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Use implicit differentiation t0 find dyldx and d2yldx2. 3) xy - X+y =4 A) d-%+dly 2y + 2 dx2 (x+ 1)2 C) d-#d-&7 dx2 (x+1)2B) dy 4d 2y _ 2 dx dx2 (x + 1)2 D) dx ...

Question

Use implicit differentiation t0 find dyldx and d2yldx2. 3) xy - X+y =4 A) d-%+dly 2y + 2 dx2 (x+ 1)2 C) d-#d-&7 dx2 (x+1)2B) dy 4d 2y _ 2 dx dx2 (x + 1)2 D) dx LX,dly _tl dx2 (x+1)24) x2 +y2 = 5,at the point (2,1) A) dx z4 dx =0 dx2 dx = -2;- dx dx =35 dx2dx =-2; dy =1 dx2 D) d-2d =5 dx2At the given point; find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve, requested. 5) x5y5 = 32, tangentat (2,1) A)y = l6x + 1 B) y = - 3x+2 Cy-Zx D)y

Use implicit differentiation t0 find dyldx and d2yldx2. 3) xy - X+y =4 A) d-%+dly 2y + 2 dx2 (x+ 1)2 C) d-#d-&7 dx2 (x+1)2 B) dy 4d 2y _ 2 dx dx2 (x + 1)2 D) dx LX,dly _tl dx2 (x+1)2 4) x2 +y2 = 5,at the point (2,1) A) dx z4 dx =0 dx2 dx = -2;- dx dx =35 dx2 dx =-2; dy =1 dx2 D) d-2d =5 dx2 At the given point; find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve, requested. 5) x5y5 = 32, tangentat (2,1) A)y = l6x + 1 B) y = - 3x+2 Cy-Zx D)y = I6x - 1 6) y6+x3 =y2 + 1Ox, normal at (0,1) A)y=- 3* B)y = 2x+1 Cy-fx+1 D)y = - 3*+1



Answers

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.
$x^{2}-\sqrt{3 x y}+2 y^{2}=5, \quad(\sqrt{3}, 2)$

We have this equation on, we have the point squared 32 And the first thing they want us to do is to confirm that this point is on the graph of the equation. So we'll put this in and we have X squared. The square root of three squared is three minus the square root of three times. The square root of three is three times the y value, which is too. So that would be three minus six and then plus two times two square to square. This four times No. Two is eight. So three minus six is negative. Three negative. Three plus eight does equal five. So, yes, that point is on our grip. Now they want us to find the equation of the tangent line and the normal well, they equate the soap of the tangent line is the value of the first derivative evaluated at our point here. So that's what we want to do. First, just find the first drink, that of this. So we'll take the derivative with respect to X. So the derivative of X squared is gonna be two X and then for the next expression, we're gonna have to use the product rule and I'm going to treat the negative. The month um minus is a negative sign and I'm gonna have square with three X as my first function. So we have the first function The minus squared three x times the derivative of the section function the derivative of why iss de wide he x or y prime And then plus the second function Why times attribute about the first which would just be negative square to three. And then finally we have the derivative of the two y squared. So that will be plus for why? To the first and then times why prime? And that will equal the derivative of five, which is zero. So now I can substitute in my square root of three and my two for the exes and the wise, And I can figure out what why Prime is and that'll be the slope with my Tangipahoa. So two times X is to Times Square with three minus scored a three times square to three is going to give me three y prime. And then why is to so that would be minus to square with three and then plus four times to will give me eight. Why prime? And that will equal zero Well, my square two threes will add out, and that gives me five y prime equals zero, which means why prime equals zero. So that's just open my tangent line, which means my tangent, Highness Horizontal. And if it's horizontal, that means the equation is why equals whatever the y coordinate iss of those points that are on the line, which is to now for my normal line if my horizontal if my tangent line is horizontal, the normal line is perpendicular to that. So that means it has to be a vertical line. And we should all remember that the equation of a vertical line is always X equals the X coordinate of the points that it goes through, which in this case, is X equal.

