Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x^2/3 +y^2/3 =4
(-3sqrt(3),1)
(astroid)
If it is an astroid, then it should be
x^(2/3) +y^(2/3) =4
the way you typed it would result in a circle
http://www.wolframalpha.com/input/?i=x%5E2%2F3+%2By%5E2%2F3+%3D4
(2/3)x^(-1/3) + (2/3)y^(-1/3) dy/dx = 0
x^(-1/3) + y^(-1/3) dy/dx = 0
dy/dx = -(x^(-1/3))/y^(-1/3)
= -(x/y)^(-1/3)
= -(y/x)^(1/3)
at the point (-3√3 , 1)
dy/dx = -(2/(-3√3))^(1/3)
equation of tangent:
y-1 = -(2/(-3√3))^(1/3)(x + 3√3)
simplify if you have to
To find the equation of the tangent line to the curve at the given point (-3√3, 1), we will need to use implicit differentiation.
Step 1: Differentiate both sides of the equation with respect to x.
Differentiating x^(2/3) + y^(2/3) = 4 with respect to x, we get:
(2/3)x^(-1/3) + (2/3)y^(-1/3)(dy/dx) = 0
Simplifying, we have:
(2/3)(x^(-1/3)) + (2/3)(y^(-1/3))(dy/dx) = 0
Step 2: Solve for dy/dx.
Rearranging the equation, we get:
(2/3)(y^(-1/3))(dy/dx) = -(2/3)(x^(-1/3))
Now divide both sides by (2/3)(y^(-1/3)):
dy/dx = -(2/3)(x^(-1/3))/(2/3)(y^(-1/3))
Simplifying further:
dy/dx = - x^(-1/3)/y^(-1/3)
Step 3: Substituting the given point into the derivative.
Substituting x = -3√3 and y = 1 into the derivative, we have:
dy/dx = - (-3√3)^(-1/3)/(1)^(-1/3)
Simplifying this expression:
dy/dx = - (-3√3)^(-1/3)
Step 4: Evaluate the derivative.
Now we need to evaluate the derivative:
dy/dx = - (-3√3)^(-1/3)
Using the negative exponent rule:
dy/dx = - 1/(-3√3)^(1/3)
Simplifying further:
dy/dx = - 1/(∛(-3√3))
Step 5: Find the equation of the tangent line.
Since we have the point (-3√3, 1) and the slope dy/dx = - 1/(∛(-3√3)), we can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Here, x1 = -3√3, y1 = 1, and m = - 1/(∛(-3√3)).
Plugging the values into the equation:
y - 1 = - 1/(∛(-3√3))(x - (-3√3))
Simplifying:
y - 1 = - 1/(∛(-3√3))(x + 3√3)
This is the equation of the tangent line to the curve x^(2/3) + y^(2/3) = 4 at the point (-3√3, 1).
To find the equation of the tangent line to the curve at a given point, we need to use implicit differentiation.
Let's start by differentiating the equation of the curve with respect to x. Remember that we treat y as a function of x, so we use the chain rule for differentiating y.
Differentiating both sides of the equation x^(2/3) + y^(2/3) = 4 with respect to x, we get:
(2/3)x^(-1/3) + (2/3)y^(-1/3)*(dy/dx) = 0
Now, let's rearrange the equation to solve for dy/dx, which represents the derivative of y with respect to x:
(2/3)y^(-1/3)*(dy/dx) = -(2/3)x^(-1/3)
Multiplying both sides by y^(1/3) and dividing by x^(1/3), we get:
dy/dx = -(x^(1/3))/(y^(1/3))
Now that we have the derivative of y with respect to x, we can find the slope of the tangent line at the given point (-3√3, 1) by substituting these values into the expression:
m = dy/dx = -((-3√3)^(1/3))/(1^(1/3)) = -((-3)^(1/3)*√3^(1/3))/1 = -(√3)/3
The slope of the tangent line at the given point is -(√3)/3.
To find the equation of the tangent line, we can use the point-slope form.
Using the point (-3√3, 1) and the slope -(√3)/3, we have:
y - y1 = m(x - x1)
y - 1 = -(√3)/3(x - (-3√3))
Simplifying the equation:
y - 1 = -(√3)/3(x + 3√3)
y - 1 = -(√3/3)x - √3
Finally, rearranging the equation to the slope-intercept form, we have:
y = -(√3/3)x + √3 - 1
Therefore, the equation of the tangent line to the curve at the point (-3√3, 1) is y = -(√3/3)x + √3 - 1.