Question

The parametric equations of a curve are x=e^-t*cost y=e^-t*sint show that dy/dx=tan(t-pi/4)

Answer 1

apply product rule to both x and y
so dx/t=-e^-t*cost-e^-tsint
=-e^-t(cost+sint)
dy/dt=-e^-tsint+e^tcost
=e^-t(-sint+cost)
dy/dx=dy/dt/dx/dt=dy/dt*dt/dx
=e^t(-sint+cost)/-e^-t(cost+sint)
=sint-cost/cost+sint

Answer 2

sint/cost+sint/sint-cost/cost-cost/sint
tant+1-1-1/tant
tant-1/tant

Answer 3

(tant^2-1)/tant=(tant+1)(tant-1)/tant

Answer 4

\[\text{ so you have } =\frac{\tan(t)-1}{1+\tan(t)} \text{ so far }\]

Answer 5

yes

Answer 6

\[x=e^{-t} \cos(t) \\ y=e^{-t} \sin(t) \\ \frac{dx}{dt}=-e^{-t}\cos(t)+e^{-t}(-\sin(t))=-e^{-t}(\cos(t)+\sin(t)) \\ \frac{dy}{dt}=-e^{-t}\sin(t)+e^{-t}\cos(t)=-e^{-t}(\sin(t)-\cos(t)) \\ \frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\sin(t)-\cos(t)}{\sin(t)+\cos(t)}=\frac{\tan(t)-1}{\tan(t)+1}\]
yeah that is the exact thing I have so far

yep so we recall
\[\tan(u-v)=\frac{\tan(u)-\tan(v)}{1+\tan(u)\tan(v)} \\ u=t \\ v=\frac{\pi}{4} \text{ since } \tan(\frac{\pi}{4})=1 \\ \tan(t-\frac{\pi}{4})=\frac{\tan(t)-1}{1+ \tan(t) \cdot 1 }\]

Answer 7

thank you @freckles

Answer 8

I had to do the top work again just to check yours
sometimes I get lost in signs if I don't