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

Question

a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)

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

Accepted Answer

Below 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. Determine the equation of the tangent and normal lines to the curve X2 + Y2 all to the 2. is equal to 25 divided by 2 multiplied by Y2 + X2 at the point 1.3 in parentheses. Awesome. So it appears for this particular problem for us to solve for two separate answers. So our first answers are trying to determine the equation of the tangent line. That's to the curve that's given to us by the problem itself. That's our first answer, and we're asked to determine what the equation of the normal line is to this curve that is given to us. So we're ultimately trying to figure out what the expressions for the tangent line and normal line equations are going to be, and those are our final two answers that we're ultimately trying to solve for. What is the tangent line equation and what is the normal line equation? So with that said, first off, in order to find the equations of the tangent and normal lines to the curve at our given point of 1.3 in parentheses, we need to recall and note that in order to determine this, we must first find the derivative of the curve, which will help us determine what the slope of the tangent line is at our specific point. So, with that said, our first step is we need to differentiate both sides of the equation with respect to X, and in order to do that, we need to recall and apply the chain rule on the left side of our equation. And then we also need to recall and use the power rule of differentiation, which, as we should recall states that D by DX of X to the power of N is equal to N multiplied by X of N minus 1. So with that said, we could go ahead and write that D by DX of X 2 plus Y 2, all the power 2 is equal to 25 divided by 2 multiplied by D by DX of Y2 minus X the power of 2. Fantastic. So when we start to differentiate, we will get 2 multiplied by X2 + Y 2 multiplied by D by DX of X2 + Y 2 is equal to 25 divided by 2. And then we need to take 25 divided by 2, and we need to multiply it by 2 Y, multiply it by DY by DX minus 2 X. Fantastic. And then, Finally, doing a little bit more simplifying, we'll get 2 multiplied by X2 plus Y2 multiplied by 2 X plus 2 Y multiplied by Dy by DX, which is gonna be all equal to 25 Y multiplied by DY by DX minus 25 X. Fantastic. So now, our second step is we need to solve for DY by DX in order to help us find what the slope of the tangent line will be. So we need to take 4 multiplied by X to the power of 3 plus X multiplied by Y 2 plus x 2 multiplied by Y multiplied by DY by DX. Fantastic. And now we need to take And we need to take this, and we need to add to the power 3 multiplied by DY by DX. Fantastic, and say that this is all equal to 25 Y multiplied by DY by DX minus 25 X. Fantastic. So now, our next order of business is we need to start simplifying this big old long expression, and we're going to be trying to rearrange it to isolate and solve for DY by DX. So we're gonna be rearranging it and I'm gonna skip a few steps, or quite a few steps, but quite a few steps, but essentially we're taking this big old long expression that we currently have. And we're going to be using algebraic techniques to rearrange it to isolate and solve for DY by DX in its most simplified form. So when we do that, we will find that DY by DX is going to be equal to negative X multiplied by 4 x 2 + 4 Y 2 + 25. All divided by Y multiplied by 4 x 2 + 4 Y 2 minus 25. Fantastic. And that is our final answer for this most simplified form for DY by DX is equal to our simplified expression. Fantastic. So now, Our third step is we need to substitute in our given point, which once again, we should recall our given point is in parentheses 1.3, knowing that this is in our XY coordinate layout. Fantastic. So we need to substitute in our given point into our derivative equation in order to find the slope of the tangent line. So meaning we need to take DY by DX is equal to negative and we need to Plug in a value of 1 in for X and a value of 3 in for Y. So when you take 1 multiplied by 4, multiplied by 1 to the power of 2, plus 4 multiplied by 3 to the power of 2. Plus 25, all divided by our value of Y, which once again is 3, multiplied by 4, multiplied by 1 to 2, plus 4 multiplied by 3 to the power of 2 minus 25. So when we plug in this big old long expression into our calculator, and we write our final answer infraction form, we will get a final answer of the slope being -13 divided by 9. Fantastic, and that is our slope. -13 divided by 9 is our slope is our answer for the slope. So now our fourth step is we need to recall and use point slope form to write our equation of the tangent line. So as we should recall, the point slope form of a line states that Y minus Y1 is equal to M or slope multiplied by X minus X1. Where once again M is our slope, and this X1.com Y1, so X1.com Y1 is our point of tangency. So this is the point on the line. So now, with that said, We now need to take this information and we need to plug in our value of M equals, since our slope we found is equal to -13 divided by 9, and we're told that our point is 1.3. We need to plug this into our point slope form equation. So when we do that, we will find that Y minus 3 is equal to -13 divided by 9 multiplied by X minus 1. So when we rearrange this equation to isolate and solve for Y all by itself, we will find that Y is equal to -13 divided by 9 X plus 40 divided by 9. So that is it's, that's when we rearrange the equation to isolate and solve for Y and we keep our answers in fraction form. Fantastic. So, with that said, So therefore, the equation of the tangent line to the given curve at 0.1.3 is going to be Y is equal to -13 divided by 9 X plus 40 divided by 9, and that is one of our final answers. Hooray, we did it. So now we found one of our answers, but we still need to find for one more answer. We still need to find the slope of the normal line in order to find our equation for the normal line. because once again, that's our second answer we're trying to solve for. We're trying to find the normal line equation. So in order to do that, we need to first determine the slope of the normal line. So with that said, We need to recall and note that the normal line to a curve at a given point is going to be the line that is perpendicular to the tangent line at that specific point. So the slope of the normal line is negative reciprocal of the slope of the tangent line, which means that, as we should recall, M the slope, that is perpendicular, which were used in upside down T to denote the perpendicular symbol. So M subscript, this perpendicular symbol. So, The perpendicular slope, so M perpendicular, is equal to -1 divided by 13 divided by 9. It's going to be equal to 9 divided by 13. Fantastic. So now our sixth step is we need to use the point slope form to write our equation of the normal line. So we need to substitute in our value of M perpendicular is equal to 9 divided by 13 and 1.3 in parentheses into our point slope equation. So we need to say that Y. Our point slope form equation, to be precise. So Y minus 3 means you plug in a value of 3 in for Y is equal to 9 divided by 13, because that's our value for the slope now, multiplied by X minus 1 is a value of 1 in for X. So like we've done before, we To rearrange the equation to isolate and solve for Y. And when we do, we will find that Y is going to be equal to 913. So 9 divided by 13, so 9/13. So 9 divided by 13 X plus 30 divided by 13. And that is our final answer for the normal line. Hooray, we did it. So therefore, the equation of the normal line to our given curve at our specific point 1.3 is Y is equal to 9 divided by 13 X plus 30 divided by 13. Hooray, we did it. So to look at our problem itself and to quickly recap what we've learned, our final answers without a doubt have to be our tangent line is going to be at Y equals -13 divided by 9 X plus 40 divided by 9. That's our tangent line, and then our normal line is going to be Y is equal to 9 divided by 13 X plus 30 divided by 13. 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 <IMAGE>
a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)
(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)

Key Concepts

Tangent Line

Normal Line

Implicit Differentiation

Watch next

79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)

Key Concepts

Tangent Line

Normal Line

Implicit Differentiation

Watch next