Pleaseee Helppp!!!! (1 Viewer)

[umm what]

Please do Question 11 and 15

THANKUUU SO MUCHH

 

[juantheron]

 

[Nooblet94]

First part's here
<a href="http://www.codecogs.com/eqnedit.php?latex=\begin{align*} A&=\int ^{3}_{-2}\left( x@plus;6-x^2\right )dx\\ &=\left[\frac{x^2}{2}@plus;6x-\frac{x^3}{3} \right ]^{3}_{-2}\\ &=\left(\frac{3^2}{2}@plus;6\cdot 3 -\frac{3^3}{3} \right )-\left(\frac{(-2)^2}{2}@plus;6\cdot (-2) -\frac{(-2)^3}{3} \right )\\ &=\frac{27}{2}@plus;\frac{22}{3}\\ &=\frac{125}{6} \end{align*}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\begin{align*} A&=\int ^{3}_{-2}\left( x+6-x^2\right )dx\\ &=\left[\frac{x^2}{2}+6x-\frac{x^3}{3} \right ]^{3}_{-2}\\ &=\left(\frac{3^2}{2}+6\cdot 3 -\frac{3^3}{3} \right )-\left(\frac{(-2)^2}{2}+6\cdot (-2) -\frac{(-2)^3}{3} \right )\\ &=\frac{27}{2}+\frac{22}{3}\\ &=\frac{125}{6} \end{align*}" title="\begin{align*} A&=\int ^{3}_{-2}\left( x+6-x^2\right )dx\\ &=\left[\frac{x^2}{2}+6x-\frac{x^3}{3} \right ]^{3}_{-2}\\ &=\left(\frac{3^2}{2}+6\cdot 3 -\frac{3^3}{3} \right )-\left(\frac{(-2)^2}{2}+6\cdot (-2) -\frac{(-2)^3}{3} \right )\\ &=\frac{27}{2}+\frac{22}{3}\\ &=\frac{125}{6} \end{align*}" /></a>

I'll do the second part once I get back from the gym if nobody's posted it by then.

 

[gurmies]

That's a very inefficient way of doing the first part. The two quarter-circles cancel out and you're left with finding the area of a trapezium. No integration necessary.

 

[Nooblet94]

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ \textrm{Length of Base of}~\triangle ABP=\sqrt{(3-(-2))^2@plus;(9-4)^2}=5\sqrt{2}\\ \textrm{Height}= \frac{|p-p^2@plus;6|}{\sqrt{1^2@plus;1^2}}=\frac{|p-p^2@plus;6|}{\sqrt{2}}\\ ~\\ A=\frac{1}{2}bh=\frac{1}{2}\cdot 5\sqrt{2} \cdot \frac{|p-p^2@plus;6|}{\sqrt{2}}\\ =\frac{5}{2}|p-p^2@plus;6|\\ ~\\ \textrm{Now, we're only concerned with values of p between -2 and 3 and for all of those values}~p-p^2@plus;6~\textrm{is positive, so the expression becomes}~\\ A=\frac{5}{2}(p-p^2@plus;6)\\ \frac{dA}{dp}=\frac{5}{2}(1-2p)\\ \frac{d^2A}{dp^2}=-5\\ \textrm{Maximum value of A occurs when}~\frac{dA}{dP}=0~\textrm{since}~\frac{d^2A}{dp^2}=-5<0~\textrm{for all values of p}\\ ~\\ \frac{5}{2}(1-2p)=0\\ 2p=1\\ p=\frac{1}{2}\\ ~\\ \therefore A_{\textrm{max}}=\frac{5}{2}(\frac{1}{2}-\left(\frac{1}{2} \right )^2@plus;6)=\frac{5}{2}\cdot \frac{25}{4}=\frac{125}{8}~\textrm{Which is }\frac{3}{4}~\textrm{the area of the parabolic segment as required.}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ \textrm{Length of Base of}~\triangle ABP=\sqrt{(3-(-2))^2+(9-4)^2}=5\sqrt{2}\\ \textrm{Height}= \frac{|p-p^2+6|}{\sqrt{1^2+1^2}}=\frac{|p-p^2+6|}{\sqrt{2}}\\ ~\\ A=\frac{1}{2}bh=\frac{1}{2}\cdot 5\sqrt{2} \cdot \frac{|p-p^2+6|}{\sqrt{2}}\\ =\frac{5}{2}|p-p^2+6|\\ ~\\ \textrm{Now, we're only concerned with values of p between -2 and 3 and for all of those values}~p-p^2+6~\textrm{is positive, so the expression becomes}~\\ A=\frac{5}{2}(p-p^2+6)\\ \frac{dA}{dp}=\frac{5}{2}(1-2p)\\ \frac{d^2A}{dp^2}=-5\\ \textrm{Maximum value of A occurs when}~\frac{dA}{dP}=0~\textrm{since}~\frac{d^2A}{dp^2}=-5<0~\textrm{for all values of p}\\ ~\\ \frac{5}{2}(1-2p)=0\\ 2p=1\\ p=\frac{1}{2}\\ ~\\ \therefore A_{\textrm{max}}=\frac{5}{2}(\frac{1}{2}-\left(\frac{1}{2} \right )^2+6)=\frac{5}{2}\cdot \frac{25}{4}=\frac{125}{8}~\textrm{Which is }\frac{3}{4}~\textrm{the area of the parabolic segment as required.}" title="\\ \textrm{Length of Base of}~\triangle ABP=\sqrt{(3-(-2))^2+(9-4)^2}=5\sqrt{2}\\ \textrm{Height}= \frac{|p-p^2+6|}{\sqrt{1^2+1^2}}=\frac{|p-p^2+6|}{\sqrt{2}}\\ ~\\ A=\frac{1}{2}bh=\frac{1}{2}\cdot 5\sqrt{2} \cdot \frac{|p-p^2+6|}{\sqrt{2}}\\ =\frac{5}{2}|p-p^2+6|\\ ~\\ \textrm{Now, we're only concerned with values of p between -2 and 3 and for all of those values}~p-p^2+6~\textrm{is positive, so the expression becomes}~\\ A=\frac{5}{2}(p-p^2+6)\\ \frac{dA}{dp}=\frac{5}{2}(1-2p)\\ \frac{d^2A}{dp^2}=-5\\ \textrm{Maximum value of A occurs when}~\frac{dA}{dP}=0~\textrm{since}~\frac{d^2A}{dp^2}=-5<0~\textrm{for all values of p}\\ ~\\ \frac{5}{2}(1-2p)=0\\ 2p=1\\ p=\frac{1}{2}\\ ~\\ \therefore A_{\textrm{max}}=\frac{5}{2}(\frac{1}{2}-\left(\frac{1}{2} \right )^2+6)=\frac{5}{2}\cdot \frac{25}{4}=\frac{125}{8}~\textrm{Which is }\frac{3}{4}~\textrm{the area of the parabolic segment as required.}" /></a>

