Author Topic: Flaw in my method (Read 553 times) Tweet Share

kenhung123 · « **on:** November 07, 2010, 10:02:17 am »

For this question this is what I did:

A=1/bh
as b=h, let b=h=x

Given hypotenuse= $4\sqrt{2}$
$x^2+x^2=(4\sqrt{2})^2$
Therefore x=4 (Take the positive root)

$\frac{\mathrm{d} A}{\mathrm{d} t}=\frac{\mathrm{d} h}{\mathrm{d} t}\frac{\mathrm{d} A}{\mathrm{d} h}$
$\frac{\mathrm{d} A}{\mathrm{d} t}=\sqrt{2}\cdot \frac{1}{2}x\cdot 2$ Since $\frac{\mathrm{d} A}{\mathrm{d} h}=\frac{\mathrm{d} (\frac{1}{2}\cdot x\cdot x)}{\mathrm{d} x}=x$

So $\frac{\mathrm{d} A}{\mathrm{d} t}=4\sqrt{2}$ (by subbing x=4) giving D

Answer is C

xZero · « **Reply #1 on:** November 07, 2010, 10:12:23 am »

I'm pretty sure u havn to diff h in respect to x and find dh/dx. Then use this to find dx/dt and then u can find da/dt by dx/dt x da/dx

Edit: ur method is wrong since u did dh/dt x da/dx

Stroodle · « **Reply #2 on:** November 07, 2010, 10:25:13 am »

The area is given by $\frac{1}{2}hcos(45)\times hsin(45)$

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Author Topic: Flaw in my method (Read 553 times) Tweet Share

kenhung123

Flaw in my method

xZero

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