# Homework Help: 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

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