anonymous
  • anonymous
When solving for the area bounded by curves, I often have trouble determing which function goes first. Can somebody help me with that?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • 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.
anonymous
  • anonymous
howabout x^2 + 2 AND x+8
anonymous
  • anonymous
Drawing a sketch helps.

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anonymous
  • 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.
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
How did you determine those intervals?
anonymous
  • anonymous
Set the two equations equal to eachother and solve for where they intersect. \[x^2 + 2 = x+8\] \[\implies x^2 -x - 6 = 0 \implies (x + 2)(x-3) = 0 \implies x \in {-2,3}\]
anonymous
  • 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.
anonymous
  • anonymous
Does that make sense?
anonymous
  • anonymous
absolutely. Thank you.
anonymous
  • anonymous
Can you assist me with a few problems? I have a thick calculus packet due tomorrow. Any assistance would be appreciated.
anonymous
  • anonymous
Certainly.
anonymous
  • 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)
anonymous
  • anonymous
Ok, so for the first part. What do you think would be a good way to find the equation of the tangent?
anonymous
  • anonymous
take the derivative of the function to find the slope. then plug it into slope intercept form?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Except I would use the point slope formula instead of slope intercept since you have a particular point, and the slope
anonymous
  • anonymous
\[y-y_1 = m(x-x_1)\]
anonymous
  • anonymous
Yes that one, sorry I was confused.
anonymous
  • anonymous
However, when I took the derivative, it turned out complicated. Can you run through that with me?
anonymous
  • 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 \[y = x\sqrt{x+7}\]
anonymous
  • anonymous
Can you take the derivative of that?
anonymous
  • anonymous
yes, by chain rule
anonymous
  • anonymous
What do you get for the derivative
anonymous
  • anonymous
Actually I'm thinking that we might want the derivative of \[-x\sqrt{x+7}\]
anonymous
  • anonymous
why did it become negative?
anonymous
  • anonymous
hey polpak the answer to my question was D right
anonymous
  • anonymous
Well there are two options. \[y^2 = x^2(x+7) \implies y = \pm\sqrt{x^2(x+7)} = \pm x\sqrt{x+7}\]
anonymous
  • anonymous
And I'm not sure how to tell which one we want. I think it's the negative one.
anonymous
  • anonymous
yes! I'm sure it's the negative one
anonymous
  • anonymous
Because y = 6, but x = -3. So that means that -x = 3
anonymous
  • 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.
anonymous
  • anonymous
\[f'(x) = \sqrt{x + 7} + {x}\ {2 sqrt {x + 7}}]
anonymous
  • 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?
anonymous
  • anonymous
Close. It should be \[\sqrt{x+7} + \frac{x}{2\sqrt{x+7}}\]
anonymous
  • anonymous
Oh, except negative.
anonymous
  • anonymous
-1 times that whole thing.
anonymous
  • anonymous
That exacty what I got.
anonymous
  • anonymous
We can simplify a bit also. \[-\sqrt{x+7} - \frac{x}{2\sqrt{x+7}} = \frac{-2x -14}{2\sqrt{x+7}} - \frac{x}{2\sqrt{x+7}} \] \[= \frac{-3x-14}{2\sqrt{x+7}}\]
anonymous
  • anonymous
So now plug in -3 for x.
anonymous
  • anonymous
how did you get -2x - 14 for your numerator?
anonymous
  • anonymous
Multiply the top and bottom of the first term by \(2\sqrt{x+7}\) \[\frac{-\sqrt{x+7}}{1} *\frac{2\sqrt{x+7}}{2\sqrt{x+7}}\]
anonymous
  • anonymous
For getting everything over the same denominator.
anonymous
  • anonymous
Gives \[\frac{-2(x+7)}{2\sqrt{x+7}}\]
anonymous
  • anonymous
allright now what?
anonymous
  • anonymous
Now plug in -3 for x and find your slope at that point.
anonymous
  • anonymous
Then plug that slope, and the point (-3,6) into the point slope formula to find the equation of the tangent line
anonymous
  • anonymous
I got y = -5/4x - 9/2
anonymous
  • anonymous
y-6 = -5/4(x+3) y = (-5/4)x -15/4 + 24/4 y = (-5/4)x - 9/4
anonymous
  • anonymous
Not -9/2
anonymous
  • anonymous
Did you use point slope?
anonymous
  • anonymous
Or did you use slope intercept?
anonymous
  • anonymous
slope intercept. my points were (-3, 6) and I plugges that into y2-y1= m (x2-x1)
anonymous
  • anonymous
Ok, 2 things. First, you don't use two points for slope intercept. Second, that's not slope intercept. =)
anonymous
  • anonymous
Just use point slope.
anonymous
  • 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.
anonymous
  • anonymous
I know where I went wrong, i reversed my x and y values
anonymous
  • anonymous
How do I find the equation of a normal line now that I found the equation of a tangent line?
anonymous
  • anonymous
Well the normal line is at that same point, but the slope is perpendicular.
anonymous
  • anonymous
Perpendicular to the tangent line that is
anonymous
  • anonymous
4/5 is my slope?
anonymous
  • anonymous
Yes, the slope of the normal line is 4/5
anonymous
  • anonymous
okayI know how to find the slope of that line. How do I find points at where the tangent line is vertical?
anonymous
  • anonymous
You don't need to. The normal at that point still touches that point (-3,6)
anonymous
  • anonymous
You just need one point and the slope to find the equation of the line.
anonymous
  • anonymous
for where the tangent line is vertical?
anonymous
  • 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.
anonymous
  • anonymous
0 in the denominator of the derivative that is.
anonymous
  • anonymous
Help? :/
anonymous
  • anonymous
What is the denominator of the derivative?
anonymous
  • anonymous
oh, set the denominator to zero?>
anonymous
  • anonymous
oh, set the denominator to zero?>
anonymous
  • anonymous
x is greater than or equal to -7
anonymous
  • 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.
anonymous
  • anonymous
Can I bother you with one more problem?
anonymous
  • anonymous
yes, but go ahead and make a new question for it. this one is getting long ;)

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