Question

the parametric equations of a curve are x=t-e^2t y=t+e^2t where take t all real values
express dy/dx in terms of t.
ii) hence find the value of t for which the gradient of the curve is 3 giving your answers in logarithmic form
iii) show that for points on the curve the greatest value of x is 1/2(ln1/2-1)

Answer 1

i)\[x=t-e ^{2t}\rightarrow dx/dt=1-(e ^{2t}\times d/dt(2t))=1-2e ^{2t}\]
\[y=t+e ^{2t}\rightarrow dy/dt=1+(e ^{2t}\times d/dt(2t))=1+2e^2t\]
\[dy/dx=dy/dt \div dx/dt=(1+2e^2t) \div (1-2e^2t)\]

Answer 2

\[3-6e ^{2t}=1+2e ^{2t}\rightarrow 8e ^{2t}=2\rightarrow e ^{2t}=1/4\rightarrow 2t*lne=\ln(1)-\ln(4)\rightarrow t=-\ln(4)/2\]

Answer 3

So far so good. =)

Answer 4

dy/dx=1+2e^2t/1-2e^2t
0=1+2e^2t
-1/2=e^2t
-(ln(1)-ln(2))=lne^2t
-(-ln(2)=2t
t=ln2/2

Answer 5

x=t-e^2t
=ln(2)/2-e^2(ln(2)/2)
=ln(2)/2-2

Answer 6

@eliassaab

Answer 7

For the first question t=-ln(2)

Answer 8

t=-ln(4)/2