#Trigonometry question
cold wolfBOT
solar widget
i think if you take the derivative and find where the slope matches with the current angle to make it perpendicular / parallel it will work
that might be wrong though
trail moth
what is the initial line?
drifting plume
you should definitely watch the live stream of grant(3b1b) lockdown lecture 4 and 5
btw what book is that
watch the second one first
frank mortar
Initial line you mean y(theta=0)=e^2*0=1?
I think it’s what they mean. So what I would do is vectorize this problem such as the position of a point M on the graph is given as:
OM=(theta, r)
echo mountain
Guys i saw all your questions but sadly i cannot give more information as i said it was a friend of mines.
frank mortar
Guys i saw all your questions but sadly i cannot give more information as i said...
It’s alright I think I got what they mean
frank mortar
I think it’s what they mean. So what I would do is vectorize this problem such a...
Nevermind I think you gotta parameterize.
theta=t
r=e^2t
You get OM(t)=(t, e^2t)
You take the derivative with respect to t. You get:
V(t)=(1, 2e^2t).
Get the tangent vector at a point t=s now which is:
T(t=s)=OM(s)+V(s) * t
=(s, e^2s)+(1, 2e^2s)t
=(s+t, e^2s(1+2t))
You might want to convert this tangent vector to its Cartesian form.
For that you eliminate t:
Tx=s+t <=>t=Tx-s (you plug that in Ty:)
Ty=e^2s(1+2t)=e^2s(1+2Tx-2s)
Tangent vector( Cartesian) is
T=(theta, e^2s(1+2(theta)-2s)
Now we’re saved because they ask you to see at which point theta the tangent is perpendicular to initial line, and at which point s’ the tangent is parallel to initial line.
You got the tangent vector above and the initial line which is OI=(0,1) (obtained by taking theta=0)
You just do scalar product T.OI=0 to find the point s such as the tangent is perpendicular to the initial line.
If it’s parallel then CROSS PRODUCT=0
TXOI=0, you get s’ the point such as the tangent is parallel to the initial line
For the two equations, you find s and s’ and you’re done!