Determine the equation ofthetangent line in slopeintercept form at x=1 of y^3 +2x^2=1 at x=1. Please show work so I can follow the process. Thanks!
y^3+2x^2=1
y^2 dy/dx+4x=0
dy/dx=-4x/y^2
at x=1, y=-1, so
dy/dx=-4/1=-4
check my thinking.
Hmm. I get
y^3 + 2x^2 = 1
3y^2 y' + 4x = 0
y' = -4x/(3y^2)
y(1) = -1, so
y'(1) = -4/3
so, now you have a point (1,-1) and a slope: -4/3
y+1 = -4/3 (x-1)
I assume that massaging that to slope-intercept form presents no problems...
To find the equation of the tangent line to the curve at a given point, we can use the concept of derivatives. The derivative of a function measures the rate at which the function is changing at a particular point. In this case, we want to find the derivative of the function y^3 + 2x^2 = 1.
Let's start by explicitly expressing y in terms of x. Rewrite the equation as:
y^3 = 1 - 2x^2.
Now take the cube root of both sides to solve for y:
y = (1 - 2x^2)^(1/3).
To find the slope of the tangent line at x = 1, we need to find the derivative of the function with respect to x and evaluate it at x = 1.
Differentiating both sides with respect to x, we get:
dy/dx = d/dx[(1 - 2x^2)^(1/3)].
To simplify the differentiation, we can rewrite the equation using the Chain Rule:
dy/dx = (1/3)(1 - 2x^2)^(-2/3)(d/dx[1 - 2x^2]).
Now find the derivative of 1 - 2x^2:
dy/dx = (1/3)(1 - 2x^2)^(-2/3)(-4x).
Now evaluate dy/dx at x = 1:
dy/dx = (1/3)(1 - 2(1)^2)^(-2/3)(-4(1)).
dy/dx = (1/3)(1 - 2)^(-2/3)(-4).
Simplify further:
dy/dx = (1/3)(-1)^(-2/3)(-4).
dy/dx = (1/3)(4)(-1)^(-2/3).
Since (-1)^(-2/3) is a complex number, we can rationalize it by multiplying the numerator and denominator by (-1)^(2/3):
dy/dx = (1/3)(4)(-1)^(-2/3) * (-1)^(2/3) / (-1)^(2/3).
Simplifying further:
dy/dx = (1/3)(4)(1) / (-1)^(2/3).
dy/dx = 4/3 * (-1)^(2/3).
Therefore, the slope of the tangent line at x = 1 is 4/3 * (-1)^(2/3).
To find the equation of the tangent line in slope-intercept form, we need to use the point-slope form of a line equation:
y - y1 = m(x - x1),
where (x1, y1) is the point on the line and m is the slope. Since the point of tangency is (1, y), the equation becomes:
y - y1 = m(x - 1).
Now substitute the known values:
y - y = (4/3 * (-1)^(2/3))(x - 1).
Simplify:
0 = (4/3 * (-1)^(2/3))(x - 1).
The equation of the tangent line at x = 1 in slope-intercept form is 0 = (4/3 * (-1)^(2/3))(x - 1).