Find the equation of the tangent and the normal lines to the curves x^2y+y^3x+3x-13=0 at the point (1,2)
To find the equation of the tangent line, we first need to find the derivative of the given curve. The equation of the curve is given by:
x^2y + y^3x + 3x - 13 = 0
Taking the derivative of both sides with respect to x:
2xy + x^2(dy/dx) + 3 + 3y^2x(dy/dx) = 0
Rearranging the terms:
(dy/dx)(x^2 + 3y^2x) = -2xy - 3
(dy/dx) = (-2xy - 3)/(x^2 + 3y^2x)
Now, to find the slope of the tangent line at the point (1,2), substitute x = 1 and y = 2 into the derivative:
(dy/dx) = (-2(1)(2) - 3)/(1^2 + 3(2)^2(1))
(dy/dx) = (-4 - 3)/(1 + 12)
(dy/dx) = -7/13
So the slope of the tangent line is -7/13.
To find the equation of the tangent line at the point (1,2), we can use the point-slope form:
y - y1 = m(x - x1)
where (x1, y1) = (1, 2) and m = -7/13:
y - 2 = (-7/13)(x - 1)
Multiply both sides by 13 to eliminate the fraction:
13y - 26 = -7(x - 1)
13y - 26 = -7x + 7
Rearrange the equation to get it in the standard form:
7x + 13y = 33
So the equation of the tangent line to the curve at the point (1,2) is 7x + 13y = 33.
To find the equation of the normal line, we need to find the negative reciprocal of the slope of the tangent line. The negative reciprocal of -7/13 is 13/7.
Using the same point-slope form as before, with m = 13/7:
y - 2 = (13/7)(x - 1)
Multiply both sides by 7 to eliminate the fraction:
7y - 14 = 13(x - 1)
7y - 14 = 13x - 13
Rearrange the equation to get it in the standard form:
13x - 7y = 1
So the equation of the normal line to the curve at the point (1,2) is 13x - 7y = 1.
To find the equation of the tangent line at the point (1,2), we need to find the slope of the curve at that point.
Given the equation: x^2y + y^3x + 3x - 13 = 0
We can differentiate implicitly with respect to x:
2xy + x^2(dy/dx) + 3 + 3y^2(dy/dx) = 0
Rearranging the equation:
(x^2 + 3y^2)(dy/dx) = -2xy - 3
To find dy/dx at the point (1,2), substitute x = 1 and y = 2 into the equation:
(1^2 + 3(2)^2)(dy/dx) = -2(1)(2) - 3
(1 + 12)(dy/dx) = -4 - 3
13(dy/dx) = -7
dy/dx = -7/13
The slope of the tangent line is -7/13.
The equation of a line in point-slope form is given by:
y - y1 = m(x - x1)
Using the point (1,2) and the slope m = -7/13:
y - 2 = (-7/13)(x - 1)
Expanding the equation:
y - 2 = (-7/13)x + 7/13
y = (-7/13)x + 7/13 + 2
y = (-7/13)x + 7/13 + 26/13
y = (-7/13)x + 33/13
Therefore, the equation of the tangent line to the curve at the point (1,2) is y = (-7/13)x + 33/13.
To find the equation of the normal line, we use the fact that the tangent and the normal lines are perpendicular to each other.
The product of the slopes of perpendicular lines is -1.
Therefore, the slope of the normal line is the negative reciprocal of the slope of the tangent line, which is 13/7.
Using the point (1,2) and the slope m = 13/7:
y - 2 = (13/7)(x - 1)
Expanding the equation:
y - 2 = (13/7)x - 13/7
y = (13/7)x - 13/7 + 14/7
y = (13/7)x + 1/7
Therefore, the equation of the normal line to the curve at the point (1,2) is y = (13/7)x + 1/7.