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Irina18 [472]
2 years ago
6

Need help on HS geometry

Mathematics
2 answers:
natita [175]2 years ago
8 0
  • Here, RY = PY.
  • Angle QRY = Angle QPY = 90°
  • QY = QY [common]
  • Therefore, ∆ QRY = ∆ QPY
  • By c.p.c.t., angle RQY = angle PQY
  • Therefore,
  • 16x - 27 = 5x + 6 \\  =  > 16x - 5x = 6 + 27 \\  =  > 11x = 33 \\  =  > x = 3
  • Therefore, the measure of angle YQR
  • = (16x - 27)° \\  = (16 \times 3 - 27)° \\  = (48 - 27)° \\  = 21°

<u>Answers:</u>

  1. 3
  2. 21°

Hope you could get an idea from here.

Doubt clarification - use comment section.

Illusion [34]2 years ago
6 0

\\ \tt\Rrightarrow 16x-27=5x+6

\\ \tt\Rrightarrow 16x-5x=6+27

\\ \tt\Rrightarrow 11x=33

\\ \tt\Rrightarrow x=3

  • m<YQR=16(3)-27=48-27=21°
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interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm
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Answer:

The curvature is \kappa=1

The tangential component of acceleration is a_{\boldsymbol{T}}=0

The normal component of acceleration is a_{\boldsymbol{N}}=1 (2)^2=4

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}

where

\boldsymbol{T}} is the unit tangent vector.

\frac{ds}{dt}=|| \boldsymbol{r}'(t)}|| is the speed of the object

We need to find \boldsymbol{r}'(t), we know that \boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k} so

\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}

Next , we find the magnitude of derivative of the position vector

|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2

The unit tangent vector is defined by

\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}

\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)

We need to find the derivative of unit tangent vector

\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})

And the magnitude of the derivative of unit tangent vector is

||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2

The curvature is

\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1

The tangential component of acceleration is given by the formula

a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}

We know that \frac{ds}{dt}=|| \boldsymbol{r}'(t)}|| and ||\boldsymbol{r}'(t)}||=2

\frac{d}{dt}\left(2\right)\: = 0 so

a_{\boldsymbol{T}}=0

The normal component of acceleration is given by the formula

a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2

We know that \kappa=1 and \frac{ds}{dt}=2 so

a_{\boldsymbol{N}}=1 (2)^2=4

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Hello there!
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(x^2+3)^2 + 7x^2+21=-10

We want to convert this equation in terms of m, where m = x^2+3

While solving this problem, I instantly noticed the x^2+3 in the first term of the equation. Replace that with m.

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Simplifying the rest is not as obvious. Using the distributive property, we can change [7x^2+21] to [7(x^2+3)]

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3 years ago
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