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.
The equation of the tangent to the curve \(x^{5/2}+y^{5/2}=33\) at point (1, 4) is determined using implicit differentiation. Both sides of the equation are differentiated with respect to \(x\).
\(\frac{d}{dx}(x^{5/2})+\frac{d}{dx}(y^{5/2})=\frac{d}{dx}(33)\)
Applying the chain and power rules yields:
\(\frac{5}{2}x^{3/2}+\frac{5}{2}y^{3/2}\frac{dy}{dx}=0\)
Solving for \(\frac{dy}{dx}\) to obtain the tangent's slope:
\(\frac{5}{2}y^{3/2}\frac{dy}{dx}=-\frac{5}{2}x^{3/2}\)
\(\frac{dy}{dx}=-\frac{x^{3/2}}{y^{3/2}}\)
Substituting \(x=1\) and \(y=4\) into the derivative gives the slope at (1, 4):
\(\frac{dy}{dx}=-\frac{1^{3/2}}{4^{3/2}}=-\frac{1}{8}\)
Using the point-slope form of a line, \(y-y_1=m(x-x_1)\), with slope \(-\frac{1}{8}\) at (1, 4):
\(y-4=-\frac{1}{8}(x-1)\)
Simplifying the equation:
\(y-4=-\frac{1}{8}x+\frac{1}{8}\)
Multiplying by 8 to eliminate fractions:
\(8y-32=-x+1\)
Rearranging terms to form the final equation:
\(x+8y-33=0\)
The equation of the tangent to the curve at point (1, 4) is \(x+8y-33=0\).