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.
Determine the equation of the tangent and normal to the graph x(xy+y^3)-13=0 at the point (1,2)
3 answers
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 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.