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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?
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?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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
 5 years ago
Best ResponseYou've already chosen the best response.0howabout x^2 + 2 AND x+8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Drawing a sketch helps.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0For 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
 5 years ago
Best ResponseYou've already chosen the best response.0This 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
 5 years ago
Best ResponseYou've already chosen the best response.0How did you determine those intervals?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Set the two equations equal to eachother and solve for where they intersect. \[x^2 + 2 = x+8\] \[\implies x^2 x  6 = 0 \implies (x + 2)(x3) = 0 \implies x \in {2,3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0err \(x \in \{2,3\}\) 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
 5 years ago
Best ResponseYou've already chosen the best response.0Does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0absolutely. Thank you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you assist me with a few problems? I have a thick calculus packet due tomorrow. Any assistance would be appreciated.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Given \[y^{2} = x^{2}(x+7)\] Find an equation for the tangent line at ( 3, 6) Find an equation for the normal line at ( 3, 6)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so for the first part. What do you think would be a good way to find the equation of the tangent?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take the derivative of the function to find the slope. then plug it into slope intercept form?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Except I would use the point slope formula instead of slope intercept since you have a particular point, and the slope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes that one, sorry I was confused.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0However, when I took the derivative, it turned out complicated. Can you run through that with me?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure. 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
 5 years ago
Best ResponseYou've already chosen the best response.0Can you take the derivative of that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you get for the derivative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually I'm thinking that we might want the derivative of \[x\sqrt{x+7}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why did it become negative?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey polpak the answer to my question was D right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well there are two options. \[y^2 = x^2(x+7) \implies y = \pm\sqrt{x^2(x+7)} = \pm x\sqrt{x+7}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And I'm not sure how to tell which one we want. I think it's the negative one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes! I'm sure it's the negative one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because y = 6, but x = 3. So that means that x = 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And that's the only way you'll get a positive y value if x is negative is to have the negative in front.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f'(x) = \sqrt{x + 7} + {x}\ {2 sqrt {x + 7}}]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Its 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
 5 years ago
Best ResponseYou've already chosen the best response.0Close. It should be \[\sqrt{x+7} + \frac{x}{2\sqrt{x+7}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 times that whole thing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That exacty what I got.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We can simplify a bit also. \[\sqrt{x+7}  \frac{x}{2\sqrt{x+7}} = \frac{2x 14}{2\sqrt{x+7}}  \frac{x}{2\sqrt{x+7}} \] \[= \frac{3x14}{2\sqrt{x+7}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So now plug in 3 for x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get 2x  14 for your numerator?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Multiply 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
 5 years ago
Best ResponseYou've already chosen the best response.0For getting everything over the same denominator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Gives \[\frac{2(x+7)}{2\sqrt{x+7}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now plug in 3 for x and find your slope at that point.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then plug that slope, and the point (3,6) into the point slope formula to find the equation of the tangent line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got y = 5/4x  9/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y6 = 5/4(x+3) y = (5/4)x 15/4 + 24/4 y = (5/4)x  9/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did you use point slope?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or did you use slope intercept?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0slope intercept. my points were (3, 6) and I plugges that into y2y1= m (x2x1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, 2 things. First, you don't use two points for slope intercept. Second, that's not slope intercept. =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just use point slope.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y  y_1 = m(xx_1)\] Plug in the slope for m, and \(x_1,y_1\) is the x, y value of one point on the line.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know where I went wrong, i reversed my x and y values

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How do I find the equation of a normal line now that I found the equation of a tangent line?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well the normal line is at that same point, but the slope is perpendicular.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Perpendicular to the tangent line that is

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, the slope of the normal line is 4/5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okayI know how to find the slope of that line. How do I find points at where the tangent line is vertical?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You don't need to. The normal at that point still touches that point (3,6)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You just need one point and the slope to find the equation of the line.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for where the tangent line is vertical?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, 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
 5 years ago
Best ResponseYou've already chosen the best response.00 in the denominator of the derivative that is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is the denominator of the derivative?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, set the denominator to zero?>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, set the denominator to zero?>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x is greater than or equal to 7

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x cannot be less than 7 or the radical becomes imaginary, but if x is greater than 7, the denominator is nonzero (positive). Only when x = 7 do you have a 0 in the denominator. Therefore at x= 7, you have a vertical tangent line.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can I bother you with one more problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but go ahead and make a new question for it. this one is getting long ;)
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