what is the slope of the line tangent to the curve y^3+x^2y^2-3x^3=9 at the point (1,2)?
To find the slope of the line tangent to the curve at a particular point, we need to differentiate the equation of the curve with respect to x and then substitute the x-coordinate of the given point into the derivative. Here's how you can do it step-by-step:
Step 1: Start with the given equation of the curve: y^3 + x^2y^2 - 3x^3 = 9.
Step 2: Differentiate both sides of the equation with respect to x using the product rule and chain rule as needed. Let's do it step-by-step:
- Differentiate y^3 with respect to x: d/dx(y^3) = 3y^2(dy/dx).
- Differentiate x^2y^2 with respect to x: d/dx(x^2y^2) = 2xy^2 + x^2 * d/dx(y^2) = 2xy^2 + x^2(2y * dy/dx).
- Differentiate -3x^3 with respect to x: d/dx(-3x^3) = -9x^2.
- Therefore, the derivative of the left-hand side of the equation is: 3y^2(dy/dx) + 2xy^2 + 2x^2y(dy/dx) - 9x^2.
Step 3: Set the derivative equal to zero, since we want to find the slope of the tangent line where the curve is tangent to the point.
3y^2(dy/dx) + 2xy^2 + 2x^2y(dy/dx) - 9x^2 = 0.
Step 4: Substitute the coordinates (1, 2) into the equation to solve for dy/dx.
3(2)^2(dy/dx) + 2(1)(2)^2 + 2(1)^2(2)(dy/dx) - 9(1)^2 = 0.
12(dy/dx) + 8 + 4(dy/dx) - 9 = 0.
Step 5: Simplify and solve for dy/dx.
12(dy/dx) + 4(dy/dx) = 9 - 8.
16(dy/dx) = 1.
dy/dx = 1/16.
Therefore, the slope of the line tangent to the curve y^3 + x^2y^2 - 3x^3 = 9 at the point (1, 2) is 1/16.
To find the slope of the line tangent to a curve at a given point, you need to find the derivative of the equation representing the curve, and then substitute the coordinates of the point into the derivative function.
In this case, the equation of the curve is y^3 + x^2y^2 - 3x^3 = 9. To find the derivative, we'll need to use implicit differentiation, treating x as the independent variable and y as the dependent variable.
Differentiating both sides of the equation with respect to x, we get:
3y^2(dy/dx) + 2xy^2(dy/dx) + 2x^2yy' - 9x^2 = 0
To find dy/dx at the point (1, 2), we substitute x = 1 and y = 2 into the equation:
3(2^2)(dy/dx) + 2(1)(2^2)(dy/dx) + 2(1)^2(2)(y') - 9(1)^2 = 0
12(dy/dx) + 8(dy/dx) + 4y' - 9 = 0
20(dy/dx) + 4y' = 9
Now, we need to isolate dy/dx (the slope of the tangent line):
dy/dx = (9 - 4y') / 20
To find the value of y' (the derivative of y with respect to x) at the point (1,2), substitute x = 1 and y = 2 into the equation of the curve:
y^3 + x^2y^2 - 3x^3 = 9
(2)^3 + (1^2)(2^2) - 3(1)^3 = 9
8 + 4 - 3 = 9
9 = 9
Therefore, the value of y' at the point (1,2) is 0.
Substituting y' = 0 into the equation dy/dx = (9 - 4y') / 20, we get:
dy/dx = (9 - 4(0)) / 20
dy/dx = 9/20
Therefore, the slope of the line tangent to the curve y^3 + x^2y^2 - 3x^3 = 9 at the point (1,2) is 9/20.
Differentiating implicitly you get
3y^2y' + 2xy^2 + 2x^2yy' - 9x^2 = 0.
Solve for y', and substitute 1 for x and 2 for y to get the slope.