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x^2 + 2xy - y^2 +5x =3 at point (1,3)

2007-03-25 22:47:53 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

x^2 + 2xy - y^2 + 5x = 3, at (1, 3)

Differentiate implicitly,

2x + 2y + (2x)(dy/dx) - 2y (dy/dx) + 5 = 0

Move everything without a (dy/dx) to the right hand side.

(2x)(dy/dx) - 2y(dy/dx) = -2x - 2y - 5

Factor (dy/dx).

(dy/dx) (2x - 2y) = -2x - 2y - 5

Divide both sides by (2x - 2y)

dy/dx = [ -2x - 2y - 5 ] / (2x - 2y)

To determine the equation of the tangent line at (1, 3), set
x = 1 and y = 3. This will be our slope m.

m = [ -2(1) - 2(3) - 5 ] / ( 2(1) - 2(3) )
m = [ -2 - 6 - 5 ] / ( 2 - 6 )
m = [ -13 ] / [- 4]
m = 13/4

Now that we have the slope, it's a matter of using high school algebra finding the equation of the line with slope m = 13/4 and through (1, 3). Using the slope formula,

(y2 - y1) / (x2 - x1) = m

Plugging in (x1, y1) = (1, 3) and (x2, y2) = (x, y) and m = 13/4,

(y - 3) / (x - 1) = 13/4

Cross multiplying,

4(y - 3) = 13(x - 1)
y - 3 = (13/4)(x - 1)
y - 3 = (13/4)x - 13/4
y = (13/4)x - 13/4 + 3
y = (13/4)x - 13/4 + 12/4
y = (13/4)x - 1/4

2007-03-25 22:56:39 · answer #1 · answered by Puggy 7 · 0 1

what's the curve equation to discover the tangent line at an element (25,5) on it? EDIT: [As in accordance to the extra desirable guidance, which i could prefer to relook basically after approximately 9 hours of earlier presentation] a million) Differentiating the given one, dy/dx = a million/(2?x) 2) At x = 25, dy/dx = a million/(2?25) = a million/10; it fairly is the slope of the tangent line on the given element; it fairly is via the geometrical definition of differentiation] 3) making use of Slope-element style, the equation of the tangent line at (25,5) is: y - 5 = (a million/10)(x - 25) increasing and simplifying, the equation is: x - 10y + 25 = 0

2016-12-08 11:26:48 · answer #2 · answered by bocklund 4 · 0 0

Differentiate both the sides with respect to x,
2x+2y+2x.dy/dx-2y.dy/dx+5=0
or, (dy/dx).(2x-2y)= - (2x+2y+5)
or, dy/dx = - (2x+2y + 5)/(2x-2y)
Put (1,3),
dy/dx = 13/4
Therefore the tangent is 13/4.

2007-03-25 23:01:54 · answer #3 · answered by Nikhil 2 · 0 1

0 = x^2 + 2xy - y^2 + 5x - 3
dy/dx = 2x + 2(dy/dx) - 2y(dy/dx) + 5
= 2x + (dy/dx)(2 - 2y) + 5
(dy/dx)(1 + y- 2) = 2x + 5
dy/dx = (2x+5)/(1 + y – 2)

thats most of it

2007-03-25 22:58:41 · answer #4 · answered by Anonymous · 1 0

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