- #1

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HOW CAN I CALCULATE ITS FINAL POINT (X2,Y2)?

like I know

(X2-X1)^2 = R^2 - (Y2-Y1)^2

tanθ= [Y2-Y1]/[X2-X1]

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- Thread starter Morgoth
- Start date

- #1

- 126

- 0

HOW CAN I CALCULATE ITS FINAL POINT (X2,Y2)?

like I know

(X2-X1)^2 = R^2 - (Y2-Y1)^2

tanθ= [Y2-Y1]/[X2-X1]

- #2

- 22,129

- 3,297

HOW CAN I CALCULATE ITS FINAL POINT (X2,Y2)?

like I know

(X2-X1)^2 = R^2 - (Y2-Y1)^2

tanθ= [Y2-Y1]/[X2-X1]

The last equation gives us

[tex]Y_2=Y_1+\tan(\theta)(X_2-X_1)[/tex]

Now plug this value for [itex]Y_2[/itex] in the equation

[tex](X_2-X_1)^2 + (Y_2-Y_1)^2=R^2[/tex]

and solve for [itex]X_2[/itex] (you will get two values, why?, which value you want depends on [itex]\theta[/itex]).

- #3

Fredrik

Staff Emeritus

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Hint: Do you also know what (Y2-Y1)/R and (X2-X1)/R are?

- #4

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Edit:I wonder how many times I've made a post only to find that micromass replied a minute earlier.

You snooze, you lose :tongue2:

But your to this question answer is way better than mine!! I didn't even think of something like that.

- #5

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thanks everyone. I think i thought of a faster way, like using:

(x2-x1)= r cosθ

(y2-y1)= r sinθ

(x2-x1)= r cosθ

(y2-y1)= r sinθ

- #6

Fredrik

Staff Emeritus

Science Advisor

Gold Member

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That's exactly what I suggested.thanks everyone. I think i thought of a faster way, like using:

(x2-x1)= r cosθ

(y2-y1)= r sinθ

- #7

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- 0

Now about the question to "why"

prolly because a vector showing to the -x or +x makes no difference for measuring its magnitude (r=+/- sqrt[Δx^2 + Δy^2, however - has no meaning, for you define r>=0). it's a problem of θ to solve.

- #8

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Knowing the initial point of a vector (X1,Y1) and its magnitude and angle (R,θ)

HOW CAN I CALCULATE ITS FINAL POINT (X2,Y2)?

An interesting variation of this problem is if θ is the bearing from North as used by surveyors and navigators.

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