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

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

Mathematics
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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.
howabout x^2 + 2 AND x+8
Drawing a sketch helps.

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

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