Section 3.7 Problem number 37. So here again, we're dealing with implicit differentiation. So we're given a curve to x y plus pie sign of why equal to pie and were given the 0.1 high over, too. And here were asked to do three things that verified at this point that were given here is actually on the curve that you see, and then find the equation of the tangent line at this point, then find the equation of the normal line at this point. So in order to do this find the equation of the change in line means you're gonna have to find the derivative. You have a point and a slope at that point. And then once we have that that slope, then we can also find the normal on. So first off, let's verify that this point that you see is actually on the curve. So let's just go ahead and do that. So to verify that we would have to times one times pi over too. Plus pie comes a sign A pirate to Sony. We need to evaluate that, innit? That's equal to two pi. Then we're good to go So when you look at this two over to this becomes pie plus pi, The sign of power over to is one. So this is equal to two pi. OK, so by proving that when I substitute this particular point into the original occur of the answer is to pie, that means that that point is actually on the curve. Okay, so the next thing I need to do is to find the value of the derivative. So we've got the equation to X Y plus pie sign. Why you called two piles? I want to differentiate this entire equation with respect to X. So when I do that. So I mean to use the product rule for the 1st 2 x y. So the derivative of two acts uses simply going to be too times why plus two x The derivative of why is d y the ex and then the derivative of the sine function is the coastline function. So this is plus pie co signer, Why do you why the X equal in the derivative two pies, simply zero. So if I saw this for do I d x so I get Do you why DX is going to be negative too. Why over to X plus High co side of why? So this is the value of do I d. X, the derivative. And now to find the slope of the line at the point. Remember the point? We're interested in the 0.0.0.1, however too. So I need to evaluate this at the 0.1 hi or two. So when I do that, I get negative too. Times pie or two. And then in the denominator, I get two times one plus high times to co sign a pie or two. Okay, I know that the co sign of pie or two this is equal to zero. And so what I end up with here When I simplify this, this becomes negative pie over too. So now I know that I've got a point one hi or two and I have the slope of negative pie or two. So using the point slope, I can come up with the equation of the tangent line. And so that's going to be why minus pi over too equal. Negative pi over two times X minus one. And if we solve this for why we're going to get why is equal to negative pie or two X plus pirate too, Plus pirate too. So that final answer in slope intercept form is negative. Pi over too X plus, Yeah, Now to find the equation of the normal line, so the normal line is gonna go through the same point. So the 0.0.1 high over too. But the slope of the normal line is gonna be negative, reciprocal off the slope of the change in line. So it's going to be too over. Hi. So this equation is going to be why minus pi over to is equal to two over pi X minus one. So solving that for why you're going to get why equal to over Pi X minus two over pi plus pi over to okay, and so what we have now. So in doing this, if we review what we did verified that the point that they give us is on the curve check um, to find the slope of the change in line, you had to find the derivatives that we use implicit differentiation to figure out what is d y. The X. And then we saw that at the 0.1 pirate to to get the slope of the curve at that point once we have the slope of the curve. At that point, we could figure out equation of the change that line in the equation of the normal line.

Were given this equation and this equation is supposed to go through the 0.1 by over two, and they want us to confirm that first. So let's just put these coordinates in and check it out. So too times one times pi over two will give us pie and then the sign off by over two. It's one that one times pie is pie and pi plus pi is to pop. So, yes, that point is on the graph. Now they want us to find the equation of the tangent line and then the equation of the normal line for this equation. It's graph and the slope of the tangent mon is the value of the first derivative evaluated at our point that we have. So I'm going to take my first expression, my two X, and I'm gonna treat that as the first function and the wise the second function. And I'm gonna do the derivative of this using the product rule. So we'll have the driven of her two x first function times the derivative of the second derivative for why is why prime or do I d X plus a second function? But why times the derivative of Hearst derivative of to access to. So there's that part taken care of. Plus, I remember pies a constant. So this is just a number times a sign off something. So the driven off that will be pi times co sign off why and then because of the chain rule, we have to take the derivative of our angle function. And again the derivative Why is why prime? And that equals the derivative off to pie, which is zero. So there's my first urine. And now putting in my one and my pi over two, I can figure out what? Why Prime Miss So two time the one give me too. Why Prime Plus putting pi over to infer why that will give me by over two times two, which is just pie. And then go sign off pi over two zero. So that all equal zero and that equals zero. So that means that why, prime, I was gonna equal negative pi over two when I saw that for Wipro subject the pie from both sides. Divide by two. So that's the slope of my Qianjin line. So for part a finding the equation of the tangent line. We can use thes hope, intercept form and plug in our Y value and are X value and are slope to find the Y intercept so we can write the equation. So why to equals r slope negative pi over two times x which is one plus b. So that's just negative. Pi over two and we had that to both sides and we get that b s by. So the equation of our tangent line will be Why equals negative by over two times X plus. Now, for the normal I Well, the normal line is perpendicular to the tangent line. So that means it's slope is the negative reciprocal of the negative pi over two. So we can do the same process, use our point and our slope and figure out the Y intercept for the normal line. So the y coordinate pi over two equals the scope of our normal line, which will be positive. Do over pie times are X coordinate of one plus b. So then we just have to subtract the two over pie from both sides to figure out that be iss hi or two minus two over pie. So the equation for the normal line will be why equals two over pi Times X plus our Y intercept the pi over two minus two over pie.

