Question 994014
I would think this is a calculus problem.
The distance to the origin as a function of {{{t}}} (time) is
{{{D(t)=sqrt(x^2+y^2)=sqrt((4cos(t))^2+(2sin(t))^2)=sqrt(16(cos(t))^2+4(sin(t))^2)=sqrt(4(4(cos(t))^2+(sin(t))^2))=2sqrt(4(cos(t))^2+(sin(t))^2)}}}
The function showing the rate of change at time {{{t}}} is
{{{dD/dt}}} , and it can be calculated using the chain rule
{{{D(t)=2*F(G(t))}}} with {{{F(u)=sqrt(u)=u^("1/2")}}} and {{{u=G(t)=4(cos(t))^2+(sin(t))^2}}}
{{{dF/du=(1/2)u^("-1/2")=1/(2u^("1/2"))=1/2sqrt(u)}}} , and
{{{d(sqrt(4(cos(t))^2+(sin(t))^2))/dt=d(4(cos(t))^2)/dt+d((sin(t))^2)/dt}}}
Applying the chain rule again, and again, to both terms:
{{{d(4(cos(t))^2)/dt=4*(2cos(t))d(cos(t))/dt=4*2*cos(t)(-sin(t))=-8cos(t)sin(t)}}} and
{{{d((sin(t))^2)/dt=(2sin(t))d(sin(t))/dt=2*sin(t)(cos(t))=2cos(t)sin(t)}}} .
So, {{{d(sqrt(4(cos(t))^2+(sin(t))^2))/dt=-8cos(t)sin(t)+2cos(t)sin(t)=-6cos(t)sin(t)}}} , and
{{{dD/dt=2(1/(2sqrt(u)))(-6cos(t)sin(t))=(-6cos(t)sin(t))/sqrt(u)}}} .
Substituting the expression for {{{u}}} ,
{{{rate(t)=dD/dt=(-6cos(t)sin(t))/sqrt(4(cos(t))^2+(sin(t))^2)))}}}
For {{{t=pi/4}}} ,
{{{t=pi/4}}}-->{{{system(cos(t)=cos(pi/4)=sqrt(2)/2,cos(t)=cos(pi/4)=sqrt(2)/2)}}}-->{{{rate(pi/4)=(-6cos(pi/4)sin(pi/4))/sqrt(4(cos(pi/4))^2+(sin(pi/4))^2)))}}}-->{{{rate(pi/4)=(-6(sqrt(2)/2)(sqrt(2)/2)/sqrt(4(sqrt(2)/2)^2+(sqrt(2)/2)^2)))}}}
-->{{{rate(pi/4)=(-6(2/4))/sqrt(4(2/4)+(2/4))))}}}-->{{{rate(pi/4)=-3/sqrt(5/2)}}}-->{{{rate(pi/4)=-3sqrt(2)/sqrt(5)}}}-->{{{rate(pi/4)=-3sqrt(10)/5}}}