Finals exam!

1. Jan 12, 2004

hodeez

Ok, so I have my Calculus Finals this coming Thursday, and I hope I receive some "tutoring" or review. The teacher sent out a review sheet via email, and I mirrored it www.hodeez.com/max3_finals.doc[/URL]

Just need a little/some help on 1,4,6,7,10 and the long problem.

Of course I know the rules and I'll show you my work first, but as of right now, I'm at work and my cable @ home keep disconnecting me. So I'll breif what I can get by today hopefully :)

Last edited by a moderator: Apr 20, 2017
2. Jan 12, 2004

himanshu121

For Prob 4

f(x)= |2x+3|-|x-4|

draw the no line
Code (Text):

------------------------------------------------------------
|           |           |
x                  -3/2          0           4
|                     |
f(x)= -(2x+3)-{-(x-4)} | 2x+3 -{-(x-4)}      | 2x+3 -(x-4)
|                     |

Code (Text):

for   x<-3/2 u have   |2x+3|=-(2x+3)  & |x-4|= -(x-4)
-3/2<=x<4             |2x+3|= (2x+3)  & |x-4|= -(x-4)
x>=4            |2x+3|= (2x+3)  & |x-4|=  (x-4)

now i believe u can recombine the above in specified interval to get f(x)

3. Jan 12, 2004

himanshu121

For problem 1 $$\frac{df(x)}{dx}= 0 \text{ as the tangent is horizontal say at point x=t}$$

u will get Quadratic Equation find the roots to 3 decimal place

Last edited: Jan 12, 2004
4. Jan 12, 2004

himanshu121

For the problem 6 find -dx/dy which will be the slope of the normal

For problem 7

Apply the formula $$\frac{d (\frac{u}{v})}{dx}=\frac{ v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$
here u and v are u=cosx and v= sqrt(2x-3)

5. Jan 12, 2004

hodeez

from what I am understanding, (6) you want me to find the reciprocal negative of the derivative of the original equation? OR find the implicit differentiation and then plugging in the (2,1)

(7) = seems like the division rule, how did i ever miss that? [zz)]

thanks for the replies, im going to look more indepth into the answers you gave me.

6. Jan 12, 2004

himanshu121

Giving Hints for long problem

lim(x->3)f(x)=f(3)
i.e f(3-h)=f(3+h) for h->0 and h>0

and $$f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$$
u can substitute x= 4 here and so on

7. Jan 12, 2004

hodeez

DOH! another simple one over looked

8. Jan 12, 2004

himanshu121

I must say all are simple

u can do it either way

9. Jan 12, 2004

hodeez

im quite consfused on the way you are approaching it. I am thinking, if its continuous the 1st piece must match the 2nd piece @ 3 and then i set it up for k/3+2=k-3-1 and then solved for k which = -9 , then the slope must equal so im finding the deriv of both pieces and then trying to make them equal again.

10. Jan 12, 2004

hodeez

May you check this (#9):

y'' = -3(sin^2(x))*(cos^2(x))
and then plug in pi/6 = 180/6 = 30, yes?

11. Jan 12, 2004

himanshu121

For long problem 2

Take the first hiker along x-axis and other at an angle of 35 deg
After time t its(1 hiker)coordinate will be(2.25t,0) and for second hiker it would be (3.8cos35 t,3.8sin35 t}.

Now apply the distance formula and and differentiate it to get ur answer substitute for t=5.4/2.25

Also it could have been done easily be relative velocity concept but i feel that u have to give Calculus exam right

12. Jan 12, 2004

himanshu121

may i know how u reach at y'' = -3(sin^2(x))*(cos^2(x))

it is incorrect Pls show me ur steps
though pi/6=30 is correct

13. Jan 12, 2004

hodeez

y'= 3(sin(x))(cos(x))
y''= 3 * product rule (sin(x))(cos(x))
= 3 * (sin(x))*(-sin(x))+(cos(x))(cos(x))
= -3(sin^2(x))*(cos^2(x))

14. Jan 12, 2004

himanshu121

look
$$y=sin^3x \Rightarrow y'=3sin^2x (cosx)$$

u got y' wrong in ur reply

15. Jan 12, 2004

hodeez

is the answer 56.52 rounded?

I used the Cosine Formula for triangles.

Also, im interested in hearing your relative velocity concept. =)

16. Jan 12, 2004

hodeez

wow, that would have caused havok on the final

17. Jan 12, 2004

himanshu121

Again how u reached the answer for hiker prob it is incorrect

18. Jan 12, 2004

hodeez

The method, I'm sure is correct, because that is the method we were taught to use. But my mechanics are probably incorrect. I'm using the Google calculator as I don't have my graphing calculator right now.

19. Jan 12, 2004

himanshu121

Yup i know the method using the cosine formula is correct u need to recheck ur steps u might be making small errors in calulating

$$cos35 = \frac{(2.25t)^2+(3.8t)^2-x^2}{2*2.25*3.8 t^2}$$

After little rearrangement

$$x^2=5.495000043t^2$$
so u will get x=2.344141643t

20. Jan 12, 2004

hodeez

My formula is C= A^2+B^2-cos(c)*A*B
Is that the correct form?

But 2.344 seems too way out of proportional