Hi all ..

problem in Calculus ( derivaives ) ( slope )

Hi all

plese I want check my answer , I don't know how i solve this queation but i try

http://store2.up-00.com/Nov10/2oP96162.jpg

thanks

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- Nov 3rd 2010, 06:54 AMr-soyproblem in Calculus ( derivaives ) ( slope )
Hi all ..

problem in Calculus ( derivaives ) ( slope )

Hi all

plese I want check my answer , I don't know how i solve this queation but i try

http://store2.up-00.com/Nov10/2oP96162.jpg

thanks - Nov 3rd 2010, 07:31 AMUnknown008
i.

The derivative of y = sin(x) is

y' = cos(x)

Then, using x = pi/4, what you you get?

iv.

Using the product rule,

$\displaystyle \dfrac{dR}{dM} = M^2\left(-\dfrac13\right) + \left(\dfrac{C}{2} - \dfrac{M}{3}\right)\cdot 2M$

And I don't get what you got.

v. I think that the 'jerk' means in some way the derivative... then yes, it's good. - Nov 3rd 2010, 10:52 PMr-soy
Q2 )

I use the product rule

I got -1/3m^2 + 2m.c/2.2m.m/3 - Nov 4th 2010, 12:23 AMEducated
How did you get that?

Unknown008 has already given you the correct answer. Maybe you should show us your working so we can identify where you went wrong. (Also remember to use brackets)

To make question (iv) easier, expand the brackets!

$\displaystyle M^2 (\frac{C}{2} - \frac{M}{3}) = \dfrac{C \cdot M^2}{2} - \dfrac{M^3}{3}$

Now that the brackets are expanded, there is no need to use the product rule and you should be able to differentiate it with no problem. - Nov 4th 2010, 12:53 AMpirateboy
haha a jerk is a derivative? i've not heard that one before.

edit: apparently it is so...well wikipedia says it is.

http://en.wikipedia.org/wiki/Jerk_%28physics%29 - Nov 4th 2010, 04:05 AMHallsofIvy
"Jerk" is a physics term. It is the

**third**derivative of the position function: first derivative is velocity, second derivative is acceleration, third derivative is jerk. But, r-soy, you have the wrong sign. If s= 5cos(t), then s'= -5sin(t), s''= -5cos(t), and s'''= 5sin(t).

Using the product rule on $\displaystyle R= M^2(\frac{C}{2}- \frac{M}{3})$, $\displaystyle R'= 2M(\frac{C}{2}- \frac{M}{3})+ M^2(-\frac{1}{3}= MC- \frac{2M^2}{3}-\frac{M^2}{3}= MC- M^2$. r-soy, you dropped the "3" in the denominator. - Nov 4th 2010, 09:01 AMUnknown008