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.
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)
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