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16. Given a function y = 3x3 + 2x2 + x ~ 1,find the equation of the tangent line passing (1, 2)17. Given fence materials in 2600 yards to fence a rectangular area w...

Question

16. Given a function y = 3x3 + 2x2 + x ~ 1,find the equation of the tangent line passing (1, 2)17. Given fence materials in 2600 yards to fence a rectangular area with one side bordering a straight river: What is the maximal area of the rectangular field it can border?

16. Given a function y = 3x3 + 2x2 + x ~ 1,find the equation of the tangent line passing (1, 2) 17. Given fence materials in 2600 yards to fence a rectangular area with one side bordering a straight river: What is the maximal area of the rectangular field it can border?



Answers

Find the maximum area of a triangle formed by the axes and a
tangent line to the graph of $y=(x+1)^{-2}$ with $x > 0$

So if we look at the graph of y equals X plus one to the negative too. Yeah, What we see is this graph right here and then if we take the derivative of it, so we'll make this a function so it could take the derivative, of course. Prime of X. What does it look like? Well, we see it looks something like this. So with this in mind, we want to determine the maximum area formed by the axes and a tangent line to the graph. So what we see is that we end up getting these are all the different possible slopes of the tangent line. But ultimately what we see is possible is we can get half a unit squared if we have one half base times height. So if we have say here, um, one unit one unit and then a slope right here of in this case we have a negative slope, but we could also have a positive slope. The highest slope we could have would be all the way up here Infinity. However, the slip that were mainly wanting to focus on would be, say, a maximum or minimum eso in that case in order to maximize our slope, we'd want right here a slope of what? This is right here, which would be approximately a negative one. So if we had a tangent line right here, we could have won 21 and, uh, that would end up giving us a maximum area of one half units squared.

This question asks us to find the tension line given a plane and a point on the plane. So as a refresher, the equation for a tangent line is T. Is equal to f sub X at a comma B times x minus a plus F sub Y at a comma b times y minus B plus Z at a comma B. So to start, I am going to factor out rz equation. Ze equation is x plus two squared minus two Times. Why -1? All squared minus five. So to start, I'm going to factor this out. So from here I have, Z is equal two, X squared plus four, X plus four minus two times Y squared minus two, Y plus one minus five. And if I simplify this even further, I have that Z is equal to X squared plus four, X -2, y squared Plus four, Y -3. So from here I can solve for F sub X and F sub Y. So my f sub x is two, x plus four is two, X plus four, two X plus four. And my F sub y is negative four. Why plus four? If we then plug in our point? So we have ffx at a comma B. We get that ffx at a comma B is eight and our f of y at a comma B is negative eight. It's taking all of these values. We can plug these in to our equation for a tangent plane. So we get that T is equal to eight times X -2 minus eight Times Y -3 plus three. And that is for our X. So from here we can simplify, We get that T is equal to eight X -16 minus eight. Why plus 24 Plus three. And if we add up all the like terms you get That T is equal to eight X -8, Y plus 11.

This question asks us to find the tangent plane given both the plane and a point on the plane. To start, we know that the tangent plane is T. Is equal to f sub X of a. Be times x minus a plus F sub Y at a B. Times Why -7 plus Z at a B. So to start we need to find the F sub X and the F sub Y given our plane. So the F sub X at our plane is one over Y squared. The F sub X at a. Come with B would then be one for it. The F. Supply For a given plane is -2 X over y cubed. So the F sub Y at a comma B is then one when you plug in the points. So from here we can plug things in to our tangent line equation. So we have that T. is equal to 1 4th x minus negative four or just plus four Plus one times y -2 minus one. And so from here we can simplify things. So we have the T. Is equal to X over four plus one plus why -2 -1. So if we take this over here we can simplify it even further, and we get that T is equal to X over four plus y minus two.

Were given a surface and appoint and rest to find an equation of the tangent plane to the surface. At this point, surface has the equation three y squared minus two x squared plus X. The point on the surface is too negative. One negative three. Yeah, well, right away. Noticed. Weaken right the as a function of X in y as Z equals three y squared minus two x squared plus X And therefore we have that the partial derivative f with respect to X is negative four x plus one and the partial derivative f with respect to why is six y so it follows that at our point to negative one negative three partial derivative with respect to X. Well, this is had to negative one which is negative. Seven. The partial derivative of F respect toe Why at two negative one is negative six. Therefore by equation two from this section. Vision for the tangent plane is Z minus and then rz value, which is negative three for this point, equals partial derivative of F with respect to x Evaluated A to negative one times X minus Our X value, which is to plus the partial derivative f with respect to why evaluated at two negative one times why minus or why value, which is negative one? Or why plus one and simplifying. We get Z plus three equals on the right hand side. We have negative seven times X minus two, minus six times Why plus one rearranging yet Z equals negative seven x minus six y and the constant term is simply five.


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