And this problem, we're given the implicit curve X sign to why, equals Why Time's the coastline of two X and were asked to consider the point pirate for come a pie or two. So first you want to show that this point does lie on the curve. So what we're gonna do is we're gonna plug in X equals pyre before and why equals pi over too. And we're going to check to see if we get a valid equation or not. So we're gonna be looking for something like zero equals zero or one equals one or pi equals pi. If we get something wrong, like zero equals one, then that means that we're not on the curves. So hopefully we get a correct solution. Plug it in pira four times the sine of to Why is two times pi over to which is pie equals? Why? So why is pi over too? Times the co sign of two ex so two times pi over four is pi over too. So the sign of pious zero. So we're checking. There's a question mark. Sorry to see if this is equal On the right hand side, the coastline of pie over two is also zero. So we get zero equals zero, Which means our point is on the curve. Great. You know, we want to start talking about slopes. So first we always know that the slope of the tangent line is given by D. Y d X, just the derivative. So we know we have to do some kind of implicit differentiation to find that and for the slope of the normal line, this is a line that is perpendicular to the tangent line. So to find the slope we take the negative reciprocal negative one divided by the slope of the tangent. Great. So let's go ahead and start by finding D Y d. X using implicit differentiation. So we need to take the derivative of both sides so ddx of X times sign of two. Why equals ddx of why Times Co sign of two x. So on the left hand side, you've got a product rule here. So it's the derivative of the first times, the second plus the first time, the dread of the second. So to take the derivative of sign of to why this is kind of like a chain rule where the outer function is sign, and the inner function is to Why so the Druid of Sinus co sign plug in the inter function and multiply by the derivative of the inside So two times D y t X now on the right side. Once again, we do have a product rule. So's the dreaded two of the first times, the second plus the first times, the derivative of the second. So that's why now again, this is a chain rule. The Druid of co sign of two exes minus two. Sign of two X. Now let's get all of our D Y D X terms on to one side. So here's our D Y D X terms, and then all of the other terms they're going to be on the other side. So on the left hand side, we have a to x co sign of two. Why times do I D X minus co sign of Jew Ex Hoops Co sign of two x time D Y d. X On the left hand side, we have the rest of the term. So sign of two. Why minus two. Why sign of two X? How in fact, out of the UAE DX two X Times Co Sign of two I minus coast to X equals signed to eye minus two. Why sign to X now? By dividing by this expression on the left, we get an expression for formula for D. Y d X Sign of two Y minus two. Why time's a sign of two X All over two ex coast to eye minus coast to X. So here's an expression for a D. Y D X, and now we just need to plug in our specific point at pi over four. Com a pie over, too. So let's plug that in and see what happens. So this is going to be the slope of the tangent line. So here, right that slope of tangent is equal to sign of. Two. Why? So that sign of two times pi over to which is pie. Mike has two times why it's pie over two times the sine of to X. So sorry hoops Sign of I made a mistake there two times. Why is two times pi over too? Which is pie times the sine of to X. So that's the sign of pi. Over four, two times higher before. Which is pi over, too. Now the dominant. We have two times x, so it's two times pi over four. Is pi over too? Times Co Sign of two. Why so the coastline of two times by or two That's co sign of pie minus co sign of two x. So that's the co sign of two times part before is co sign of pie over too. Now we just need to use our knowledge of trick values. Sign of pious zero sign of pie over to is one. So the numerator is minus pi, and in the denominator co sign of pie is negative. One and co sign of pie over to is zero. So we just have a minus pi over, too. So minus pi divided by minus by over two is two. So here is the slope of the tangent line. Let's open up a new page. Now we want to find the equation for the tangent line, so we know that the slope is too. And our point is pi over four com a pie over to so we have our point. We have a slope. Let's use the point slope form. Why might aspire for two equals the slope times X minus the X value, which is X minus pyre before. This is a perfectly fine way to write this. But if we wanted to, we could also turn it into a slope. Intercept form. So why equals X minus pi over two plus pie over too. So why equals X? I hope so. I think one of these should be a pull loss. Did we, um, make ah Oh, sorry. There should be two acts. That was the problem. There we go. So why equals two X? Sorry about that. So here is our attention line. And for the normal line, we know the slope is one divided, a negative one divided by the slope of the tangent line because it's perpendicular. So negative one half. So again, point slope. Why minus pi over two equals negative one half times x minus pirate for So this is a perfectly good answer, just like this was. But if we want to write it in slope intercept form, which again, we don't have to. Why? Equal native one half x plus pie over eight plus supply over too. So why equals negative one half times x plus five pie over eight. So here's the slope intercept form for writing this answer


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