# Where can ı fiind thomas calculus solution manual

1. Nov 5, 2006

### oahsen

i am searching that book thomas calculus solution manual. is there anybody who can tell me from where can i doenload. thanks

2. Nov 5, 2006

### dimachka

you really should do your homework....

3. Nov 6, 2006

### oahsen

it is not for hw

i would like to check my answers and look some solutions that i couldnt solved

4. Nov 6, 2006

5. Nov 6, 2006

### oahsen

thanks

very very thanks for your useful tips.

6. Nov 6, 2006

### arildno

BUY the damn manual instead of acting like a thief trying to filch it from the net somewhere. :grumpy:

Shame on you!

7. Nov 6, 2006

### oahsen

thanks for your useful tips. i am not a short of person that will make his hw's with such a illegal way.ok it could be stay.

8. Nov 6, 2006

### oahsen

i couldnt find it in my country and i have not a credit card to buy it fron net

9. Nov 6, 2006

### oahsen

ok i changed my opininon. could you help m with only one problem. i havent solve it. that is in the chapter 3.additional problem 28 ;
i would try to write a summary of the problem : assume an ice cube retains its cubical shape as it melts. if we call edge lenth s its vlume is v=s^3 and the surface area is = 6*s^2. we also assume that the cube's volume decreases at a rate that is proportional to its surface area. in math terms : dv/dt=-k(6*s^2) ; assume that the cube lost 1/4 of its volume during the first hour and that the volume is Vo at t=0. how long will it take the ice cube to melt...

i tried to solve that with integrating(as t goes to 0 V goes to 3V/4) and tried to find k but then k is going to be a strange value.also the chapter is about derivative applications i can not solve it with integral. then what should i do please help me with this problem...

10. Nov 6, 2006

### arildno

Okay, first of all:
Here, it is smart to express the surface S in terms of the volume V:
$$S=6V^{\frac{2}{3}}$$
Thus, the differential equation for the rate of change of the volume is:
$$\frac{dV}{dt}=-6kV^{\frac{2}{3}}$$
This is a separable equation:
$$\frac{dV}{V^{\frac{2}{3}}}=-6kdt$$
or, integrating both sides from t=0 and and t=T:
$$3(V(T)^{\frac{1}{3}}-V(0)^{\frac{1}{3}})=-6kT$$
or simply, for arbitrary T:
$$V(T)=(V(0)^{\frac{1}{3}}-2kT)^{3}$$
Now you should be able to do the last steps on your own!

11. Nov 6, 2006

### oahsen

yes i know first put V(t)=3v/4 t=1 find k than put v(t)=0 put k and find t. this what i must do isn't it?. but this is a problem from derivative chapter. but we found the answer with integration is it true?

12. Nov 6, 2006

### oahsen

one more question: i have found the answer t=1/(1-(3/4)^1/3))) is it true?

13. Nov 6, 2006

### mathwonk

i will solve your problems for \$50 apiece.

14. Nov 6, 2006

### chroot

Staff Emeritus
Come on, guys.

- Warren