## anonymous one year ago I seriously need help with question with polynomials. Please help me? Medal and fan!

1. anonymous

A bucket of paint has spilled on a tile floor. The paint flow can be expressed with the function r(t) = 3t, where t represents time in minutes and r represents how far the paint is spreading. The flowing paint is creating a circular pattern on the tile. The area of the pattern can be expressed as A(r) = πr2. Part A: Find the area of the circle of spilled paint as a function of time, or A[r(t)]. Show your work. (6 points) Part B: How large is the area of spilled paint after 10 minutes? You may use 3.14 to approximate π in this problem. (4 points)

2. anonymous

Hello?

3. anonymous

Um, Haseeb? Are you there?

4. anonymous

@Hero If you have the time can you help me out with this?

5. beginnersmind

Can you calculate A for a given value of t? Say t = 2, what will be the painted area?

6. anonymous

Yes.

7. beginnersmind

Ok, so you can solve part B.

8. anonymous

Yes, but Part A is what I'm having trouble with.

9. beginnersmind

Cool. Take the same steps as for part B, but instead of using 10 just keep t. You won't get a number but an expression involving t.

10. anonymous

So it would A[r(t)]=πr2+3t?

11. beginnersmind

Just to check, what result did you get for part B?

12. anonymous

I haven't done Part B yet.

13. beginnersmind

You should. It will help with part A. It also makes it easier to understand someone else's explaination, if you'll still need help.

14. anonymous

Okay, I'll try.

15. anonymous

Wait, but what is the radius of the spilled paint?

16. beginnersmind

r(t) = 3t, where t represents time in minutes and r represents how far the paint is spreading.

17. beginnersmind

so r is meant to be the radius

18. anonymous

Oh, so the paint is spreading equivalent to the radius of the circle?

19. beginnersmind

The radius of the painted circle is increasing as time goes on. So it's 3 after 1 minute, 6 after 2 minutes, etc.

20. anonymous

So it's 282.6?

21. beginnersmind

what did you get for the radius after 10 minutes?

22. anonymous

30.

23. anonymous

Oh.

24. anonymous

295.788?

25. beginnersmind

Let me check how I can write pretty expressions real quick. :)

26. beginnersmind

$A =\pi r ^{2}$

27. beginnersmind

Is that what you used?

28. anonymous

Yes

29. anonymous

With 30 as r

30. anonymous

Because 30 is the radius after 10 minutes.

31. beginnersmind

When I plug in r = 30 I get$\pi (30)^{2} = 900\pi= 900x3.14 = 2826$

32. beginnersmind

Hm, seems like your first answer was almost right, just off by a factor of 10 for some reason.

33. anonymous

Oh.

34. anonymous

So part B is 2826

35. beginnersmind

Yes. Now for part A.

36. beginnersmind

The idea is the same, but now instead of pluging in 30 into A = πr2 you plug in 3t

37. anonymous

A[r(t)] = pi*3t^2?

38. beginnersmind

More like pi*(3t)^2

39. beginnersmind

Probably should be simplified to 9*pi*t^2

40. anonymous

Oh.

41. anonymous

And that's part A?

42. beginnersmind

Yes. It gives you an expression to calculate the area directly from time. Without having to calculate the radius first.

43. anonymous

Thank you. :)

44. beginnersmind

No problem :)

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