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equation of horizontal tangent line

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Question:How would you solve this?

Answers:A horizontal line will have a slope equal to zero. To find the equation for the horizontal tangent line to the equation y = f(x), differentiate y with respect to x and set the derivative equal to zero. 2y^3 + 6x^2y - 12x^2 + 6y = 1 Differentiating this equation with respect to x, 6y^2 dy/dx + 12xy + 6x^2 dy/dx - 24x + 6 dy/dx = 0 6(y^2 + x^2 + 1) dy/dx + 12xy - 24x = 0 6(y^2 + x^2 + 1) dy/dx = 12x(2 - y) dy/dx = 12x(2 - y) / (y^2 + x^2 + 1) Set dy/dx = 0 and solve for y, 12x(2 - y) / (y^2 + x^2 + 1) = 0 12x(2 - y) = 0 y = 2 The equation for the horizontal tangent line to the curve 2y^3 + 6x^2y - 12x^2 + 6y = 1 is the line y = 2

Question:what are the equations for the horizontal tangents to the curve y= (x^3) - 3x - 2? what are the equations for the lines that are perpendicular to these tangents at the points of tangency? what is the smallest slope on the curve? at what point on the curve does the curve have this slope? and what is an equation for the line that is perpendicular to the curve's tangent at this point?

Answers:y=(x^3) - 3x - 2 dy/dx= 3x -3 = slope horizontal tangent means m = 0 = 3x -3 =3(x -1) x = 1 y = (x^3) - 3x - 2 y = -4 and y = 3 equations for the lines that are perpendicular to these tangents at the points of tangency would be x = 1 The function increases from - to , the smallest slope on the curve is at x = 1 where slope = 0

Question:A. Find an equation for the line tangent to the circle x^2 +y^2 = 25 at the point (3,-4). B. At what other point on the circle will a tangent line be parallel to the tangent line in part A?

Answers:x + y = 5 The circle is centered over the origin. The slope of the radius through (3, -4) is -4/3. The tangent through (3, -4) is perpendicular to the radius, so the tangent has slope 3/4. point-slope equation of the tangent: y + 4 = (3/4)(x - 3) The other point is (-3, 4). http://www.flickr.com/photos/dwread/4975012943/

Question:Consider the circle of radius 5 centered at (0,0). Find an equation of the line tangent to the circle at the point (3,4). Please explain this problem to me step by step. I haven't entered calculus yet, so I would know nothing about any calculus concept.

Answers:Calculus isn't needed here, just remembering some properties of circles. The line tangent to a circle is also perpendicular to the radius drawn to the point of tangency. So finding the slope of the radius out to the tangent line is easy in this case, since you're starting at the origin and going out to (3.4): rise/run, or 4/3. The tangent line will be perpendicular to the radius, so it's slope will be a negative reciprocal: -3/4 Now use y = mx + b, substitute in -3/4 for m, and (3,4) for x and y to find b: 4 = -3/4(3) + b 4 = -9/4 + b b = 25/4 So the equ'n of the tangent line is y = -(3/4)x + 25/4.

From Youtube

Calculus: The Equation of a Tangent Line :Watch full lesson here: www.mindbites.com This lesson is part of a series: Calculus In this lesson, you will cover how to find the equation of a line tangent to a curve, equations of tangent lines, how to find the derivative, and how to find the point where the tangent line is horizontal. To find the equation of the tangent line, you'll start by taking the derivative of the curve and then evaluate that derivative at the point of tangency. You'll then substitute the coordinates of the point of tangency as well as the calculated slope into the point-slope form of the line: y-y1 = m*(x-x1) to get the equation of the tangent line. To find where the tangent line is horizontal, you'll set the derivative of the function equal to zero and solve for x. Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at www.thinkwell.com The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'H pital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Horizontal and Vertical Tangent Lines to Polar Curves :This video explains how to determine the points on a polar curve where there are horizontal and vertical tangent lines. mathispower4u.yolasite.com