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The slope of the tangent line is equal to the derivative of the curve at the point of tangency.

Using implicit differentiation on the equation of the curve,
3x^2 + 3y^2 dy/dx = 2x dy/dx + 2y

The point of tangency is (1,1). So just plug in 1 for each x and y in the above equation to get the derivative at the point of tangency:

3(1) + 3(1^2) dy/dx = 2(1) dy/dx + 2(1)
3 + 3 dy/dx = 2 dy/dx + 2
1 + dy/dx = 0
dy/dx = -1

Now to get the equation of the line, use the point-slope form. We now know the slope is -1. We also know that ( 1,1) is a point on this line:

( y -1) = m ( x -1)
y - 1 = -1 (x-1)
y - 1 = -x + 1
y = -x + 2

2007-11-03 19:17:35 · answer #1 · answered by BB 2 · 1 0

what's the curve equation to come across the tangent line at a factor (25,5) on it? EDIT: [As in step with the added information, which i could desire to relook in common terms after approximately 9 hours of till now presentation] one million) Differentiating the given one, dy/dx = one million/(2?x) 2) At x = 25, dy/dx = one million/(2?25) = one million/10; this is the slope of the tangent line on the given factor; this is thru the geometrical definition of differentiation] 3) using Slope-factor variety, the equation of the tangent line at (25,5) is: y - 5 = (one million/10)(x - 25) increasing and simplifying, the equation is: x - 10y + 25 = 0

2016-10-03 07:28:34 · answer #2 · answered by Anonymous · 0 0

Differentiate with respect to x at (1,1),
3+3y' = 2+2y'
Solve for y',
y' = -1
The equation is,
y = -(x-1) + 1
Simiplify,
y = -x+2

2007-11-03 19:14:15 · answer #3 · answered by sahsjing 7 · 0 0

Greetings,

Differentiate with respect to x,

3x^2 + 3y^2dy/dx = 2xdy/dx + 2y

dy/dx = (3x^2-2y)/(2x - 3y^2)

evaluating at (1,1)

dy/dx = -1

Hence the equation of the tangent line is y - 1 = -(x -1)

or y = -x + 2

Regards

2007-11-03 19:13:50 · answer #4 · answered by ubiquitous_phi 7 · 0 0

3x^2+3y^2 y'=2y+2xy'
x=1
y=1
3+3y'=2+2y'
y' = 1

y-1=1(x-1)
y=x-1+1

The tangent equation is
y=x

2007-11-03 19:14:52 · answer #5 · answered by iyiogrenci 6 · 0 0

3x^2 + 3y^2y' = 2( xy' + y)
(3y^2 - x) y' = 2y - 3x^2
y' = (2y - 3X^2)/(3y^2 - x)
at (1,1)
y' = -1/2
y = mx + b
so, 1 = -1/2 + b
b = 1 + 1/2 = 3/2
the tangent line is
y = -x/2 + 3/2

2007-11-03 19:16:26 · answer #6 · answered by Anonymous · 0 0

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