Question

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).

Answer 1

please see the answer below

Explanation:

we have

`r(t)=<e^tcost,e^tsint>`

(b) the unit tangential vector is computed by using

`T(t)=(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√(cos^2t-2sintcost+sin^2t+sin^2t+2sintcost+cos^2t)\\\\|r'(t)|=√(2)e^t\\\\T(r)=(1)/(√(2))<cost-sint,sint+cost>`

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

(c) the arc length is given by

`L=\int_(-\inf)^(0)\sqrt{((dx)/(dt))^2+((dy)/(dt))^2+((dz)/(dt))^2}dt\\\\L=\int_(-inf)^0e^t√(2)=√(2)[e^(0)-e^(-inf)]=√(2)`

(d)

`f(x,y)=-12ln(x^2+y^2)-arctan(xy)\\\\\bigtriangledown f(x,y)=<(-24)/(x^2+y^2)-(y)/(1+x^2y^2),(-24)/(x^2+y^2)-(x)/(1+x^2y^2)>`

(e)

`|\bigtriangledown f(x,y)|=\sqrt{((-24)/(x^2+y^2)-(y)/(1+x^2y^2))^2+((-24)/(x^2+y^2)-(x)/(1+x^2y^2))^2}\\\\\theta=cos^(-1)[(\bigtriangledown f(x,y) \cdot r'(t))/(|\bigtriangledown f(x,y)||r'(t)|)]`

hope this helps!!