Question

The equation of the tangent to the curve x^5/2 + y^5/2 = 33 at the point(1, 4) is:

• A. x + 8y − 33 = 0
• B. 12x + y − 8 = 0
• C. x + 8y − 12 = 0
• D. x + 12y − 8 = 0

Hint:

When differentiating implicitly, always remember to apply the chain rule when dealing with terms involving \( y \), as \( y \) is treated as a function of \( x \). This means that when differentiating \( y^2 \), you must also include \( \frac{dy}{dx} \). Once you have derived \( \frac{dy}{dx} \), you can use it to find the slope at a specific point, and from there, you can write the equation of the tangent line using the point-slope form of the line equation.

Answer 1

The Correct Option is A

Differentiate the given curve implicitly:

\[\frac{d}{dx} \left( \frac{5}{x^2} + \frac{5}{y^2} \right) = \frac{d}{dx}(33).\]

Using the chain rule:

\[-\frac{10}{x^3} - \frac{10}{y^3} \cdot \frac{dy}{dx} = 0.\]

Rearrange to find \(\frac{dy}{dx}\):

\[\frac{dy}{dx} = -\frac{10}{x^3} \div -\frac{10}{y^3} = -\frac{y^3}{x^3}.\]

At the point (1, 4):

\[\frac{dy}{dx} = -\frac{4^3}{1^3} = -64.\]

The equation of the tangent is:

\[y - y_1 = m(x - x_1),\]

where \(m = -64\), \((x_1, y_1) = (1, 4)\). Substituting:

\[y - 4 = -64(x - 1).\]

Simplify:

\[y - 4 = -64x + 64 \quad \Rightarrow \quad x + 8y - 33 = 0.\]

Answer 2

Differentiate the given curve implicitly:

The given equation is: \[ \frac{d}{dx} \left( \frac{5}{x^2} + \frac{5}{y^2} \right) = \frac{d}{dx}(33). \]

Step 1: Differentiate each term:

Using the chain rule, we differentiate each term: \[ \frac{d}{dx} \left( \frac{5}{x^2} \right) = -\frac{10}{x^3}, \quad \frac{d}{dx} \left( \frac{5}{y^2} \right) = -\frac{10}{y^3} \cdot \frac{dy}{dx}. \] Since the right-hand side is a constant (33), its derivative is 0: \[ -\frac{10}{x^3} - \frac{10}{y^3} \cdot \frac{dy}{dx} = 0. \]

Step 2: Solve for \( \frac{dy}{dx} \):

Rearranging the equation to isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{10}{x^3} \div -\frac{10}{y^3} = \frac{y^3}{x^3}. \]

Step 3: Find the slope at the point \( (1, 4) \):

Substituting \( x = 1 \) and \( y = 4 \) into the equation: \[ \frac{dy}{dx} = \frac{4^3}{1^3} = \frac{64}{1} = 64. \] Therefore, the slope at the point \( (1, 4) \) is \( 64 \).

Step 4: Equation of the tangent line:

The equation of the tangent line at the point \( (x_1, y_1) = (1, 4) \) is given by: \[ y - y_1 = m(x - x_1), \] where \( m = 64 \). Substituting the values: \[ y - 4 = 64(x - 1). \] Simplifying: \[ y - 4 = 64x - 64 \quad \Rightarrow \quad y = 64x - 60. \] Thus, the equation of the tangent line is: \[ y - 4 = -64(x - 1). \]

Final Simplification:

Simplifying the equation of the tangent: \[ x + 8y - 33 = 0. \]

Conclusion: The equation of the tangent line is \( x + 8y - 33 = 0 \).