Numerical methods

I've got to show that something is a local maximum but my answer shows that it is a local minimum.

p(t)=0.1 ln (t+1) - cos (0.5t) + 0.1 t^3/2 + 199.3

a) Figure 1 is a graph of the price of a stock during a 12-hour trading window. The equation of the curve is given above. Show that the price reaches a local maximum in the interval 8.5<t<8.6.

I have completed all of the steps except for the last one in showing that the price is a local maximum. Any help would be appreciated.

Reply 1

ThomH97

21

What does your answer say that shows you have a local minimum? There are two turning points of that graph 'quite' close together. You may have looked at the wrong one, or muddled up the way you checked things.

Reply 2

Ellierebecca

OP

4

Original post by ThomH97

What does your answer say that shows you have a local minimum? There are two turning points of that graph 'quite' close together. You may have looked at the wrong one, or muddled up the way you checked things.

P’(t)= (10t+10)^-1 + 0.5sin(0.5t) + 0.15t^1/2

I tested p’(8.5)=3.530*10^-4 (3dp) and p’(8.6)=-7.780*10^-3 (3dp) which had a sign change showing that there is at least one sign change between 8.5<t<8.6.

Then I found p’’(t)= -10(10t+10)^-2 + 0.25cos(0.5t) + 0.075t^-1/2. I subbed 8.5 and 8.6 into this. P’’(8.5)=0.274 (3dp) and p’’(8.6)=0.274 (3dp) which means p’’(t)>0 which is a local minimum not a local maximum.

Hopefully you can spot something that I’ve done wrong!!

Reply 3

mqb2766

21

Original post by Ellierebecca

P’(t)= (10t+10)^-1 + 0.5sin(0.5t) + 0.15t^1/2

I tested p’(8.5)=3.530*10^-4 (3dp) and p’(8.6)=-7.780*10^-3 (3dp) which had a sign change showing that there is at least one sign change between 8.5<t<8.6.

Then I found p’’(t)= -10(10t+10)^-2 + 0.25cos(0.5t) + 0.075t^-1/2. I subbed 8.5 and 8.6 into this. P’’(8.5)=0.274 (3dp) and p’’(8.6)=0.274 (3dp) which means p’’(t)>0 which is a local minimum not a local maximum.

Hopefully you can spot something that I’ve done wrong!!

I get -0.087 when evaluating the 2nd derivative (expression is right).
Are you using degrees rather than radians?

Reply 4

Ellierebecca

OP

4

Original post by mqb2766

I get -0.087 when evaluating the 2nd derivative (expression is right).
Are you using degrees rather than radians?

Oh I think I am!! That makes a lot more sense now as p''(t)<0 and so it is a local maximum. How did you know to use radians rather than degrees?

Reply 5

mqb2766

21

Original post by Ellierebecca

Oh I think I am!! That makes a lot more sense now as p''(t)<0 and so it is a local maximum. How did you know to use radians rather than degrees?

I think at A-level you use radians by default. Check with your teacher though.
TBH, the basic trig differentiation relationships generally assume you're using radians.

Reply 6

kevdownes

1

Original post by mqb2766

I get -0.087 when evaluating the 2nd derivative (expression is right).
Are you using degrees rather than radians?

there is no sign change - plot it in Desmos. There's something wrong with the q

Reply 7

kevdownes

1

Original post by Ellierebecca

P’(t)= (10t+10)^-1 + 0.5sin(0.5t) + 0.15t^1/2
I tested p’(8.5)=3.530*10^-4 (3dp) and p’(8.6)=-7.780*10^-3 (3dp) which had a sign change showing that there is at least one sign change between 8.5<t<8.6.
Then I found p’’(t)= -10(10t+10)^-2 + 0.25cos(0.5t) + 0.075t^-1/2. I subbed 8.5 and 8.6 into this. P’’(8.5)=0.274 (3dp) and p’’(8.6)=0.274 (3dp) which means p’’(t)>0 which is a local minimum not a local maximum.
Hopefully you can spot something that I’ve done wrong!!

p'(8.6) is positive

Reply 8

mqb2766

21