# Math Help - trig function - proof

1. ## trig function - proof

i have the equation

r(t) = sin(2t)i + 2cos(t)j where 0<t<2pi

I have to prove that using this equation that the object it describes passes through the origin twice in the time limit. Does this just mean I have to show the values of t that give r(t) = 0? i.e. pi/2 and 3pi/2?

cheers for the help

2. Originally Posted by thermalwarrior
i have the equation

r(t) = sin(2t)i + 2cos(t)j where 0<t<2pi

I have to prove that using this equation that the object it describes passes through the origin twice in the time limit. Does this just mean I have to show the values of t that give r(t) = 0? i.e. pi/2 and 3pi/2?

cheers for the help
Yes.

Having said that, I hope you got those values in the following way:

sin(2t) = 0 .... (1)
2 cos(t) = 0 .... (2)

Solve (1) and (2) simultaneously over the domain 0 < t < 2 pi: t = pi/2, 3pi/2.

3. i pretty much used that way of doing it but question is only 2 marks so i imagine very little detail is needed. Thanks for the post!!!

4. a further part of this question is asking when are the times that the displacement is perpendicular to the velocity. Clearly one time would be t=0 as the i and j directions are mutually perpendicular to each other. How would i find the others? Is there some way involving dot products?

5. Originally Posted by thermalwarrior
a further part of this question is asking when are the times that the displacement is perpendicular to the velocity. Clearly one time would be t=0 as the i and j directions are mutually perpendicular to each other. How would i find the others? Is there some way involving dot products?
Solve $\frac{d\vec{r}}{dt} \cdot \vec{r} = 0$ (over the domain 0 < t < 2 pi ?)

This will boil down to solving $2 \sin (2t) (\cos (2t) - 1) = 0$ (over the domain 0 < t < 2 pi ?).....

Note: Some of the resulting solutions might not be wanted - why?

6. how did you reach that final equation to solve?? have you just multiplied the velocity with the displacement???

7. Originally Posted by thermalwarrior
how did you reach that final equation to solve?? have you just multiplied the velocity with the displacement???
I took the dot product.

If you're stuck, show the working you've got and I'll get you over the hump.

8. okay,

r(t) = sin(2t) + 2cos(t)

r(t)' = 2cos(2t) - 2sin(t)

r(t)' * r(t) =

2sin cos(2t)^2 - 4 sin cos(t)^2

this is clearly the bit thats wrong but not sure why.....

9. Originally Posted by thermalwarrior
okay,

r(t) = sin(2t) i + 2cos(t) j

r(t)' = 2cos(2t) i - 2sin(t) j

r(t)' * r(t) =

2sin cos(2t)^2 - 4 sin cos(t)^2 Mr F says: Here is the error.

this is clearly the bit thats wrong but not sure why.....
Note my edits in red ....

When you take the dot product:

sin(2t) [2 cos(2t)] - [2 cos t] [2 sin (t)] = 0

and you know that 2 sin(t) cos(t) is sin(2t), right?

Have you learned about the dot product (also called the scalar product)? Do you understand that it's very different to multiplying two scalars ......?

10. im okay with trig identities but the dot product hasnt really been explained very well to me.

11. Originally Posted by thermalwarrior
im okay with trig identities but the dot product hasnt really been explained very well to me.

i.i = j.j = k.k = 1.

i.j = i.k = j.k = 0.

12. ive got that bit written down but not sure how to interpret in relation to this quesiton.

with regards dot product

i now have that in the case of these equations

a*b=a1b1 + a2b2

taking a to be the velocity components and b to be displacement components

which leads to 2cos(2t)*sin(2t) and -2sin(t)*2cps(t)

is this correct??

13. if i follow this through I get

2cos(2t)sin(2t) - 4sin(t)cos(t)

cos(2t)sin(2t) - 2sin(t)cos(t)

cos(2t)sin(2t) - sin(t)

any better???

14. Originally Posted by thermalwarrior
if i follow this through I get

2cos(2t)sin(2t) - 4sin(t)cos(t) = 0

cos(2t)sin(2t) - 2sin(t)cos(t) = 0

cos(2t)sin(2t) - sin(2t) = 0

any better???
Note the red edits (especially in the last line).

15. okay so now i understand how you got to the final part apart from one thing - where does the -1 come from???

thanks

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