Also, this should really be in the MX1 forum.

 

[cutemouse]

View attachment 24360View attachment 24361

Please do Question 11 and 15

THANKUUU SO MUCHH

Where did you get the q's from? They looks familiar...

 

[Timske]

Im guessing past hsc papers "HSC Mathematics by topic"

 

[umm what]

That's a very inefficient way of doing the first part. The two quarter-circles cancel out and you're left with finding the area of a trapezium. No integration necessary.

So u just integrate the equation of y=2 and y= -x+6 ??????and whats the diff b/w part (i) and (ii)????

 

[umm what]

Where did you get the q's from? They looks familiar...

My teacher gave me LOL most of the questions are MX1....and its two unit lol

 

[umm what]

First part's here
<a href="http://www.codecogs.com/eqnedit.php?latex=\begin{align*} A&=\int ^{3}_{-2}\left( x@plus;6-x^2\right )dx\\ &=\left[\frac{x^2}{2}@plus;6x-\frac{x^3}{3} \right ]^{3}_{-2}\\ &=\left(\frac{3^2}{2}@plus;6\cdot 3 -\frac{3^3}{3} \right )-\left(\frac{(-2)^2}{2}@plus;6\cdot (-2) -\frac{(-2)^3}{3} \right )\\ &=\frac{27}{2}@plus;\frac{22}{3}\\ &=\frac{125}{6} \end{align*}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\begin{align*} A&=\int ^{3}_{-2}\left( x+6-x^2\right )dx\\ &=\left[\frac{x^2}{2}+6x-\frac{x^3}{3} \right ]^{3}_{-2}\\ &=\left(\frac{3^2}{2}+6\cdot 3 -\frac{3^3}{3} \right )-\left(\frac{(-2)^2}{2}+6\cdot (-2) -\frac{(-2)^3}{3} \right )\\ &=\frac{27}{2}+\frac{22}{3}\\ &=\frac{125}{6} \end{align*}" title="\begin{align*} A&=\int ^{3}_{-2}\left( x+6-x^2\right )dx\\ &=\left[\frac{x^2}{2}+6x-\frac{x^3}{3} \right ]^{3}_{-2}\\ &=\left(\frac{3^2}{2}+6\cdot 3 -\frac{3^3}{3} \right )-\left(\frac{(-2)^2}{2}+6\cdot (-2) -\frac{(-2)^3}{3} \right )\\ &=\frac{27}{2}+\frac{22}{3}\\ &=\frac{125}{6} \end{align*}" /></a>

I'll do the second part once I get back from the gym if nobody's posted it by then.

Thanks bro! Can u do the second part as well?

 

[Nooblet94]

Thanks bro! Can u do the second part as well?

