Math Problem Statement

but now you're on board the spaceship #218811928, and the captain is asking to solve tricky "ZP" (Z Planet) problem (you know what it means when captain is "asking"... that's an order):

(!!! Do Not use decimals/calculator. Present fractions in an exact format, for example, 7910779107 , 2292422924)

Now, find the slope of the tangent line to the curve

3x2+4xy4y3=35-3x^2+4xy-4y^3=-35

at the point (3,2)

  

!!! The challenge is that the easiest differentiation rule on Earth, d\differentialDxxn=nxn1\frac{d}{\differentialD x}x^{n}=nx^{n-1} , on Z planet is "twisted" as the following:

 d\differentialDxxn=n2xn1\frac{d}{\differentialD x}x^{n}=n^2x^{n-1}  (ddxconst=0ddxconst=0 , and the other rules have not been changed).

Solution

We are tasked with finding the slope of the tangent line to the curve given by:

3x2+4xy4y3=35-3x^2 + 4xy - 4y^3 = -35

at the point (3,2)(3, 2), under a modified differentiation rule on Z Planet. The differentiation rule on Z Planet states:

ddxxn=n2xn1\frac{d}{dx}x^n = n^2x^{n-1}

This differs from the usual Earth rule where ddxxn=nxn1\frac{d}{dx}x^n = nx^{n-1}.

Step 1: Differentiate implicitly

We need to apply implicit differentiation to the given equation 3x2+4xy4y3=35-3x^2 + 4xy - 4y^3 = -35, using the Z Planet rule for differentiating terms involving xx.

Differentiating each term with respect to xx:

  1. First term: 3x2-3x^2

    • Using the Z Planet rule for differentiating x2x^2, we get: ddx(3x2)=322x21=12x\frac{d}{dx}(-3x^2) = -3 \cdot 2^2 x^{2-1} = -12x
  2. Second term: 4xy4xy

    • For the product 4xy4xy, we use the product rule: ddx(4xy)=4(ddx(x)y+xdydx)\frac{d}{dx}(4xy) = 4 \left( \frac{d}{dx}(x) \cdot y + x \cdot \frac{dy}{dx} \right) Here, ddx(x)=1\frac{d}{dx}(x) = 1, so this becomes: 4(y+xdydx)4(y + x \frac{dy}{dx})
  3. Third term: 4y3-4y^3

    • Since this term only involves yy, we differentiate it using the chain rule: ddx(4y3)=43y2dydx=12y2dydx\frac{d}{dx}(-4y^3) = -4 \cdot 3y^2 \cdot \frac{dy}{dx} = -12y^2 \cdot \frac{dy}{dx}
  4. Constant term: 35-35

    • The derivative of a constant is 0: ddx(35)=0\frac{d}{dx}(-35) = 0

Step 2: Combine the results

Now, we put everything together. The differentiated equation becomes:

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

Simplifying this:

12x+4y+4xdydx12y2dydx=0-12x + 4y + 4x \frac{dy}{dx} - 12y^2 \frac{dy}{dx} = 0

Step 3: Solve for dydx\frac{dy}{dx}

Rearrange the equation to solve for dydx\frac{dy}{dx}:

4xdydx12y2dydx=12x4y4x \frac{dy}{dx} - 12y^2 \frac{dy}{dx} = 12x - 4y

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

dydx(4x12y2)=12x4y\frac{dy}{dx} (4x - 12y^2) = 12x - 4y

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

dydx=12x4y4x12y2\frac{dy}{dx} = \frac{12x - 4y}{4x - 12y^2}

Step 4: Evaluate at the point (3,2)(3, 2)

Substitute x=3x = 3 and y=2y = 2 into the expression for dydx\frac{dy}{dx}:

dydx=12(3)4(2)4(3)12(22)=3681212(4)=281248=2836=79\frac{dy}{dx} = \frac{12(3) - 4(2)}{4(3) - 12(2^2)} = \frac{36 - 8}{12 - 12(4)} = \frac{28}{12 - 48} = \frac{28}{-36} = -\frac{7}{9}

Final Answer:

The slope of the tangent line to the curve at the point (3,2)(3, 2) is:

79\boxed{-\frac{7}{9}}


Would you like further details or have any questions on this?

Relative Questions:

  1. How do we handle product rules under different differentiation rules?
  2. Why is implicit differentiation useful in cases like this?
  3. What are the implications of modifying basic differentiation rules like ddxxn\frac{d}{dx} x^n?
  4. How would the solution change if we were given a different modified differentiation rule?
  5. What techniques are useful for solving implicit differentiation problems more generally?

Tip:

When applying modified differentiation rules, always focus on ensuring consistency in applying the rule throughout, especially in product and chain rule cases.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Tangent Line
Algebraic Curves

Formulas

Z Planet differentiation rule: $$ \frac{d}{dx} x^n = n^2 x^{n-1} $$
Product Rule
Chain Rule

Theorems

Implicit Differentiation

Suitable Grade Level

Grades 11-12 or College Level