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equation of horizontal tangent line

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.