What is the slope of the line tangent to the curve y3−xy2+x3=5 at the point (1,2) ?
To find the slope of the line tangent to a curve at a specific point, we can use the concept of derivatives
To find the slope of the line tangent to a curve at a specific point, we can use the concept of derivatives.
First, let’s differentiate the given equation to find the derivative of y with respect to x.
Differentiating y^3 − xy^2 + x^3 = 5 with respect to x, we get:
3y^2 * (dy/dx) – y^2 + 3x^2 = 0
Now, we need to find the value of dy/dx at the point (1,2).
Substitute x = 1 and y = 2 into the equation:
3(2^2) * (dy/dx) – 2^2 + 3(1^2) = 0
12(dy/dx) – 4 + 3 = 0
12(dy/dx) = 1
dy/dx = 1/12
So, the slope of the line tangent to the curve y^3 − xy^2 + x^3 = 5 at the point (1,2) is 1/12.
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