some more problems!!

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May 24th 2007, 01:22 PM
carlasader

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some more problems!!

i have not able to answer any of that at all , i feel that i will never pass the final year exam with all that stress on my head ...need please ,but only if u guys know any them .:D
May 24th 2007, 01:41 PM
ecMathGeek

1. Find $\frac {dy}{dx}$ at $x=2$ if $y=2e^x+x^3$

$y^\prime(x)=2e^x + 3x^2$

Plugging in $x=2$ we get:

$y^\prime(2)=2e^{(2)}+3(2)^2=2e^2+12$

Do you need this explained more?

I'm not sure what method you need to use to solve number 2 of the first page. Are you expected to use Newton's method or something else?
May 24th 2007, 01:43 PM
Jhevon

Quote:

Originally Posted by ecMathGeek

1. Find $\frac {dy}{dx}$ at $x=2$ if $y=2e^x+x^3$

$y^\prime(x)=2e^x + 3x^2$

Plugging in $x=2$ we get:

$y^\prime(2)=2e^{(2)}+3(2)^2=2e^2+12$

Do you need this explained more?

i see you're getting the hang of LaTex already. I still use ' instead of typing ^\prime though, it looks the same
May 24th 2007, 02:05 PM
ecMathGeek

3. a) Prove that $9cos(x)+6sin(x)=10$

Notice that $\angle QBA$ , $\angle ABC$ and $\angle CBR$ are suplementary (they sum up to equal $180^\circ$ ), and that $\angle ABC=90^\circ$ . Therefore, $\angle QBA$ and $\angle CBR$ are complementary. This means that $\angle CBR \cong \angle QAB$ and thus that $\triangle AQB \cong \triangle BRC$ .

With all of that said. Here's where I can start the proof:

Notice that $x = \angle CBR = \angle QAB$ , Therefore: $sin(x) = \frac {QB}{AB}$ and $cos(x)=\frac {BR}{BC}$ . However, $QB = 30 - BR$ . Therefore:

$sin(x) = \frac {30-BR}{18} \rightarrow 18sin(x) = 30-BR$

$cos(x) = \frac {BR}{27} \rightarrow 27cos(x) = BR$

If we add these two equations, we get:

$18sin(x) + 27cos(x) = 30 -BR+BR=30$

Now divide everything by 3 to get:

$6sin(x) + 9cos(x) = 10$
May 24th 2007, 02:09 PM
ecMathGeek

Quote:

Originally Posted by Jhevon

i see you're getting the hang of LaTex already. I still use ' instead of typing ^\prime though, it looks the same

I'm learning it... and still making plenty of mistakes. But it's a nice program.
May 24th 2007, 02:35 PM
ecMathGeek

3. b) Express $9cos(x)+6sin(x)$ in the form $Rcos(x-a)$ and find the values of $R$ and $a$ to 2 decimal places.

Notice that $Rcos(x-a)=Rcos(x)cos(a)+Rsin(x)sin(a)$ but we want $9cos(x)+6sin(x)=Rcos(x-a)$ . So we want $9cos(x)+6sin(x)=Rcos(x)cos(a)+Rsin(x)sin(a)$ . Therefore, we need $9=Rcos(a)$ and $6=Rsin(a)$ .

To solve this will take some creativity.

We have:
$9=Rcos(a) \rightarrow cos(a) = \frac {9}{R}$
$6=Rsin(a) \rightarrow sin(a) = \frac {6}{R}$

Let's square both sides of each equation:
$cos^2(a)= \frac {81}{R^2}$
$sin^2(a)= \frac {36}{R^2}$

Now, we'll use the identity $cos^2(a) = 1-sin^2(a)$ to get:
$1-sin^2(a) = \frac {81}{R^2} \rightarrow sin^2(a) = 1-\frac{81}{R^2}$

Setting the two $sin^2(a)$ equal, we get:

$\frac {36}{R^2} = 1-\frac{81}{R^2}$

$36 = R^2-81 \rightarrow R^2 = 117 \rightarrow R \approx 10.82$

Using this, we can find $a$ :

$cos(a) = \frac {9}{10.817}$

$a = cos^{-1}(\frac{9}{10.817}) \rightarrow a \approx 33.69$
May 24th 2007, 02:42 PM
ecMathGeek

3. c) Find the value of $x$ .

$10.82cos(x-33.69)=10$
$cos(x-33.69)=\frac {10}{10.82}$
$x-33.69= cos^{-1}(\frac {10}{10.82})$
$x = 33.69+cos^{-1}(\frac {10}{10.82})$
$x= 56.1$