Question
Check the continuity and differentiability of the function at (0, 0) where
(b) Find the domain and range of the function f, defined by . Also find two level curves of this function. Give a rough sketch of them.
Answer :
Word Count : 354
First, we check the continuity and differentiability of [ f(x, y) = \begin{cases} \dfrac{2x^3y}{x^2 + y^2}, & (x, y) \neq (0,0) [2mm] 0, & (x, y) = (0,0) \end{cases} ] at ((0,0)). Continuity at ((0,0)): We compute the limit: [ \lim_{(x,y)\to(0,0)} \frac{2x^3y}{x^2 + y^2}. ] Try converting to polar coordinates: (x = r\cos\theta, y = r\sin\theta). Then [ f(r,\theta) = \frac{2 (r\cos\theta)^3 (r\sin\theta)}{r^2} = \frac{2 r^4 \cos^3\theta \sin\theta}{r^2} = 2 r^2 \cos^3\theta \sin\theta. ] As (r \to 0), (2 r^2 \cos^3\theta \sin\theta \to 0) for all (\theta). Hence, ________ ________ ______ ___ ___ ____.
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First, we check the continuity and differentiability of [ f(x, y) = \begin{cases} \dfrac{2x^3y}{x^2 + y^2}, & (x, y) \neq (0,0) [2mm] 0, & (x, y) = (0,0) \end{cases} ] at ((0,0)). Continuity at ((0,0)): We compute the limit: [ \lim_{(x,y)\to(0,0)} \frac{2x^3y}{x^2 + y^2}. ] Try converting to polar coordinates: (x = r\cos\theta, y = r\sin\theta). Then [ f(r,\theta) = \frac{2 (r\cos\theta)^3 (r\sin\theta)}{r^2} = \frac{2 r^4 \cos^3\theta \sin\theta}{r^2} = 2 r^2 \cos^3\theta \sin\theta. ] As (r \to 0), (2 r^2 \cos^3\theta \sin\theta \to 0) for all (\theta). Hence, ________ ________ ______ ___ ___ ____.
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