Consider The Equation Xy^2 + X^2 Y = 6a

Consider The Equation Xy2 X2y 6a

Consider the equation xy² − x²y = 6 and perform the following tasks: (a) Use implicit differentiation to determine dy/dx. (b) Calculate the instantaneous rate of change dy/dx at the point (2, −1). (c) Find the equation of the line tangent to the graph at the point (2, −1).

Paper For Above instruction

Understanding implicit differentiation and its applications is crucial in calculus, especially when dealing with equations where y is not explicitly isolated. The equation xy² − x²y = 6 presents such a case, requiring differentiation with respect to x while treating y as a function of x. This paper will systematically address each part of the task, providing the methodological steps and mathematical reasoning involved.

(a) Implicit Differentiation of the Equation xy² − x²y = 6

The primary goal here is to differentiate both sides of the equation with respect to x, applying the product rule and chain rule where necessary. Starting with the original equation:

xy² − x²y = 6

Differentiate term-by-term:

For the first term xy², applying the product rule, we get:

d/dx [xy²] = (dy/dx)x + y² d/dx[x] = y² + x 2y dy/dx

Similarly, for the second term −x²y:

d/dx [−x²y] = −(2x y + x² dy/dx)

Putting this together, the differentiated form becomes:

y² + 2xy dy/dx − (2xy + x² dy/dx) = 0

Expanding and grouping like terms:

y² + 2xy dy/dx − 2xy − x² dy/dx = 0

Group the dy/dx terms and constant terms:

(2xy dy/dx − x² dy/dx) + (y² − 2xy) = 0

Factor dy/dx out of the terms:

dy/dx (2xy − x²) + (y² − 2xy) = 0

Solve for dy/dx:

dy/dx = [2xy − y²] / [2xy − x²]

This expression gives the derivative dy/dx in terms of x and y.

(b) Instantaneous Rate of Change at the Point (2, −1)

Substituting x=2 and y=−1 into the expression for dy/dx:

dy/dx = [2 2 (−1) − (−1)²] / [2 2 (−1) − (2)²]

Calculate numerator:

2 2 (−1) = −4; (−1)² = 1; Numerator: −4 − 1 = −5

Calculate denominator:

2 2 (−1) = −4; (2)² = 4; Denominator: −4 − 4 = −8

Thus:

dy/dx = (−5) / (−8) = 5/8

The instantaneous rate of change dy/dx at the point (2,−1) is 5/8.

(c) Equation of the Tangent Line at (2, −1)

The tangent line at a point (x₀, y₀) has a slope m = dy/dx evaluated at that point. Using the previous part, m = 5/8.

Applying the point-slope form of a line:

y − y₀ = m (x − x₀)

Substituting x₀=2, y₀=−1, and m=5/8:

y + 1 = (5/8)(x − 2)

Expressed explicitly:

y = (5/8)(x − 2) − 1

Therefore, the equation of the tangent line is y = (5/8)(x − 2) − 1.

This comprehensive analysis demonstrates the application of implicit differentiation techniques to find derivatives, evaluate the rate of change at a specific point, and determine the tangent line equation, integral concepts in calculus analysis.

References

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