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

To find the equation of the tangent to the curve given by \(x^{5/2}+y^{5/2}=33\) at the point (1, 4), we will use implicit differentiation. Differentiate both sides with respect to \(x\). 

\(\frac{d}{dx}(x^{5/2})+\frac{d}{dx}(y^{5/2})=\frac{d}{dx}(33)\)

Using the chain rule and power rule, we have:

\(\frac{5}{2}x^{3/2}+\frac{5}{2}y^{3/2}\frac{dy}{dx}=0\)

Solve for \(\frac{dy}{dx}\) to find the slope of the tangent:

\(\frac{5}{2}y^{3/2}\frac{dy}{dx}=-\frac{5}{2}x^{3/2}\)

\(\frac{dy}{dx}=-\frac{x^{3/2}}{y^{3/2}}\)

Now substitute \(x=1\) and \(y=4\) into the derivative to find the slope at (1, 4):

\(\frac{dy}{dx}=-\frac{1^{3/2}}{4^{3/2}}=-\frac{1}{8}\)

With the slope \(-\frac{1}{8}\) at (1, 4), use the point-slope form of the line equation \(y-y_1=m(x-x_1)\):

\(y-4=-\frac{1}{8}(x-1)\)

Simplify this equation:

\(y-4=-\frac{1}{8}x+\frac{1}{8}\)

Multiply through by 8 to clear fractions:

\(8y-32=-x+1\)

Rearrange to form the equation:

\(x+8y-33=0\)

The equation of the tangent to the curve at point (1, 4) is \(x+8y-33=0\).

Answer 2

To find the equation of the tangent to the curve \(x^{5/2} + y^{5/2} = 33\) at the point (1, 4), we follow these steps: 

• First, differentiate the given equation implicitly with respect to x. The differentiation leads to: \(\frac{5}{2}x^{3/2} + \frac{5}{2}y^{3/2}\frac{dy}{dx} = 0\).
• Solve for \(\frac{dy}{dx}\):
  \(\frac{dy}{dx} = -\frac{x^{3/2}}{y^{3/2}}\).
• Substitute the point (1, 4) into the derivative to find the slope of the tangent:
  \(\frac{dy}{dx} = -\frac{1^{3/2}}{4^{3/2}} = -\frac{1}{8}\).
• Using the point-slope form of the tangent line equation: \(y - y_1 = m(x - x_1)\), where \(m = -\frac{1}{8}\) and \((x_1, y_1) = (1, 4)\), we have:
  \(y - 4 = -\frac{1}{8}(x - 1)\).
• Simplify to form the equation in standard form:
  Multiply through by 8 to eliminate the fraction: \(8(y - 4) = -(x - 1)\)
  \(8y - 32 = -x + 1\)
  Rearrange: \(x + 8y = 33\).

Thus, the equation of the tangent is \(x + 8y - 33 = 0\).