Determine the equation of the tangent and normal to the graph x(xy+y^3)-13=0 at the point (1,2)
To find the equation of the tangent and normal at a given point on a curve, we need to find the slope of the tangent and normal lines.
Given the equation of the curve, x(xy+y^3)-13=0, we can find the slope of the tangent line by taking the derivative of the equation with respect to x.
Differentiating both sides of the equation, we get:
d/dx [x(xy+y^3)-13] = d/dx [0]
xy+y^3 + x(dy/dx) = 0
To find the slope of the tangent line at the point (1,2), we substitute x=1 and y=2 into the equation above and solve for dy/dx.
(1)(2)(2)+(2)^3 + 1(dy/dx) = 0
4 + 8 + dy/dx = 0
dy/dx = -12
Therefore, the slope of the tangent line at the point (1,2) is -12.
The equation of the tangent line can be written in the point-slope form:
y - y1 = m(x - x1)
where (x1, y1) is the point of tangency and m is the slope.
Substituting (1,2) and -12 into the equation, we get:
y - 2 = -12(x - 1)
Simplifying, we have:
y - 2 = -12x + 12
y = -12x + 14
Therefore, the equation of the tangent line to the graph at the point (1,2) is y = -12x + 14.
To find the equation of the normal line, we use the fact that the product of the slopes of perpendicular lines is equal to -1. Since the tangent line has a slope of -12, the normal line will have a slope of 1/12.
Using the point-slope form again, we get:
y - 2 = (1/12)(x - 1)
Simplifying, we have:
y - 2 = (1/12)x - 1/12
y = (1/12)x + 23/12
Therefore, the equation of the normal line to the graph at the point (1,2) is y = (1/12)x + 23/12.
first off, the point (1,2) is not on the graph!
Apologies for the mistake. Let's correct the solution.
To find the equation of the tangent and normal at a given point on the curve, we need to find the slope of the tangent and normal lines.
The equation of the curve is x(xy + y^3) - 13 = 0.
To find the slope of the tangent line at a point (x0, y0), we can use implicit differentiation. Taking the derivative of the equation with respect to x, we get:
d/dx [x(xy + y^3) - 13] = 0
Applying the product rule and chain rule, the derivative becomes:
y + xy + x(dy/dx)y + 3y^2 = 0
To find the slope at (1,2), we substitute x = 1 and y = 2 into the equation and solve for dy/dx:
2 + (1)(2) + (1)(dy/dx)(2) + 3(2)^2 = 0
2 + 2 + 2(dy/dx) + 12 = 0
2(dy/dx) = -16
dy/dx = -8/2 = -4
Therefore, the slope of the tangent line at the point (1,2) is -4.
The equation of the tangent line can be written in the point-slope form:
y - y1 = m(x - x1)
where (x1, y1) is the point of tangency and m is the slope.
Substituting (1,2) and -4 into the equation, we get:
y - 2 = -4(x - 1)
y - 2 = -4x + 4
y = -4x + 6
Therefore, the equation of the tangent line to the graph at the point (1,2) is y = -4x + 6.
To find the equation of the normal line, we use the fact that the product of the slopes of perpendicular lines is equal to -1. Since the tangent line has a slope of -4, the normal line will have a slope of 1/4.
Using the point-slope form again, we get:
y - 2 = (1/4)(x - 1)
Simplifying, we have:
y - 2 = (1/4)x - 1/4
y = (1/4)x + 7/4
Therefore, the equation of the normal line to the graph at the point (1,2) is y = (1/4)x + 7/4.
To find the equation of the tangent and normal to the graph at the point (1,2), we will first differentiate the given equation implicitly with respect to x.
Differentiating both sides of the equation:
d/dx [x(xy+y^3)-13] = d/dx [0]
By applying the product rule and chain rule, we have:
x * d/dx(xy+y^3) + (xy+y^3) * d/dx(x) = 0
Now let's find the derivatives:
d/dx(xy+y^3) = y + 3y^2 * dy/dx + x * dy/dx
d/dx(x) = 1
Substituting the derivatives back into the equation:
x * (y + 3y^2 * dy/dx + x * dy/dx) + (xy+y^3) * 1 = 0
Simplifying the equation, we have:
xy + 3xy^2 * dy/dx + x^2 * dy/dx + xy + y^3 = 0
Rearranging the terms, we get:
2xy + (3xy^2 + x^2) * dy/dx + y^3 = 0
Now, we can substitute the values of x and y from the given point (1,2):
2*1*2 + (3*1*(2^2) + 1^2) * dy/dx + 2^3 = 0
Simplifying further:
4 + (3*4 + 1) * dy/dx + 8 = 0
4 + 13 * dy/dx + 8 = 0
13 * dy/dx = -12
dy/dx = -12/13
Since the derivative dy/dx represents the slope of the tangent line, the slope of the tangent line at the point (1,2) is -12/13.
To find the equation of the tangent line at (1,2), we can use the point-slope form:
y - y1 = m(x - x1)
Substituting the values, we get:
y - 2 = (-12/13)(x - 1)
Simplifying the equation:
13y - 26 = -12x + 12
13y = -12x + 38
Therefore, the equation of the tangent line to the graph at the point (1,2) is 13y = -12x + 38.
To find the equation of the normal line, we will use the fact that the slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is 13/12.
Using the point-slope form again, substituting the values, we have:
y - 2 = (13/12)(x - 1)
Simplifying the equation:
12y - 24 = 13x - 13
13x - 12y = 11
Therefore, the equation of the normal line to the graph at the point (1,2) is 13x - 12y = 11.