Find the equation of the normal lines to the curves
y=2x^3-x^2+3x+2 at the point x=1
x^2y+y^3x+3x-13=0 at the point (1,2
To find the equation of the normal line to a curve at a given point, we need to find the slope of the tangent line to the curve at that point, and then find the negative reciprocal of that slope to get the slope of the normal line.
1) Curve: y = 2x^3 - x^2 + 3x + 2
To find the slope of the tangent line, we take the derivative of the equation:
y' = 6x^2 - 2x + 3
Evaluating y' at x = 1, we get:
y'(1) = 6(1)^2 - 2(1) + 3 = 6 - 2 + 3 = 7
So, the slope of the tangent line at x = 1 is 7.
The negative reciprocal of 7 is -1/7, so the slope of the normal line is -1/7.
Now we use the point-slope form of a line (y - y1) = m(x - x1), where (x1, y1) is the point of tangency (1, f(1)).
y - 2 = (-1/7)(x - 1)
Multiplying through by 7 to eliminate fractions, we get:
7y - 14 = -x + 1
Rearranging the equation, we obtain the equation of the normal line:
x + 7y = 15
2) Curve: x^2y + y^3x + 3x - 13 = 0
To find the slope of the tangent line, we differentiate implicitly with respect to x:
2xy + x^2(dy/dx) + y^3 + 3 - x(dy/dx) = 0
Simplifying and solving for dy/dx, we find:
(dy/dx) = (y^3 - 2xy - 3) / (x^2 - x)
Substituting the point (1, 2) into the equation, we get:
(dy/dx)(1) = (2^3 - 2(1)(2) - 3) / (1^2 - 1) = (8 - 4 - 3) / 0 = -7/0 (undefined)
In this case, the derivative does not exist at x = 1, so we need to find the slope of the normal line in a different way.
To find the slope of the normal line at a given point on the curve, we can find the slope of the tangent line to the curve at that point, and then find the negative reciprocal of that slope to get the slope of the normal line.
Since the derivative is undefined at x = 1 for this curve, we cannot find the slope of the tangent line. Therefore, we cannot determine the slope of the normal line at this point, and consequently, we cannot find the equation of the normal line.
To find the equation of the normal line to a curve at a given point, we need to find the derivative of the function at that point and then find the negative reciprocal of the slope of the tangent line.
1. For the curve y = 2x^3 - x^2 + 3x + 2 at the point x = 1:
a) Find the derivative of the function y with respect to x, denoted as dy/dx:
dy/dx = d/dx(2x^3 - x^2 + 3x + 2)
= 6x^2 - 2x + 3
b) Substitute x = 1 into the derivative to find the slope of the tangent line at x = 1:
slope = dy/dx (at x = 1)
= 6(1)^2 - 2(1) + 3
= 6 - 2 + 3
= 7
c) The slope of the normal line would be the negative reciprocal of the slope of the tangent line:
slope_normal = -1/slope
= -1/7
d) The equation of the normal line passing through point (1, y) would be:
y - y1 = m(x - x1)
where x1 = 1, y1 = f(x1), and m is the slope of the normal line.
Substitute the values and simplify to find the equation of the normal line:
y - y1 = -1/7(x - x1)
y - f(1) = -1/7(x - 1)
y - (2(1)^3 - (1)^2 + 3(1) + 2) = -1/7(x - 1)
y - (2 - 1 + 3 + 2) = -1/7(x - 1)
y - 6 = -1/7(x - 1)
y - 6 = -(1/7)x + 1/7
y = -(1/7)x + 1/7 + 6
y = -(1/7)x + 1/7 + 42/7
y = -(1/7)x + 43/7
Therefore, the equation of the normal line to the curve y = 2x^3 - x^2 + 3x + 2 at the point x = 1 is y = -(1/7)x + 43/7.
2. For the curve x^2y + y^3x + 3x - 13 = 0 at the point (1, 2):
a) Find the derivative of the function x^2y + y^3x + 3x - 13 with respect to x, denoted as dy/dx:
Differentiating implicitly, we get:
2xy + x^2(dy/dx) + y^3 + 3 = 0
b) Substitute x = 1 and y = 2 into the derivative to find the slope of the tangent line at (1, 2):
2(1)(2) + (1)^2(dy/dx) + (2)^3 + 3 = 0
4 + dy/dx + 8 + 3 = 0
dy/dx = -15
c) The slope of the normal line would be the negative reciprocal of the slope of the tangent line:
slope_normal = -1/slope
= -1/(-15)
= 1/15
d) The equation of the normal line passing through point (x, y) = (1, 2) would be:
y - y1 = m(x - x1)
where x1 = 1, y1 = 2, and m is the slope of the normal line.
Substitute the values and simplify to find the equation of the normal line:
y - y1 = (1/15)(x - x1)
y - 2 = (1/15)(x - 1)
y - 2 = (1/15)x - 1/15
y = (1/15)x - 1/15 + 2
y = (1/15)x - 1/15 + 30/15
y = (1/15)x + 29/15
Therefore, the equation of the normal line to the curve x^2y + y^3x + 3x - 13 = 0 at the point (1, 2) is y = (1/15)x + 29/15.