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Learn how to solve Linea approximation equation in this sixth example.
Tags:Linear Approximation Equation Example 6 Part 1/2,approximation,calculus,help,linear,integralCALC
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Transcript
How to Solve a Linear Approximation Equation Example 6 Part 1/2
Hi everyone. Welcome back to integralcalc.com. We’re going to be doing another linear approximation problem today. This one is asking us to find the equation of the tangent line and it was submitted by one of our viewers. The function that was given or the equation I should say that’s given is exy + x2 - y2 = 5 and they asked us to find the equation of the tangent plane at the point, or sorry, at the tangent at the point two zero.
Okay. So, when we’re dealing with linear approximation, we’re asked to find the tangent line or the tangent plane. When we have two variables as we do here with x and y, another way for finding the tangent line and there is one variable, we just take the derivative and then do some work. In this case, since we have two variables, we have to take partial derivatives.
So, we’ll first need to take partial derivatives with respect to X and Y. Actually before I do that, let me go ahead and write out the linear approximation equation, the equation that we’re going to use or the formula that we’re going to use to build our equation of the tangent line. It looks like this, z - z1 = a(x - x1) + b(y -y1). When you’re dealing with two variables as we are here, x and y, this is always the equation that you want to use to find the equation of the tangent line.
So, each of these component is represented by something we’re going to solve for. You’re going to end up leaving z, x and y intact and we’re going to be solving for z1, x1 and y1 as well as a and b. And I’ll let you know what each of those represent as we go through.
So, like I said, first thing we do, partial derivatives with respect to x and y. So, first, the partial derivative with respect to x. we go term by term. So, partial derivative with respect to x of exy; remember when we’re taking partial derivatives or holding the opposite variable constant, so in this case since we’re taking it with respect to x, we’re treating y as a constant.
So, I like to think of it as the number two or the number three, a constant. So, if y = 3, then this term here, instead of exy, it ends up looking like e3x, right, because that, if we replace that with three, it would become the coefficient on this x term and we have e3x. well, the derivative of e3x is 3e3x. Well, in this case, what does three represent? It represents y. So, it’s not 3e3x, it's yeyx. In that case, we would reorganize the variable you always want to have your variables in alphabetic order. So, instead of 3e3x, it would be yexy. So, yexy is the derivative of this first term, the partial derivative of this first term with respect to x.
Here on the second term, we know that is much simpler. It’s just 2x and then this term here, the y2 is actually going to be 0 because if we plug in three for y, again, we’re treating it as a constant like 5 here.
If we plug in 3, we’ll get 9 which is still a constant and the derivative of any constant is just 0. So we end up with 0 for this term. And then the other thing I did want to do before we moved on, this equation here when you’re doing linear approximation, what I should have done before I withdrew all of these steps, was actually subtracted. If you can imagine, if you take this equation and you subtract five from both sides, right, this will become zero equals and then this whole thing right, instead of =5, the 5 used to be over here. We subtracted it for both sides and you end up with 0 on your side and then you’ll 5 on this side, we want to do that because it’s important that we, at least acknowledge that the 5 is in here and like I need to make sure that I’m dealing with the 5 and I’m taking partial derivatives.
Of course in this case, it’s going to go away because it’s a constant and he derivative of any constant is zero. So, it doesn’t factor into this partial derivative. But you should always first subtract that constant from both sides and get this equal to zero.
So, let’s go ahead and continue. This is the entire partial derivative with respect to x. So now, let’s go ahead and take the partial derivative with respect to y, same as before except now we’re holding x constant. So, in this case that we hold x constant, we treat x as, let’s call it the number four. So, pretend =4. e4y, we’re taking the derivative with respect to y when y is our variable.
So, the derivative of e4y would be 4e4y. So, in this case, four represents x, so we’re looking at xe4y for our derivative. Then we move on to x2. The derivative of x2 with respect to y is zero. With respect to x, of course it’s 2x, but with respect to y, we’re holding x constant. So, this is treated like four, 42 = 16 and 16 is a constant, so the derivative is zero. Of course then, this y2, y is actually the variable we’re dealing with here. So this becomes -2y and then of course the five is still a constant so the derivative there is zero.
So, these are the partial derivatives with respect to x and y. And now, what we need to do is three things. We need to plug in our point to zero to three different things, the original equation and both partial derivatives. So, let’s go ahead and plug it into our original equation first. So, two is representing x here obviously and zero y. So, let’s go ahead in saying 0=ex, so that’s two times y w= 0 + x2. So, 22 - y2 -0 - 5, okay. So, 2 x 0 = 0, so we end up with e0 + 4 - 0 - 5. You do the zero, in fact anything raised to the zero power is one. So we end up with one plus 4 - 5, which is 5 - 5=0.
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