Question

@shubhamsrg's problem

Answer 1

\(\vec{v} = \langle 2,1\rangle \)

Answer 2

forgot to mention, I was more interested in the second part.

Answer 3

forgot to mention, I was more interested in the second part.

Answer 4

for second part,
just negative gradient direction?

Answer 5

also a bit confused in the first part..so in case anyone could throw some light on that too?

Answer 6

negative gradient direction? How does that work out?

Answer 7

Imagine a plane parallel to z axis through the line d = 2v

Answer 8

That plane cuts the given surface and the intersection is some curve, yes ?

Answer 9

yo

Answer 10

you're asked to study that curve

Answer 11

when you walk on xy plane along the vector (2, 1), how that intersection curve changes ?

Answer 12

directional derivative gives the derivative of the intersection curve that you see as you walk on xy plane in the direction (2, 1)

Answer 13

that is part a

Answer 14

yes.so far so good

Answer 15

\(\nabla R\) : direction of greatest increase of infection
\(-\nabla R\): direction of greatest decrease of infection

Answer 16

Thw ratio comes out to be negative. What does that imply?

Answer 17

what ever is negative, it means those quantities need to be reduced

Answer 18

delR/delV is negative. Not too sur3 if I follow?

Answer 19

for example
if you get < -1 , -2 >
then the best option is to reduce d and v such that the reduction in v is twice that of d

Answer 20

aha..so we must increase both in this case d increases at twice the rate..?

Answer 21

such that*

Answer 22

with respect to <2,1>

Answer 23

am I still misinterpreting?

Answer 24

i'm not sure what you are saying

\[\nabla R = < \frac{\partial R}{\partial d},\frac{\partial R}{\partial v}>\]

Answer 25

is that what you computed?

Answer 26

you need to find \(\nabla R\) at (2 , 0.1)

Answer 27

then take the negative of the resulting vector

Answer 28

so I was misinterpreting.
I now have <0.5,-3.33>

Answer 29

why would I take the negative of this?

Answer 30

<0.5 , -3.33> is the direction in which R( number of infected people) increases.

the health officials want to decrease number of infections.

Answer 31

I see.I think now I got the gist

Answer 32

hence decreasing d by 0.5 or doubling it and factoring v by 3.33 is optimum. Is that right?

Answer 33

if we were climbing a mountain,
the steepest way upward, is also the steepest way downward if we were coming down the same mountain....

same way, gradient vector tells us direction of steepest increase.
its negative tells us direction of steepest decrease.

Answer 34

very insightful sir. Thank you.

Answer 35

Was my final interpretation correct?

Answer 36

after taking the negative ,
we get <-0.5 , 3.33 >

we computed this at d=2 and v=0.1

the best strategy is to decrease d by an amount \(\delta d\) and increase v by \(\delta v\) such that \(\frac{\delta d }{\delta v}\)= 0.5/3.33

Answer 37

aha..the change needs to be a factor. That's more sound.

Answer 38

yes! :)

Answer 39

cool.. thank you sir.

Answer 40

Here is a contour plot, with the blue arrow pointing "downhill" and the red arrow pointing in the direction <2,1>
we can see that though the red arrow is not close to optimal, it does point a bit downhill, and so improves (minimizes) R(d,v). If we calculate the angle between the red and blue we get about 71 degrees, which means the red arrow helps (a little)