Solve this question using my solutions---**Question:**
The curve C has equation $4y^3 + (x + y)^6 = 96(1 + x)$ where $y > 0$
(a) Find the y-coordinate of the point on C where $x = 0$.
(b) Find the value of $\frac{dy}{dx}$ at $x = 0$.
(c) Find the Maclaurin series expansion of y, in ascending powers of x up to and including the term in $x^2$, giving each coefficient in simplest form.
**Solution:**
(a) $x = 0 \Rightarrow 4y^3 + y^6 = 96 \Rightarrow (y^3 - 8)(y^3 + 12) = 0 \Rightarrow y^3 = 8$ only $\Rightarrow y = 2$
(b) $12y^2 \frac{dy}{dx} + 6(x + y)^5 (1 + \frac{dy}{dx}) = 96$
$x = 0, y = 2 \Rightarrow 12 \times 2^2 \times \frac{dy}{dx} + 6 \times 2^5 \times (1 + \frac{dy}{dx}) = 96 \Rightarrow \frac{dy}{dx} = -\frac{2}{5}$
(c) $24y(\frac{dy}{dx})^2 + 12y^2 \frac{d^2y}{dx^2} + 30(x + y)^4 (1 + \frac{dy}{dx})^2 + 6(x + y)^5 \frac{d^2y}{dx^2} = 0$
$24 \times 2 \times (-\frac{2}{5})^2 + 12 \times 2^2 \times \frac{d^2y}{dx^2} + 30 \times 2^4 \times (1 - \frac{2}{5})^2 + 6 \times 2^5 \times \frac{d^2y}{dx^2} = 0 \Rightarrow \frac{d^2y}{dx^2} = -\frac{94}{125}$
$y = 2 - \frac{2}{5}x - \frac{94}{125} \frac{x^2}{2!} = 2 - \frac{2}{5}x - \frac{47}{125}x^2$
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We have an implicit curve equation with y as a function of x. For part a, we substitute x equals 0 into the equation. This gives us 4y cubed plus y to the sixth equals 96. Factoring out y cubed, we get y cubed times 4 plus y cubed equals 96. This simplifies to y cubed equals 8, so y equals 2.
For part b, we need to find dy dx at x equals 0. We differentiate the original equation implicitly. The derivative of 4y cubed is 12y squared dy dx. For the term x plus y to the sixth, we use the chain rule to get 6 times x plus y to the fifth times 1 plus dy dx. The right side becomes 96. Substituting x equals 0 and y equals 2, we get 48 dy dx plus 192 times 1 plus dy dx equals 96. This simplifies to 240 dy dx equals negative 96, giving us dy dx equals negative 2 fifths.
For part c, we need the second derivative to construct the Maclaurin series. We differentiate the first derivative equation implicitly. This gives us a complex expression involving both first and second derivatives. Substituting our known values x equals 0, y equals 2, and dy dx equals negative 2 fifths, we get an equation in terms of the second derivative. After simplification, we find that d squared y dx squared equals negative 94 over 125.
Now we construct the Maclaurin series using the formula y of x equals y of 0 plus y prime of 0 times x plus y double prime of 0 over 2 factorial times x squared. We have y of 0 equals 2, y prime of 0 equals negative 2 fifths, and y double prime of 0 equals negative 94 over 125. Substituting these values and simplifying, we get y of x equals 2 minus 2 fifths x minus 47 over 125 x squared.
Let me summarize our final answers. For part a, when x equals 0, the y-coordinate is 2. For part b, the derivative dy dx at x equals 0 is negative 2 fifths. For part c, the Maclaurin series expansion up to the x squared term is y equals 2 minus 2 fifths x minus 47 over 125 x squared. This polynomial approximation helps us understand the local behavior of the implicit curve near the origin.