I did:

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ \textrm{Length of Base of}~\triangle ABP=\sqrt{(3-(-2))^2@plus;(9-4)^2}=5\sqrt{2}\\ \textrm{Height}= \frac{|p-p^2@plus;6|}{\sqrt{1^2@plus;1^2}}=\frac{|p-p^2@plus;6|}{\sqrt{2}}\\ ~\\ A=\frac{1}{2}bh=\frac{1}{2}\cdot 5\sqrt{2} \cdot \frac{|p-p^2@plus;6|}{\sqrt{2}}\\ =\frac{5}{2}|p-p^2@plus;6|\\ ~\\ \textrm{Now, we're only concerned with values of p between -2 and 3 and for all of those values}~p-p^2@plus;6~\textrm{is positive, so the expression becomes}~\\ A=\frac{5}{2}(p-p^2@plus;6)\\ \frac{dA}{dp}=\frac{5}{2}(1-2p)\\ \frac{d^2A}{dp^2}=-5\\ \textrm{Maximum value of A occurs when}~\frac{dA}{dP}=0~\textrm{since}~\frac{d^2A}{dp^2}=-5<0~\textrm{for all values of p}\\ ~\\ \frac{5}{2}(1-2p)=0\\ 2p=1\\ p=\frac{1}{2}\\ ~\\ \therefore A_{\textrm{max}}=\frac{5}{2}(\frac{1}{2}-\left(\frac{1}{2} \right )^2@plus;6)=\frac{5}{2}\cdot \frac{25}{4}=\frac{125}{8}~\textrm{Which is }\frac{3}{4}~\textrm{the area of the parabolic segment as required.}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ \textrm{Length of Base of}~\triangle ABP=\sqrt{(3-(-2))^2+(9-4)^2}=5\sqrt{2}\\ \textrm{Height}= \frac{|p-p^2+6|}{\sqrt{1^2+1^2}}=\frac{|p-p^2+6|}{\sqrt{2}}\\ ~\\ A=\frac{1}{2}bh=\frac{1}{2}\cdot 5\sqrt{2} \cdot \frac{|p-p^2+6|}{\sqrt{2}}\\ =\frac{5}{2}|p-p^2+6|\\ ~\\ \textrm{Now, we're only concerned with values of p between -2 and 3 and for all of those values}~p-p^2+6~\textrm{is positive, so the expression becomes}~\\ A=\frac{5}{2}(p-p^2+6)\\ \frac{dA}{dp}=\frac{5}{2}(1-2p)\\ \frac{d^2A}{dp^2}=-5\\ \textrm{Maximum value of A occurs when}~\frac{dA}{dP}=0~\textrm{since}~\frac{d^2A}{dp^2}=-5<0~\textrm{for all values of p}\\ ~\\ \frac{5}{2}(1-2p)=0\\ 2p=1\\ p=\frac{1}{2}\\ ~\\ \therefore A_{\textrm{max}}=\frac{5}{2}(\frac{1}{2}-\left(\frac{1}{2} \right )^2+6)=\frac{5}{2}\cdot \frac{25}{4}=\frac{125}{8}~\textrm{Which is }\frac{3}{4}~\textrm{the area of the parabolic segment as required.}" title="\\ \textrm{Length of Base of}~\triangle ABP=\sqrt{(3-(-2))^2+(9-4)^2}=5\sqrt{2}\\ \textrm{Height}= \frac{|p-p^2+6|}{\sqrt{1^2+1^2}}=\frac{|p-p^2+6|}{\sqrt{2}}\\ ~\\ A=\frac{1}{2}bh=\frac{1}{2}\cdot 5\sqrt{2} \cdot \frac{|p-p^2+6|}{\sqrt{2}}\\ =\frac{5}{2}|p-p^2+6|\\ ~\\ \textrm{Now, we're only concerned with values of p between -2 and 3 and for all of those values}~p-p^2+6~\textrm{is positive, so the expression becomes}~\\ A=\frac{5}{2}(p-p^2+6)\\ \frac{dA}{dp}=\frac{5}{2}(1-2p)\\ \frac{d^2A}{dp^2}=-5\\ \textrm{Maximum value of A occurs when}~\frac{dA}{dP}=0~\textrm{since}~\frac{d^2A}{dp^2}=-5<0~\textrm{for all values of p}\\ ~\\ \frac{5}{2}(1-2p)=0\\ 2p=1\\ p=\frac{1}{2}\\ ~\\ \therefore A_{\textrm{max}}=\frac{5}{2}(\frac{1}{2}-\left(\frac{1}{2} \right )^2+6)=\frac{5}{2}\cdot \frac{25}{4}=\frac{125}{8}~\textrm{Which is }\frac{3}{4}~\textrm{the area of the parabolic segment as required.}" /></a>

Also, this should really be in the MX1 forum.

 

[umm what]

That's a very inefficient way of doing the first part. The two quarter-circles cancel out and you're left with finding the area of a trapezium. No integration necessary.

Whats the difference between part (i) and (ii) ????