the boundaries?
anonymous
  • anonymous
I'm also having trouble with the rest though... So if you can help, I'd be obliged :D

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anonymous
  • anonymous
For the first part you need the slope m and y-intercept. \[y=mx+b\] For the slope \[m=(y_1-y_2)/(x_1-x_2)\] \[m=(2-4)/(1-4)\] \[m=2/3\] Plug that in and one of the points to find the y-intercept \[y=2/3x+b\] \[2=2/3*1+b\] \[b=4/3\] \[y=2/3x+4/3\]
anonymous
  • anonymous
The boundaries for ii are 1 and 4, according to your diagram
anonymous
  • anonymous
Hmm. Ok... but I thought you couldn't get the curve of a slope with that method?
anonymous
  • anonymous
Sorry, I am asleep today, that's the line AB
anonymous
  • anonymous
But, AB is a curve... isn't there a different method to find the slope for a curve? dx/dy?
anonymous
  • anonymous
I mean dy/dx?
anonymous
  • anonymous
dy/dx is it \[dy/dx=1/\sqrt{x}\]
anonymous
  • anonymous
and to find the normal...?
anonymous
  • anonymous
Do you equate it to -1?
anonymous
  • anonymous
jeez my head is spinning
anonymous
  • anonymous
\[y=2\sqrt{x},y'=\frac{1}{\sqrt{x}}\] slope of tangent line is \[\frac{1}{\sqrt{4}}=\frac{1}{2}\] slope of normal line is \[-2\] equation of the line is \[y-4=-2(x-4)\] \[y=-2x+12\]
anonymous
  • anonymous
Yeah :) I got it.. And the second one has boundaries of 1, and 4? For the integration?
anonymous
  • anonymous
\[\int_1^42\sqrt{x}dx\] is second one
anonymous
  • anonymous
great :) Thanks!
anonymous
  • anonymous
Is the answer 28/3?

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