Question

r(t)=(50e^(-t)cos(t))i + (50e^(-t))sin(t))j + (5-5e^(-t))k

Eliminate the parameter t to show that z=5-r/10, where r^2 = x^2+y^2.

Answer 1

ok first note that\[x=50e^{-t}\cos t\]\[y=50e^{-t}\sin t\]\[z=5-5e^{-t}\]now evaluate r and see wat happens :)

Answer 2

The easiest way I can think of solving such an equation is diving y/x

\[\Large \frac{y}{x}=\tan t \]
that would mean that:

\[\Large t=\tan^{-1}\left(\frac{y}{x}\right)=\phi \]
But that doesn't really seem to be what they want to see here, so besides that. (Inspired by the question itself) -> Square x and y, will make the trigonometric expressions disappear, divide by ten and back substitute.

Answer 3

I get it. If r = 50e^-t the sum of x^2 and y^2 gives r^2. They should have used a variable besides r

Answer 4

I agree