1. Let r(t) and y(t) be define as z(t) = so it<0 t2 :t> 0 { g(t) = { سر +2 :t<0 0 :t> 0 Consider the path c(t): R + R2 :

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answerhappygod
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1. Let r(t) and y(t) be define as z(t) = so it<0 t2 :t> 0 { g(t) = { سر +2 :t<0 0 :t> 0 Consider the path c(t): R + R2 :

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1. Let r(t) and y(t) be define as z(t) = so it<0 t2 :t> 0 { g(t) = { سر +2 :t<0 0 :t> 0 Consider the path c(t): R + R2 :(t) + (x(t), y(t)). (a) Using the limit definition of differentability of multivariate functions show that c(t) is differentable for all t > 0. (b) Using the limit definition of differentability of multivariate functions show that ct) is differentable for all t < 0. (c) Using the limit definition of differentability of multivariate functions show that c(t) is differentable for t = 0. (Hint: since the domain is 1-dimensional you can solve the limit using left and right hand limits.] (d) Graphing the curve on the plane we can see that ct) makes a sharp turn, but is still differentable at that point, why is that the case?
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