Find the slope of the tangent to x^2 +y^3 =5 at the point where x = 2
taking the derivative, we have
2x + 3y^2 y' = 0
y' = -(2x)/(3y^2)
so, now we need to find y when x=2:
4+y^3 = 5
y = 1
So, at (2,1), the tangent has slope -4/3
To find the slope of the tangent to the curve at a given point, we need to differentiate the equation with respect to x and then substitute the x-coordinate of the given point into the derivative. Let's differentiate the equation x^2 + y^3 = 5 with respect to x using implicit differentiation:
d/dx(x^2 + y^3) = d/dx(5)
2x + 3y^2 * dy/dx = 0
Now, substitute x = 2 into the equation:
2(2) + 3y^2 * dy/dx = 0
4 + 3y^2 * dy/dx = 0
Now, we need to solve for dy/dx. Rearranging the equation gives us:
3y^2 * dy/dx = -4
Divide both sides by 3y^2:
dy/dx = -4 / (3y^2)
To find the slope of the tangent at the point where x = 2, we need to find the corresponding y-coordinate. Plugging x = 2 into the original equation:
2^2 + y^3 = 5
4 + y^3 = 5
y^3 = 1
Taking the cube root of both sides, we get y = 1. So, at the point where x = 2, y = 1. Substituting these values into the derived equation for dy/dx:
dy/dx = -4 / (3(1)^2)
dy/dx = -4 / 3
Therefore, the slope of the tangent line to the curve x^2 + y^3 = 5 at the point (2, 1) is -4/3.
To find the slope of the tangent to the curve at a given point, we need to first find the derivative of the equation and then substitute the x-coordinate of the point into the derivative.
Given the equation x^2 + y^3 = 5, we'll find the derivative by implicitly differentiating with respect to x. Let's do the differentiation step by step:
Step 1: Differentiate both sides of the equation with respect to x:
d/dx (x^2 + y^3) = d/dx (5)
Step 2: Apply the power rule for differentiation:
2x + 3y^2 * dy/dx = 0
Step 3: Now we can solve for dy/dx, which represents the derivative of y with respect to x:
dy/dx = -2x / (3y^2)
Step 4: Substitute x = 2 into the derivative equation to find the slope of the tangent at the point where x = 2:
dy/dx = -2(2) / (3y^2)
= -4 / (3y^2)
So, the slope of the tangent to the curve x^2 + y^3 = 5 at the point where x = 2 is given by -4 / (3y^2).