Studdy AI Logo
Math  /  Calculus

Question2 3- جد مركبات شعاع الوحدة المماسي

Studdy Solution

STEP 1

1. The point M M is described by polar coordinates ρ=2ρ0e−θ \rho = \sqrt{2} \rho_0 e^{-\theta} and θ=ωt \theta = \omega t .
2. ρ0 \rho_0 and ω \omega are positive constants.
3. We need to find the components of the unit tangent vector UTundefined \overrightarrow{U_T} .

STEP 2

1. Express the position vector OMundefined \overrightarrow{OM} in polar coordinates.
2. Derive the velocity vector Vundefined \overrightarrow{V} in polar coordinates.
3. Derive the acceleration vector γundefined \overrightarrow{\gamma} in polar coordinates.
4. Determine the unit tangent vector UTundefined \overrightarrow{U_T} .

STEP 3

The position vector in polar coordinates is given by: OMundefined=ρe^ρ \overrightarrow{OM} = \rho \hat{e}_\rho where ρ=2ρ0e−θ \rho = \sqrt{2} \rho_0 e^{-\theta} and θ=ωt \theta = \omega t .
Substitute the expression for ρ \rho : OMundefined=2ρ0e−ωte^ρ \overrightarrow{OM} = \sqrt{2} \rho_0 e^{-\omega t} \hat{e}_\rho

STEP 4

The velocity vector in polar coordinates is given by: Vundefined=ρ˙e^ρ+ρθ˙e^θ \overrightarrow{V} = \dot{\rho} \hat{e}_\rho + \rho \dot{\theta} \hat{e}_\theta
Calculate ρ˙ \dot{\rho} and θ˙ \dot{\theta} : ρ˙=ddt(2ρ0e−ωt)=−ω2ρ0e−ωt \dot{\rho} = \frac{d}{dt}(\sqrt{2} \rho_0 e^{-\omega t}) = -\omega \sqrt{2} \rho_0 e^{-\omega t} θ˙=ddt(ωt)=ω \dot{\theta} = \frac{d}{dt}(\omega t) = \omega
Substitute these into the velocity vector: Vundefined=(−ω2ρ0e−ωt)e^ρ+(2ρ0e−ωt⋅ω)e^θ \overrightarrow{V} = (-\omega \sqrt{2} \rho_0 e^{-\omega t}) \hat{e}_\rho + (\sqrt{2} \rho_0 e^{-\omega t} \cdot \omega) \hat{e}_\theta

STEP 5

The acceleration vector in polar coordinates is given by: γundefined=(ρ¨−ρθ˙2)e^ρ+(ρθ¨+2ρ˙θ˙)e^θ \overrightarrow{\gamma} = (\ddot{\rho} - \rho \dot{\theta}^2) \hat{e}_\rho + (\rho \ddot{\theta} + 2 \dot{\rho} \dot{\theta}) \hat{e}_\theta
Calculate ρ¨ \ddot{\rho} and substitute known values: ρ¨=ddt(−ω2ρ0e−ωt)=ω22ρ0e−ωt \ddot{\rho} = \frac{d}{dt}(-\omega \sqrt{2} \rho_0 e^{-\omega t}) = \omega^2 \sqrt{2} \rho_0 e^{-\omega t}
Substitute into the acceleration vector: γundefined=(ω22ρ0e−ωt−2ρ0e−ωt⋅ω2)e^ρ+(2(−ω2ρ0e−ωt)⋅ω)e^θ \overrightarrow{\gamma} = (\omega^2 \sqrt{2} \rho_0 e^{-\omega t} - \sqrt{2} \rho_0 e^{-\omega t} \cdot \omega^2) \hat{e}_\rho + (2(-\omega \sqrt{2} \rho_0 e^{-\omega t}) \cdot \omega) \hat{e}_\theta
Simplify: γundefined=0e^ρ+(−2ω22ρ0e−ωt)e^θ \overrightarrow{\gamma} = 0 \hat{e}_\rho + (-2 \omega^2 \sqrt{2} \rho_0 e^{-\omega t}) \hat{e}_\theta

STEP 6

The unit tangent vector UTundefined \overrightarrow{U_T} is in the direction of the velocity vector Vundefined \overrightarrow{V} .
Calculate the magnitude of Vundefined \overrightarrow{V} : ∣Vundefined∣=(−ω2ρ0e−ωt)2+(ω2ρ0e−ωt)2 |\overrightarrow{V}| = \sqrt{(-\omega \sqrt{2} \rho_0 e^{-\omega t})^2 + (\omega \sqrt{2} \rho_0 e^{-\omega t})^2} =2ω2ρ02e−2ωt = \sqrt{2 \omega^2 \rho_0^2 e^{-2\omega t}} =ω2ρ0e−ωt = \omega \sqrt{2} \rho_0 e^{-\omega t}
The unit tangent vector is: UTundefined=Vundefined∣Vundefined∣ \overrightarrow{U_T} = \frac{\overrightarrow{V}}{|\overrightarrow{V}|} =(−ω2ρ0e−ωt)e^ρ+(ω2ρ0e−ωt)e^θω2ρ0e−ωt = \frac{(-\omega \sqrt{2} \rho_0 e^{-\omega t}) \hat{e}_\rho + (\omega \sqrt{2} \rho_0 e^{-\omega t}) \hat{e}_\theta}{\omega \sqrt{2} \rho_0 e^{-\omega t}} =−e^ρ+e^θ = -\hat{e}_\rho + \hat{e}_\theta
The components of the unit tangent vector UTundefined \overrightarrow{U_T} are:
(−1,1) \boxed{(-1, 1)}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord