Differentiation

To find the length of the tangent at the point $P \left( \frac{\pi}{4} \right)$ on the curve $x^{2/3} + y^{2/3} = 2^{2/3}$, we can proceed as follows:The given equation of the curve can be parameterized as:$x = 2 \cos^3 \theta, \quad y = 2 \sin^3 \theta$For the point $P$, we substitute $\theta = \frac{\pi}{4}$ :$x = 2 \cos^3 \frac{\pi}{4}, \quad y = 2 \sin^3 \frac{\pi}{4}$Calculating the coordinates:$x = 2 \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}, \quad y = 2 \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$So the point is $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$.Next, differentiate the curve equation $x^{2/3} + y^{2/3} = 2^{2/3}$ with respect to $x$ :$\frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0$Solving for $\frac{dy}{dx}$ :$\frac{dy}{dx} = - \left( \frac{y}{x} \right)^{1/3}$At the point $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$,$\frac{dy}{dx} = -1$The formula to find the length of the tangent at a point is:$L = y \sqrt{1 + \left( \frac{dx}{dy} \right)^2}$Given $\frac{dx}{dy} = -1$, the length of the tangent at the point is calculated as:$L = \frac{1}{\sqrt{2}} \sqrt{1 + (-1)^2} = \frac{1}{\sqrt{2}} \times \sqrt{2} = 1$Thus, the length of the tangent is 1 .