# some more problems!!

• May 24th 2007, 01:22 PM
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 $\displaystyle \frac {dy}{dx}$ at $\displaystyle x=2$ if $\displaystyle y=2e^x+x^3$

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

Plugging in $\displaystyle x=2$ we get:

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

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

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

Plugging in $\displaystyle x=2$ we get:

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

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 $\displaystyle 9cos(x)+6sin(x)=10$

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

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

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

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

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

If we add these two equations, we get:

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

Now divide everything by 3 to get:

$\displaystyle 6sin(x) + 9cos(x) = 10$
• 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 $\displaystyle 9cos(x)+6sin(x)$ in the form $\displaystyle Rcos(x-a)$ and find the values of $\displaystyle R$ and $\displaystyle a$ to 2 decimal places.

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

To solve this will take some creativity.

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

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

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

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

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

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

Using this, we can find $\displaystyle a$:

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

$\displaystyle a = cos^{-1}(\frac{9}{10.817}) \rightarrow a \approx 33.69$
• May 24th 2007, 02:42 PM
ecMathGeek
3. c) Find the value of $\displaystyle x$.

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