# Having Lots Of Calculus III Problems

• Nov 27th 2006, 07:09 AM
amanda1410
Having Lots Of Calculus III Problems
My brain has gone dead, got a few days to do the following questions:

http://img.photobucket.com/albums/v1...cer/maths4.jpg
http://img.photobucket.com/albums/v1...cer/maths3.jpg
http://img.photobucket.com/albums/v1...cer/maths2.jpg
http://img.photobucket.com/albums/v1...cer/maths1.jpg

would appreciate any help with any of the above question and methods

thank you
• Nov 27th 2006, 08:03 AM
CaptainBlack
Quote:

Originally Posted by amanda1410
My brain has gone dead, got a few days to do the following questions:

would appreciate any help with any of the above question and methods

thank you

Q4a. Let:

$\displaystyle f(x)=x^3-2x^2+x-3$

then $\displaystyle f(2)=-1$ and $\displaystyle f(3)=9$. Also as $\displaystyle f(x)$ is continuous by the intermediate value theorem it takes all values between $\displaystyle f(2)$ and $\displaystyle f(3)$ between $\displaystyle x=2$ and $\displaystyle x=3$, but as $\displaystyle f(x)$ changes sign in this interval one of these values is $\displaystyle 0$. Hence we conclude that $\displaystyle f(x)$ has a root in $\displaystyle (2,3)$.

RonL
• Nov 27th 2006, 08:07 AM
amanda1410
cheers for that, appreciate all the help possbile with the rest, at uni pullin my hair out trying to do this and my electronics at the same time
• Nov 27th 2006, 09:02 AM
ThePerfectHacker
First I would like to mention that $\displaystyle f''_{xy}$ and $\displaystyle f''_{yx}$ are the same! Because the function is continous on some open disk.
(But since you are in mechanical engineering you probably do not care about that).

$\displaystyle f(x,y)=x\ln y+e^{xy}+\cos (2x-3y)$
Thus,
$\displaystyle f'_x(x,y)=\ln y+ye^{xy}-2\sin (2x-3y)$
$\displaystyle f'_y(x,y)=\frac{x}{y}+xe^{xy}+3\sin (2x-3y)$

Now take the derivative along y on the second one,
$\displaystyle f''_{yy}(x,y)=-\frac{x}{y^2}+x^2e^{xy}-9\cos (2x-3y)$

And finally,
$\displaystyle f''_{xy}=\frac{1}{y}+e^{xy}+xye^{xy}+6\cos (2x-3y)$
• Nov 27th 2006, 01:03 PM
topsquark
Question 1 a) I presume.

i) $\displaystyle y = (2x - 1)3^{5x+1} - tan(2x)$

First note that $\displaystyle \frac{d}{dx}a^x = ln(a) \cdot a^x$. The rest is an exercise in the chain rule:

So
$\displaystyle y' = 2 \cdot 3^{5x+1} + (2x - 1) \cdot ln(3) \cdot 3^{5x+1} \cdot 5 - sec^2(2x) \cdot 2$

$\displaystyle y' = [ 2 + 5(2x - 1)ln(3) ] 3^{5x+1} - 2 sec^2(2x)$

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ii) $\displaystyle y = ln \left ( 2x + e^{2x} \right ) + sin( \pi x)$

$\displaystyle y' = \frac{1}{2x + e^{2x}} \cdot (2 + e^{2x} \cdot 2 ) + cos( \pi x) \cdot \pi$

$\displaystyle y' = \frac{2(1 + e^{2x})}{2x + e^{2x}} + \pi cos( \pi x)$

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iii) $\displaystyle y = x + cos^{-1}(y)$

First, this needs to be done by implicit differentiation.
Second: $\displaystyle \frac{d}{dx}cos^{-1}(x) = - \frac{1}{\sqrt{1-x^2}}$.

So:
$\displaystyle y' = 1 + -\frac{1}{\sqrt{1-y^2}} \cdot y'$

$\displaystyle y' + \frac{1}{\sqrt{1-y^2}} \cdot y' = 1$

$\displaystyle y' \left ( 1 + \frac{1}{\sqrt{1-y^2}} \right ) = 1$

$\displaystyle y' = \frac{1}{1 + \frac{1}{\sqrt{1-y^2}}}$

This needs to be neated up a bit:
$\displaystyle y' = \frac{1}{1 + \frac{1}{\sqrt{1-y^2}}} \cdot \frac{\sqrt{1-y^2}}{\sqrt{1-y^2}}$

$\displaystyle y' = \frac{\sqrt{1-y^2}}{\sqrt{1-y^2} + 1}$

$\displaystyle y' = \frac{\sqrt{1-y^2}}{\sqrt{1-y^2} + 1} \cdot \frac{\sqrt{1-y^2} - 1}{\sqrt{1-y^2} - 1}$

$\displaystyle y' = \frac{(\sqrt{1-y^2})(\sqrt{1-y^2} - 1)}{1 - y^2 - 1}$

$\displaystyle y' = -\frac{1 - y^2 - \sqrt{1-y^2}}{y^2}$

I'll leave it like this. :)

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iv) Ah, a practically straightforward problem!
$\displaystyle y = \frac{\sqrt{2x+1}}{x}$

Using the quotient rule:
$\displaystyle y' = \frac{\left ( \frac{1}{2} \frac{1}{\sqrt{2x+1}} \cdot 2 \right ) x - \sqrt{2x+1} \cdot 1}{x^2}$

$\displaystyle y' = \frac{\left ( \frac{1}{\sqrt{2x+1}} \right ) x - \sqrt{2x+1}}{x^2}$

$\displaystyle y' = \frac{\left ( \frac{1}{\sqrt{2x+1}} \right ) x - \sqrt{2x+1}}{x^2} \cdot \frac{\sqrt{2x+1}}{\sqrt{2x+1}}$

$\displaystyle y' = \frac{x - (2x+1)}{x^2 \sqrt{2x+1}}$

$\displaystyle y' = \frac{-x - 1}{x^2 \sqrt{2x+1}} \cdot \frac{\sqrt{2x+1}}{\sqrt{2x+1}}$

$\displaystyle y' = -\frac{(x+1)\sqrt{2x+1}}{x^2 (2x+1)}$

-Dan