# Math Help - some more problems!!

1. ## 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 .

2. 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?

3. 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

4. 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:

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

5. 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.

6. 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 $a$:

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

$a = cos^{-1}(\frac{9}{10.817}) \rightarrow a \approx 33.69$

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

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