45–50. Tangent lines Carry out the following steps.

Question

b. Determine an equation of the line tangent to the curve at the given point.

x⁴-x²y+y⁴=1; (−1, 1)

Accepted Answer

Hello. In this video, we are going to be finding the equation to a tangent line. The problem asks us to find the equation of the tangent line to the curve defined by the equation X2 minus 4 XY plus 4 Y2 equal to 49 at the point -3.2. So how do we start this problem? Well, since we are trying to find the equation of the tangent line, there are two pieces of information we need. First, we need a defined point, which is given to us at -3.2, but then we also need to find the slope at that defined point. In order to find the slope, we are going to need to take the derivative of the given equation. Now, the equation given to us, again, is X2 minus 4 XY plus 4Y2 equal to 49. In order to take the derivative, we will need to take the derivative of the entire equation. With respect to X. And of course, since we have both X and Y variables in our equation, we will take the derivative using implicit differentiation. So let's go ahead and start by taking the derivative of the left-hand side of the equation. The derivative of X2 will be 2 X by the power rule. For the second term for XY, since we have a product of X and Y, we will need to take the derivative using the product rule. So, that will be -4, multiplied by the product rule of X multiplied by Y. That will be X multiplied by the derivative of Y. Now, the derivative of just Y will be 1 by the power rule, but since we are taking a derivative of Y with respect to X, that is going to generate a DYDX term through implicit differentiation. And one multiplied by DYDX is just going to leave us with DYDX, and this will be the derivative of Y with respect to X. Following the product rule, we will then add Y multiplied by the derivative of X, which is just going to be 1. Then for the final term, that will be 4 multiplied by the derivative of Y2d, which will be 2 Y by the power rule, but again, this will also generate a DYDX term through implicit differentiation. And for the right hand side, the derivative of 49 will be 0, because the derivative of any constant is 0. Now what we want to do is we want to simplify the left hand side. We will start by distributing -4 to the quantity XDYDX plus Y. That will allow us to rewrite the left-hand side as 2X minus 4 XDYDX minus 4Y. And for the last term, 4 multiplied by 22YDYDX will give us 8YDYDX, and this will all be equal to 0. Now notice that we have multiple terms, where some terms have factors of DYDX, and the others don't. What we want to do is we want to keep all of the DYDX terms to one side of the equation and move all the terms without a DYDX factor to the other side of the equation. So, by reordering and moving terms around, we can rewrite the equation as 8 Y D YDX minus 4 XDYDX equal to 4 Y minus 2 X. Since we want to solve for the derivative DYDX, what we can go ahead and do is we can factor out a DYDX from the left-hand side of the equation. That will allow us to rewrite the equation as DYDX multiplied by 8 Y minus 4 X, and that will equal to 4 Y minus 2 X. Finally, in order to solve for the derivative DYDX, we will divide both sides of the equation by 8 Y minus 4 X. That will give us the derivative DYDX equal to 4 Y minus 2X divided by 8 Y minus 4 X. So what we have is a derivative for the original equation. Now, in order to obtain the slope for the tangent line, we are going to take the given point, which again, is going to be -3.2, and plug it into our derivative. That is going to allow us to rewrite the derivative as 4 multiplied by 2, minus 2 multiplied by -3, divided by 8 multiplied by 2, minus 4 multiplied by -3. Now we will go ahead and simplify. The numerator will simplify to be 8 + 6, and the denominator will simplify to be 16 + 12. This will further simplify to be 14 divided by 28, which is then simplified to be 12. So at the point, -3.2, the slope of the tangent line is going to be positive 1/2. Now that we have the two pieces of information we need, we can now solve for the tangent line. So, in order to solve for the tangent line, we will use the point slope form. We have Y minus Y1 equal to the slope M multiplied by X minus X1. Now we will plug in the information. That will allow us to rewrite this equation as Y minus 2 equal to 12 multiplied by X minus -3, which will be 3. We will distribute 1/2 to X + 3, allowing us to rewrite the tangent line as Y minus 2, equal to 1/2 X plus 3 divided by 2. Finally, what we will go ahead and do is we will add 2 to both sides of the equation, and that will allow us to rewrite the tangent line as Y is equal to 1/2 X plus 7 divided by 2. And this is going to be the equation to the tangent line at the point -3.2. And with that being said, the overall solution is going to be D. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

45–50. Tangent lines Carry out the following steps. <IMAGE>
b. Determine an equation of the line tangent to the curve at the given point.
x⁴-x²y+y⁴=1; (−1, 1)

Key Concepts

Implicit Differentiation

Tangent Line Equation

Slope of the Tangent Line

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