Find the equation of the tangent and 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 tangent line to the curve y=2x^3-x^2+3x+2 at the point x=1, we first need to find the slope of the tangent line.
To find the slope, we can take the derivative of the function y=2x^3-x^2+3x+2:
dy/dx = 6x^2 - 2x + 3.
Then evaluate the derivative at x=1:
dy/dx = 6(1)^2 - 2(1) + 3 = 6 - 2 + 3 = 7.
So the slope of the tangent line at x=1 is 7.
Next, we need to find the y-coordinate of the point on the curve at x=1:
y = 2(1)^3 - (1)^2 + 3(1) + 2 = 2 - 1 + 3 + 2 = 6.
So the coordinate of the point on the curve at x=1 is (1, 6).
Now we can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1),
where (x1, y1) is the point (1, 6) and m is the slope 7:
y - 6 = 7(x - 1).
Simplifying, we get:
y - 6 = 7x - 7.
The equation of the tangent line to the curve y=2x^3-x^2+3x+2 at the point x=1 is y = 7x - 1.
For the second question, to find the equation of the tangent line to the curve x^2y+y^3x+3x-13=0 at the point (1,2), we first need to find the slope of the tangent line.
Since the equation is not given in the form y = f(x), we need to implicitly differentiate using the chain rule:
d/dx (x^2y) + d/dx (y^3x) + d/dx (3x) - d/dx (13) = 0.
Using the product rule for each term, we get:
2xy + x^2(dy/dx) + 3 + y^3 + 3x(dy/dx) = 0.
Now we can solve for dy/dx by plugging in the values of x=1 and y=2:
2(1)(2) + (1)^2(dy/dx) + 3 + (2)^3 + 3(1)(dy/dx) = 0.
4 + dy/dx + 3 + 8 + 3(dy/dx) = 0.
4 + 3 + 8 + 7(dy/dx) = 0.
15 + 7(dy/dx) = 0.
7(dy/dx) = -15.
dy/dx = -15/7.
So the slope of the tangent line at the point (1,2) is -15/7.
Using the point-slope form of a line, we get:
y - y1 = m(x - x1),
where (x1, y1) is the point (1, 2) and m is the slope -15/7:
y - 2 = (-15/7)(x - 1).
Multiplying through by 7 to clear the fraction, we get:
7y - 14 = -15(x - 1).
Simplifying, we get:
7y - 14 = -15x + 15.
7y = -15x + 29.
The equation of the tangent line to the curve x^2y+y^3x+3x-13=0 at the point (1,2) is 7y = -15x + 29.
To find the equation of the tangent and normal lines to a curve at a given point, we need to find the slope of the curve at that point.
For the curve y = 2x^3 - x^2 + 3x + 2, we can find the slope by taking the derivative of the equation with respect to x:
dy/dx = 6x^2 - 2x + 3
To find the slope at x = 1, we substitute x = 1 into the derivative:
dy/dx = 6(1)^2 - 2(1) + 3
dy/dx = 6 - 2 + 3
dy/dx = 7
The slope of the curve at x = 1 is 7.
Now, we have the slope, and we can find the equation of the tangent line using the point-slope form:
y - y1 = m(x - x1)
Substituting the values x1 = 1, y1 = 2, and m = 7:
y - 2 = 7(x - 1)
Simplifying the equation:
y - 2 = 7x - 7
y = 7x - 5
Therefore, the equation of the tangent line to the curve y = 2x^3 - x^2 + 3x + 2 at the point x = 1 is y = 7x - 5.
To find the equation of the normal line, we know that the slope of the normal line is the negative reciprocal of the slope of the tangent line. So the slope of the normal line is -1/7.
Using the point-slope form again, we can find the equation of the normal line:
y - 2 = (-1/7)(x - 1)
Simplifying the equation:
7y - 14 = -x + 1
x + 7y = 15
Therefore, the equation of the normal line to the curve y = 2x^3 - x^2 + 3x + 2 at the point x = 1 is x + 7y = 15.
Now, let's move on to the second curve.
For the curve x^2y + y^3x + 3x - 13 = 0, we need to find the slope at the point (1, 2).
To find the slope, we can take the derivative of the equation with respect to x:
d/dx (x^2y + y^3x + 3x - 13) = 0
Differentiating term by term using the product rule, we get:
2xy + x^2(dy/dx) + 3 = 0
Substituting x = 1 and y = 2 into the equation, we can solve for the derivative dy/dx:
2(1)(2) + (1)^2(dy/dx) + 3 = 0
4 + dy/dx + 3 = 0
dy/dx = -7
The slope of the curve at (1, 2) is -7.
Using the point-slope form, we can find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting x1 = 1, y1 = 2, and m = -7:
y - 2 = -7(x - 1)
Simplifying the equation:
y - 2 = -7x + 7
y = -7x + 9
Therefore, the equation of the tangent line to the curve x^2y + y^3x + 3x - 13 = 0 at the point (1, 2) is y = -7x + 9.
To find the equation of the normal line, we know that the slope of the normal line is the negative reciprocal of the slope of the tangent line. So the slope of the normal line is 1/7.
Using the point-slope form again, we can find the equation of the normal line:
y - 2 = (1/7)(x - 1)
Simplifying the equation:
7y - 14 = x - 1
x - 7y = -13
Therefore, the equation of the normal line to the curve x^2y + y^3x + 3x - 13 = 0 at the point (1, 2) is x - 7y = -13.