## anonymous 5 years ago When solving for the area bounded by curves, I often have trouble determing which function goes first. Can somebody help me with that?

1. anonymous

It depends on which variable you're integrating, but generally you are solving the integral of the upper function minus the integral of the lower function.

2. anonymous

howabout x^2 + 2 AND x+8

3. anonymous

Drawing a sketch helps.

4. anonymous

I agree, draw a quick sketch and see which one is above. The only other thing you can do is solve for their intersections, then plug in values before and after they intersect to see which one is below and which is above.

5. anonymous

For example, we can see that they intersect at x = -2, and x = 3. So when you look in this interval at a few values (x=0 for example) you can see that x^2 +2 = 2, but x + 8 = 8 So the linear function is above over that interval.

6. anonymous

This is a nice way to figure it out also because you'll need the intersecting points in order to know your bounds of integration anyway.

7. anonymous

How did you determine those intervals?

8. anonymous

Set the two equations equal to eachother and solve for where they intersect. $x^2 + 2 = x+8$ $\implies x^2 -x - 6 = 0 \implies (x + 2)(x-3) = 0 \implies x \in {-2,3}$

9. anonymous

err $$x \in \{-2,3\}$$ rather. So the area you are looking for is in the interval $$x \in [-2,3]$$. So I just picked an easy point in that interval to plug in for x and see which curve was higher than the other.

10. anonymous

Does that make sense?

11. anonymous

absolutely. Thank you.

12. anonymous

Can you assist me with a few problems? I have a thick calculus packet due tomorrow. Any assistance would be appreciated.

13. anonymous

Certainly.

14. anonymous

Given $y^{2} = x^{2}(x+7)$ Find an equation for the tangent line at ( -3, 6) Find an equation for the normal line at ( -3, 6)

15. anonymous

Ok, so for the first part. What do you think would be a good way to find the equation of the tangent?

16. anonymous

take the derivative of the function to find the slope. then plug it into slope intercept form?

17. anonymous

Yes.

18. anonymous

Except I would use the point slope formula instead of slope intercept since you have a particular point, and the slope

19. anonymous

$y-y_1 = m(x-x_1)$

20. anonymous

Yes that one, sorry I was confused.

21. anonymous

However, when I took the derivative, it turned out complicated. Can you run through that with me?

22. anonymous

Sure. Since it's y^2 we'll have a $$\pm$$, but the point they gave us is postive in y, so we'll rewrite the function as $y = x\sqrt{x+7}$

23. anonymous

Can you take the derivative of that?

24. anonymous

yes, by chain rule

25. anonymous

What do you get for the derivative

26. anonymous

Actually I'm thinking that we might want the derivative of $-x\sqrt{x+7}$

27. anonymous

why did it become negative?

28. anonymous

hey polpak the answer to my question was D right

29. anonymous

Well there are two options. $y^2 = x^2(x+7) \implies y = \pm\sqrt{x^2(x+7)} = \pm x\sqrt{x+7}$

30. anonymous

And I'm not sure how to tell which one we want. I think it's the negative one.

31. anonymous

yes! I'm sure it's the negative one

32. anonymous

Because y = 6, but x = -3. So that means that -x = 3

33. anonymous

And that's the only way you'll get a positive y value if x is negative is to have the negative in front.

34. anonymous

$f'(x) = \sqrt{x + 7} + {x}\ {2 sqrt {x + 7}}] 35. anonymous Its difficult for me to write out what Igot for the derivative, but if you please will show me what you got for the derivative then Ill tell you if I got the same thing? 36. anonymous Close. It should be \[\sqrt{x+7} + \frac{x}{2\sqrt{x+7}}$

37. anonymous

Oh, except negative.

38. anonymous

-1 times that whole thing.

39. anonymous

That exacty what I got.

40. anonymous

We can simplify a bit also. $-\sqrt{x+7} - \frac{x}{2\sqrt{x+7}} = \frac{-2x -14}{2\sqrt{x+7}} - \frac{x}{2\sqrt{x+7}}$ $= \frac{-3x-14}{2\sqrt{x+7}}$

41. anonymous

So now plug in -3 for x.

42. anonymous

how did you get -2x - 14 for your numerator?

43. anonymous

Multiply the top and bottom of the first term by $$2\sqrt{x+7}$$ $\frac{-\sqrt{x+7}}{1} *\frac{2\sqrt{x+7}}{2\sqrt{x+7}}$

44. anonymous

For getting everything over the same denominator.

45. anonymous

Gives $\frac{-2(x+7)}{2\sqrt{x+7}}$

46. anonymous

allright now what?

47. anonymous

Now plug in -3 for x and find your slope at that point.

48. anonymous

Then plug that slope, and the point (-3,6) into the point slope formula to find the equation of the tangent line

49. anonymous

I got y = -5/4x - 9/2

50. anonymous

y-6 = -5/4(x+3) y = (-5/4)x -15/4 + 24/4 y = (-5/4)x - 9/4

51. anonymous

Not -9/2

52. anonymous

Did you use point slope?

53. anonymous

Or did you use slope intercept?

54. anonymous

slope intercept. my points were (-3, 6) and I plugges that into y2-y1= m (x2-x1)

55. anonymous

Ok, 2 things. First, you don't use two points for slope intercept. Second, that's not slope intercept. =)

56. anonymous

Just use point slope.

57. anonymous

$y - y_1 = m(x-x_1)$ Plug in the slope for m, and $$x_1,y_1$$ is the x, y value of one point on the line.

58. anonymous

I know where I went wrong, i reversed my x and y values

59. anonymous

How do I find the equation of a normal line now that I found the equation of a tangent line?

60. anonymous

Well the normal line is at that same point, but the slope is perpendicular.

61. anonymous

Perpendicular to the tangent line that is

62. anonymous

4/5 is my slope?

63. anonymous

Yes, the slope of the normal line is 4/5

64. anonymous

okayI know how to find the slope of that line. How do I find points at where the tangent line is vertical?

65. anonymous

You don't need to. The normal at that point still touches that point (-3,6)

66. anonymous

You just need one point and the slope to find the equation of the line.

67. anonymous

for where the tangent line is vertical?

68. anonymous

Oh, sorry I misunderstood. The tangent line will be vertical when the slope is something of the form $\frac{a}{0}$ So you are looking for the point where you will have a 0 in the denominator.

69. anonymous

0 in the denominator of the derivative that is.

70. anonymous

Help? :/

71. anonymous

What is the denominator of the derivative?

72. anonymous

oh, set the denominator to zero?>

73. anonymous

oh, set the denominator to zero?>

74. anonymous

x is greater than or equal to -7

75. anonymous

x cannot be less than -7 or the radical becomes imaginary, but if x is greater than -7, the denominator is non-zero (positive). Only when x = -7 do you have a 0 in the denominator. Therefore at x= -7, you have a vertical tangent line.

76. anonymous

Can I bother you with one more problem?

77. anonymous

yes, but go ahead and make a new question for it. this one is getting long ;)