Answer:
please see the answer below
Step-by-step explanation:
we have
[tex]r(t)=<e^tcost,e^tsint>[/tex]
(b) the unit tangential vector is computed by using
[tex]T(t)=\frac{r'(t)}{|r'(t)|}\\\\r'(t)=<e^tcost-e^tsint,e^tsint+e^tcost>\\\\|r'(t)|=\sqrt{e^{2t}(cost-sint)^2+e^{2t}(sint+cost)^2}}\\\\|r'(t)|=e^t\sqrt{cos^2t-2sintcost+sin^2t+sin^2t+2sintcost+cos^2t}\\\\|r'(t)|=\sqrt{2}e^t\\\\T(r)=\frac{1}{\sqrt{2}}<cost-sint,sint+cost>[/tex]
where we have used cos^2t+sin^2t=1
(c) the arc length is given by
[tex]L=\int_{-\inf}^{0}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt\\\\L=\int_{-inf}^0e^t\sqrt{2}=\sqrt{2}[e^{0}-e^{-inf}]=\sqrt{2}[/tex]
(d)
[tex]f(x,y)=-12ln(x^2+y^2)-arctan(xy)\\\\\bigtriangledown f(x,y)=<\frac{-24}{x^2+y^2}-\frac{y}{1+x^2y^2},\frac{-24}{x^2+y^2}-\frac{x}{1+x^2y^2}>[/tex]
(e)
[tex]|\bigtriangledown f(x,y)|=\sqrt{(\frac{-24}{x^2+y^2}-\frac{y}{1+x^2y^2})^2+(\frac{-24}{x^2+y^2}-\frac{x}{1+x^2y^2})^2}\\\\\theta=cos^{-1}[\frac{\bigtriangledown f(x,y) \cdot r'(t)}{|\bigtriangledown f(x,y)||r'(t)|}][/tex]
hope this helps!!
