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ipn [44]
4 years ago
15

n exponential spiral has parametric equationr(t) =〈etcost, etsint〉.(a) Plotr(t). Make sure your plot includeslimt→−[infinity]r(t

).(b) Find the unit tangent vectorT(t)tor(t).(c) Find the length ofr(t)for−[infinity]< t≤0.(d) Letf(x, y) =−12ln(x2+y2)−arctanxy. Find∇f(x, y).(e) Compute the angle between∇f(r(t))andr′(t).
Mathematics
1 answer:
romanna [79]4 years ago
6 0

Answer:

please see the answer below

Step-by-step explanation:

we have

r(t)=

(b) the unit tangential vector is computed by using

T(t)=\frac{r'(t)}{|r'(t)|}\\\\r'(t)=\\\\|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}}

where we have used cos^2t+sin^2t=1

(c) the arc length is given by

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}

(d)

f(x,y)=-12ln(x^2+y^2)-arctan(xy)\\\\\bigtriangledown f(x,y)=

(e)

|\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)|}]

hope this helps!!

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