for the curve y^3+(1/2)xy=1 find a formula for dy/dx in terms of x and y and find the equation of the tangent line to the curve at (-7,2)
To find the formula for dy/dx in terms of x and y, we can use implicit differentiation. Implicit differentiation involves differentiating both sides of the equation with respect to x, treating y as a function of x.
Given the equation y^3 + (1/2)xy = 1, let's differentiate both sides with respect to x:
d/dx(y^3) + d/dx((1/2)xy) = d/dx(1)
To differentiate y^3, we can use the chain rule since y is a function of x. We get:
3y^2 * dy/dx + (1/2)(x)(dy/dx) + (1/2)y = 0
Next, we can rearrange the terms to solve for dy/dx:
3y^2 * dy/dx + (1/2)x * dy/dx = -(1/2)y
Now, factoring out dy/dx:
dy/dx (3y^2 + (1/2)x) = -(1/2)y
Finally, dividing both sides by (3y^2 + (1/2)x):
dy/dx = -(1/2)y / (3y^2 + (1/2)x)
Now, to find the equation of the tangent line to the curve at (-7,2), we'll substitute x = -7 and y = 2 into the derived equation of dy/dx:
dy/dx = -(1/2)(2) / [3(2)^2 + (1/2)(-7)]
dy/dx = -1 / [12 + (-7/2)]
dy/dx = -1 / [12 - 7/2]
dy/dx = -1 / (24/2 - 7/2)
dy/dx = -1 / (17/2)
dy/dx = -2/17
So, the slope of the tangent line is -2/17 at the point (-7,2).
Using the point-slope form of a line, we can find the equation of the tangent line:
y - y1 = m(x - x1)
where (x1, y1) is the point on the curve (-7,2) and m is the slope (-2/17):
y - 2 = (-2/17)(x - (-7))
y - 2 = (-2/17)(x + 7)
Multiplying through by 17 to eliminate the fraction, we get:
17y - 34 = -2x - 14
Rearranging, we get the equation of the tangent line:
2x + 17y = 20