Recent content by melifaro

anyone?

I tried that, then R(t) = t i - t2j + t2k R'(t) = i - 2t j + 2t k and at given point P(1,-1,1) from equation of R(t) I get t = 1 So R'(1) = i - 2j + 2k , which is also velocity But then speed, ||R'(1)|| = sqrt{1 + 4 + 4} = 3.. And according to the problem description, it should...

Homework Statement A particle moves along the curve of intersection of shapes y = -x2 and z = x2 in the direction in which x increases. At the instant when the particle is at the point P(1,-1,1), its speed is 9cm/s and that speed is increasing at a rate of 3cm/s2. Find the velocity and...

Thank you!

Just to make sure, is it because t is always positive from the problem description? Edit Yes, thanks. As I said I am not familiar with hyperbolic functions so SammyS' solution better for me

I'm not very familiar with hyperbolic functions but does that mean that S = 2 sinh(t) sinh-1(\frac{S}{2}) = t ? So R(s) is defined by x = esinh-1(\frac{S}{2}), y = \sqrt{2}sinh-1(\frac{S}{2}), z = e-sinh-1(\frac{S}{2}) EDIT This is so simple and so smart. Thanks a lot!

Reparametrize the curve R(t) in terms of arc length measured from the point where t = 0 R(t) is defined by x = et, y = \sqrt{2}t, z = -e-t Arc length S = ∫ ||R'(t)||dt ||R'(t)||= sqrt{\dot{x}2 + \dot{y}2 + \dot{z}2}The attempt at a solution Getting R'(t) ==> x = et, y = \sqrt{2}, z = e-t...