A Parametric/differentiation question (1 Viewer)

[pi-ka-chew]

Given the parametric equations x = e^t.cos(t) and y = e^t.sin(t), show that dy/dx = tan(t+pi/4).

What i did was:

e^t = x/(cos(t)
so y=x.sin(t) / cos(t)

dy/dx = tan t

i'm not sure how to get tan(t + pi/4)

please help. thanks.

 

[random-1005]

Given the parametric equations x = e^t.cos(t) and y = e^t.sin(t), show that dy/dx = tan(t+pi/4).

What i did was:

e^t = x/(cos(t)

so y=x.sin(t) / cos(t)

dy/dx = tan t

i'm not sure how to get tan(t + pi/4)

please help. thanks.

dy/dx= dy/dt *dt/dx

dx/dt= -e^tsint +e^tcost

dy/dt= e^tcost +e^t sint


dy/dx= e^t(sint +cost) x 1/(e^t(cost-sint))
= (sint+cost)/(cost-sint)
= (1 +sin2t) /cos2t
= etc