Function problemDate: Wed, 7 Dec 1994 14:48:43 -0500 (EST) From: Aklilu Habte - ASNG/F94 Subject: Help, Math Problem Hi...there, I have been working on the problem written below and am not able to solve it, need help. Q- Find the slope of the tangent line for the function y2 = x2 - x4 at (0, 0), also explain what is going on at the origin. NB: The function read as " y SQUARE = x SQUARE - x RAISE 4" I appreciate your help, thank you. Aklilu Habte. Ryerson Polytechnic University. Aero-Space Engineering. Date: Thu, 8 Dec 1994 12:28:46 -0500 (EST) From: Dr. Ken Subject: Re: Help, Math Problem Hello there! This problem is one that makes use of the concept of implicit differentiation. I'll go ahead and use the concept, and please write back if you don't understand what I've done. The standard way to write your equation is y^2 = x^2 - x^4. Note that this is not really a function in the strictest sense of the word, since you get two values of y for almost every value of x you plug in. So we'll take the derivative with respect to x of both sides: we use the chain rule on the left side, and it spits out a dy/dx. 2*y*dy/dx = 2*x - 4*x^3. So solving for dy/dx (the derivative of y with respect to x, i.e. the slope of the tangent line), we get dy/dx = (x - 2x^3)/y This is the form in which most mathematicians would leave the answer. It's kind of a strange answer, since it has both x's and y's in it, but it's the best we can do. So to find the value of dy/dx at (0,0), we'd like to just plug in (0,0) for x and y. But that would be really really bad, since then we'd divide by zero. So we'll look at the limit as (x,y) approaches (0,0) instead. Ordinarily, we'd solve for y in the original equation, and plug that value into the derivative equation, and then take the limit. I love to take limits. The tricky part here, though, is that we can't really solve for y, since there are two values for y. But we can get around that by using both of the values y = Sqrt{x^2 - x^4} and y = -Sqrt{x^2 - x^4}. So substituting into the derivative equation, we see that dy/dx = +- (x - 2x^3)/Sqrt{x^2 - x^4} Where the +- means "positive or negative." Note at this point that our only hope of having a well-defined tangent line is if its slope is zero, since that's the only time the limit as x->0 will exist. This makes sense since the graph of our original equation is symmetric about the y-axis. So I'll leave it to you to verify that this limit as x->0 is indeed zero. Here's a hint: try squaring (if the square of something goes to zero, the something must go to zero too). Good luck! Please write back if you have more questions! -Ken "Dr." Math Date: Thu, 8 Dec 1994 14:06:04 -0500 (EST) From: Aklilu Habte - ASNG/F94 Subject: Re: Help, Math Problem Hello Dr. I have tried in the same way too. However, when I set up, the "limit" still didn't make any sense to me. Your hint: "Try squaring (if the square of something goes to zero, the something must go to zero too)." Can you be a little more specific, please! I also tried to approach by graphing the equation, and at the origin (0,0) the slope of the tangent line seems to be "vertical", i.e. the slope is undefined! How does this sound to you? Thank you for your help. Aklilu Habte. "Student" Date: 9 Dec 1994 15:41:46 GMT From: Dr. Math Subject: Re: Help, Math Problem Hello there! When we last spoke, we had gotten to the formula dy/dx = +- (x - 2x^3)/Sqrt{x^2 - x^4}. Now we want to take the limit as x->0. I must apologize for misleading you about what to expect this limit to be. I think I suggested that it was going to be zero, and in fact it's not. But let's find out what it is. Notice that we can pull an x^2 out of the square root sign. So we can cancel an x with the top, and we get dy/dx = +- (1 - 2x^2)/Sqrt{1 - x^2}. If we let x go to zero, then we see that this limit approaches either 1 or -1, depending on what direction we come from. So in answer to the original question, there are _two_ tangent lines at the origin, one that has slope 1, and one that has slope -1. If you plot the picture, you should get something that looks like a figure eight turned on its side, something that looks kind of like an infinity sign. The figure eight has two lines that cross at the origin, one that has slope 1, and one that has slope -1. I hope this helps you! Please write back again if you need more help. |
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