Math Problem Statement

By implicit differentiation, find the equation of the tangent line to the curve x^2 + 2xy^2 - y = 11 at the point (1, -2).

Solution

The problem asks for the equation of the tangent line to the curve x2+2xy2y=11x^2 + 2xy^2 - y = 11 at the point (1,2)(1, -2), using implicit differentiation.

We need to implicitly differentiate both sides of the given equation with respect to xx and then solve for the slope at the point (1,2)(1, -2).

Step 1: Differentiate both sides of the equation

We start with: x2+2xy2y=11x^2 + 2xy^2 - y = 11 Differentiate each term with respect to xx, using the product rule where necessary:

  • ddx(x2)=2x\frac{d}{dx} (x^2) = 2x
  • For 2xy22xy^2, apply the product rule: ddx(2xy2)=2(y2+x2ydydx)\frac{d}{dx}(2xy^2) = 2 \left( y^2 + x \cdot 2y \frac{dy}{dx} \right) This simplifies to 2y2+4xydydx2y^2 + 4xy \frac{dy}{dx}.
  • ddx(y)=dydx\frac{d}{dx} (-y) = -\frac{dy}{dx}.

So the differentiated equation is: 2x+(2y2+4xydydx)dydx=02x + (2y^2 + 4xy \frac{dy}{dx}) - \frac{dy}{dx} = 0

Step 2: Substitute the point (1,2)(1, -2)

Substitute x=1x = 1 and y=2y = -2 into the equation to find dydx\frac{dy}{dx}:

  • For 2x2x, we get 2(1)=22(1) = 2,
  • For 2y22y^2, we get 2(2)2=82(-2)^2 = 8,
  • For 4xydydx4xy \frac{dy}{dx}, we get 4(1)(2)dydx=8dydx4(1)(-2) \frac{dy}{dx} = -8 \frac{dy}{dx},
  • For dydx- \frac{dy}{dx}, we get dydx- \frac{dy}{dx}.

Substituting into the differentiated equation gives: 2+88dydxdydx=02 + 8 - 8 \frac{dy}{dx} - \frac{dy}{dx} = 0 Simplify: 109dydx=010 - 9 \frac{dy}{dx} = 0 Solve for dydx\frac{dy}{dx}: 9dydx=10dydx=1099 \frac{dy}{dx} = 10 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{10}{9}

Step 3: Equation of the tangent line

The slope of the tangent line is 109\frac{10}{9}, and the point on the curve is (1,2)(1, -2). Using the point-slope form of the equation of a line:

y(2)=109(x1)y - (-2) = \frac{10}{9}(x - 1) Simplify: y+2=109x109y + 2 = \frac{10}{9}x - \frac{10}{9} y=109x1092y = \frac{10}{9}x - \frac{10}{9} - 2 y=109x109189y = \frac{10}{9}x - \frac{10}{9} - \frac{18}{9} y=109x289y = \frac{10}{9}x - \frac{28}{9}

Thus, the equation of the tangent line is: y=109x289y = \frac{10}{9}x - \frac{28}{9}

Let me know if you'd like to go over any steps or explore related concepts!

Here are some follow-up questions:

  1. What is implicit differentiation, and why is it used here?
  2. How is the product rule applied when differentiating terms like 2xy22xy^2?
  3. Why do we substitute the point (1,2)(1, -2) after differentiating?
  4. How do we simplify expressions involving fractions, like in the final equation of the tangent line?
  5. What are other methods to find the equation of a tangent line, and when do they apply?

Tip: When differentiating implicitly, remember to apply the chain rule whenever you differentiate terms involving yy, since yy is a function of xx.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Tangent Line
Calculus

Formulas

Implicit Differentiation
Product Rule: d(uv)/dx = u'(v) + u(v')
Equation of a Line: y - y1 = m(x - x1)

Theorems

Chain Rule: d(f(g(x)))/dx = f'(g(x))g'(x)
Implicit Differentiation Theorem

Suitable Grade Level

Grades 11-12 or Introductory College Calculus