79–82. {Use of Tech} Visualizing tangent and normal lines

Question

b. Graph the tangent and normal lines on the given graph.

(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Graph the tangent line with the equation Y equals 13 divided by 9 X plus 40 divided by 9 and the normal line to the following curve at the given point. X2 + Y2 to the power of 2 is equal to 25 divided by 2, all multiplied by Y2 minus X2. 1.3 in parentheses. So this -1.3 in parenthesis is our given point. So this is our coordinate point, as we should recall, which is our XY coordinate point. Awesome. And let us also note, once again, we're ultimately trying to graph two separate curves on this graph that is provided to us, and the graph that is provided to us is a shape that has a curve that is shaped like a figure 8. Or it looks almost like an hourglass. And as you can see, we have our given point plotted on our graph, which is at -1.3. And what we're doing is we're gonna be using this graph to plot two separate lines. We're trying to graph two lines. We're trying to graph the tangent line. That's our first answer. And our second and final answer is the normal line. So we're trying to graph the tangent line and the normal line on this graph, and that is our final answer that we're ultimately trying to solve for. So, with that said, First off, in order to solve this problem, we need to recall and note once again that the equation of the tangent line to the given curve at the point -1.3 will be Y is equal to 13 divided by 9 X plus 40 divided by 9. And let us use this equation for Y to help us determine a point on our tangent line. So let us say in order to find a point on our tangent line as we should recall, we need to give x some value and then we need to plug in that value of X into our equation. So let us say that X is equal to 0. Let's make things easy on ourselves. So when X is equal to 0, Y will be equal to 13 9 or 13 divided by 9 X. Which in this case we need to plug in a value of 0 in for X, it's going to be 13 divided by 9, multiplied by 0, plus 40 divided by 9, which will give us 40 divided by 9. So that will mean when we take 40 divided by 9, that will mean that Y is approx. Ultimately equal to 4.44, and we plug that into our calculator and round to two decimal places. So that will mean that we now have a point on our tangent line and we need to plot this point 0.4.44, and we need to plot this. On our graph, and we need to take this point, we're going to plot it on our graph, and then we're going to draw a line that passes through this point on the tangent line through our given point, which is -1.3, and that will produce our tangent line. So when we do that, We will get the following graph. And as you can see, we have our point that we found on our tangent line, which is 0.4.44, which we just rounded to 4/4, but it's like 44444 when you type it into your calculator to be more precise. And as you can see when we plot that, and we plot it with our given point, we have a red diagonal line that starts in the positive quadrant and then works its way down into the negative quadrant, forming a diagonal line. Awesome, and it looks like it intersects the Y axis at 0.4.444, and it strikes the negative X axis at a little bit over -3, so it's like -3.2 something or more or less. So, our next step now, since we saw for one part of our Essentially, we solved for one answer for this problem that we're trying to solve for. Now we need to determine the slope of the normal line. That is our next answer that we're trying to solve for is we're trying to graph our normal line on our graph. So in order, in order to do that, we need to determine the slope of the normal line in order to help us figure out a way to graph it. So finding the slope of the normal line, we need to note and recall that a normal line to a curve at a given point will be the line that is perpendicular to the tangent line at that specific point. So the slope of the normal line, as we should recall, will be the negative reciprocal of the slope of the tangent line. So that will mean That M perpendicular, which it's an upside down T indicates the perpendicular symbol. So M perpendicular is equal to -1 divided by 13 divided by 9. Which is going to be equal to -9 divided by 13. And now our next step is we need to recall and use the point slope form to write our equation of the normal line. So in order to do that, we need to substitute the value of M equals -9 divided by 13 besides our value for the slope, and we need to substitute in our coordinate point, which is -1.3. So when we do that, we're calling and using the point slope, form, equation, and plugging in our known variables. We will find that. Why Minus 3, because that is our value for Y, so Y minus 3 is equal to our slope -9 divided by 13, all multiplied by X minus -1 means that our -1 is our value or X. So essentially, we're going to be using this equation, we need to rearrange it to isolate and solve for Y. And when we do that, when we do a little bit of simplifying, I'm going to skip a few steps, but all you need to do is just use a little bit of algebra to shuffle things around to get Y on one side of your equation. And when we do, we will find that Y is equal to -9 divided by 13 X plus. 30 divided by 13, fantastic. So therefore, the equation of the normal line to the given curve at our given point of -1.3 will be Y is equal to -9 divided by 13 X plus 30 divided by 13. So now, let us find a point on the normal line, and like we did for our point on the tangent line, let us just simply state that X is equal to 0 to keep things simplistic. So we will use our equation for Y and Y is equal to, which we just found, -9 divided by 13. So we need to take -9 divided by 13, and since it's multiplied by X, except X in this case is 0, so we need to multiply it by 0. Plus 30 divided by 13. So when we solve for Y, we will find that Y is simply equal to 30 divided by 13, which when we plug that into our calculator will mean that it's approximately equal to 2.31 when we round to two decimal places. So now we need to plot the point 0.2.31. And we're gonna be plot, we need to plot this point on our graph, and then we need to draw a line passing through this point. And now we also through our given point to help us draw the normal line. So when we do that, we will we will obtain the following final graphs. So this is our final answer that we were ultimately trying to solve for. And as you can see, All what we were trying to do. we were trying to get our two lines, essentially, we were trying to get our tangent line, which we did. That was what's represented in red, and now we're trying to find our normal line which is represented in blue. And what we needed to do is we needed to plot our point that we found together, which was 0.2.31. So we plot that point and then we have to draw a line that passes through our given point and the point that we found on the normal line. So when we do that, we will achieve our blue diagonal line, which starts in the negative quadrant and works its way down into the negative quadrant. So it goes from the 2nd quadrant to the 4th quadrant quadrant, whereas our tangent line starts in the 1st quadrant and goes to the 3rd quadrant. Awesome. So, with that said, that is it. So, essentially, once again, we were trying to figure out, once again, how to graph the tangent line and the normal line, which we did both. So that means we've officially solved for this problem. Hooray, we did it. So thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

79–82. {Use of Tech} Visualizing tangent and normal lines
b. Graph the tangent and normal lines on the given graph.
(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)

Key Concepts

Tangent Line

Normal Line

Implicit Differentiation

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