Finding Tangents to Curves Using DerivativesDate: 09/29/1999 at 22:58:04 From: Emon Eshaque Subject: Calculus 1. Let T be the point on the graph y = k/x whose x-coordinate is t. Prove that the x-intercept of the tangent at T is 2t. 2. The line y = mx+1 meets the parabola y = (1/4)x^2 at A and B. Prove that the tangents at A and B are perpendicular. 3. A curve known as a strophoid is defined implicitly by the equation x3 + 4x2 + xy2 - 4y2 = 0 (where -4 < x < 4) and is shown in the diagram. Find the points at which the tangent line is horizontal. Round the coordinates off to 2 decimal places. I will be glad if you send me some hints. Date: 10/24/1999 at 22:16:41 From: Doctor Jaffee Subject: Re: Calculus Hi Emon, Here are some hints that should help you. In problem 1 we are concerned about the point where the x-coordinate is t. If you substitute t for x in the equation, you will have the y-coordinate at that point. The derivative of the function will tell you the slope of the tangent line, so calculate the derivative and then substitute t for x to find the slope at your point. Now you know the slope and a point on your line, so use the slope-point form of the equation to determine the particular equation. Finally, the x-intercept is the point where y = 0. Therefore, substitute 0 for y and solve for x and you should get 2t. In problem 2, remember that two lines are perpendicular if the product of their slopes is -1. Problem 3: d/dx(x^3) = 3x^2 d/dx(4x^2) = 8x d/dx(xy^2) = y^2 + 2xyy' d(4y^2) = 8yy' Take the derivative of both sides of the equation and solve for y'. Since the slope of a horizontal line is 0, set y' = 0. I hope these hints are helpful. Give them a try, and if you need more help, write back. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/ |
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