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)

1 answer

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.