Math Problem Statement

Find the slope of the tangent line drawn to the graph of -3x^2 + 4xy + 4y^3 = -99 at the point (-3, -3). Enter exact values.

Solution

The problem is asking to find the slope of the tangent line to the implicitly defined curve at a specific point. The equation given is:

3x2+4xy+4y3=99-3x^2 + 4xy + 4y^3 = -99

at the point (3,3)(-3, -3).

Steps to solve:

  1. Implicit Differentiation: To find the slope of the tangent line, we need to implicitly differentiate the given equation with respect to xx, treating yy as a function of xx, i.e., y=y(x)y = y(x).

ddx(3x2+4xy+4y3)=ddx(99)\frac{d}{dx} \left(-3x^2 + 4xy + 4y^3 \right) = \frac{d}{dx}(-99)

Differentiating each term:

  • For 3x2-3x^2, we get: ddx(3x2)=6x\frac{d}{dx}(-3x^2) = -6x

  • For 4xy4xy, we use the product rule: ddx(4xy)=4(ddx(x)y+xdydx)=4(y+xdydx)\frac{d}{dx}(4xy) = 4 \left( \frac{d}{dx}(x) \cdot y + x \cdot \frac{dy}{dx} \right) = 4(y + x \frac{dy}{dx})

  • For 4y34y^3, we use the chain rule: ddx(4y3)=43y2dydx=12y2dydx\frac{d}{dx}(4y^3) = 4 \cdot 3y^2 \cdot \frac{dy}{dx} = 12y^2 \frac{dy}{dx}

Now, the derivative of the constant 99-99 is 00, so putting it all together:

6x+4(y+xdydx)+12y2dydx=0-6x + 4(y + x \frac{dy}{dx}) + 12y^2 \frac{dy}{dx} = 0

  1. Solving for dydx\frac{dy}{dx}: Now, solve for dydx\frac{dy}{dx} (which represents the slope of the tangent line):

6x+4y+4xdydx+12y2dydx=0-6x + 4y + 4x \frac{dy}{dx} + 12y^2 \frac{dy}{dx} = 0

Isolate dydx\frac{dy}{dx}:

4xdydx+12y2dydx=6x4y4x \frac{dy}{dx} + 12y^2 \frac{dy}{dx} = 6x - 4y

Factor out dydx\frac{dy}{dx}:

dydx(4x+12y2)=6x4y\frac{dy}{dx} (4x + 12y^2) = 6x - 4y

Solve for dydx\frac{dy}{dx}:

dydx=6x4y4x+12y2\frac{dy}{dx} = \frac{6x - 4y}{4x + 12y^2}

  1. Substitute the point (3,3)(-3, -3): Now, substitute x=3x = -3 and y=3y = -3 into the equation:

dydx=6(3)4(3)4(3)+12(3)2\frac{dy}{dx} = \frac{6(-3) - 4(-3)}{4(-3) + 12(-3)^2}

Simplify the numerator:

6(3)4(3)=18+12=66(-3) - 4(-3) = -18 + 12 = -6

Simplify the denominator:

4(3)+12(3)2=12+12(9)=12+108=964(-3) + 12(-3)^2 = -12 + 12(9) = -12 + 108 = 96

Thus, the slope is:

dydx=696=116\frac{dy}{dx} = \frac{-6}{96} = -\frac{1}{16}

Final Answer:

The slope of the tangent line at the point (3,3)(-3, -3) is:

m=116m = -\frac{1}{16}

Would you like further clarification on any part of the solution?

Here are 5 related questions for further exploration:

  1. How do you apply implicit differentiation to other types of functions?
  2. What is the product rule, and how is it used in implicit differentiation?
  3. How do you determine if a function is implicitly defined or explicitly defined?
  4. How can you find the equation of the tangent line, given the slope?
  5. What happens when you differentiate higher-order terms like y3y^3?

Tip: When using implicit differentiation, always remember to apply the chain rule to any yy terms, as they are functions of xx.

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Math Problem Analysis

Mathematical Concepts

Calculus
Implicit Differentiation
Tangent Line

Formulas

Implicit Differentiation: d/dx[f(x, y)] = df/dx + (df/dy) * (dy/dx)
Product Rule: d(uv)/dx = u(dv/dx) + v(du/dx)
Chain Rule: d[f(g(x))]/dx = f'(g(x)) * g'(x)

Theorems

-

Suitable Grade Level

Grades 11-12