What is a parameter and how to solve? (1 Viewer)

InteGrand · Mar 28, 2015

bubb said:
View attachment 31967

View attachment 31968

For the first picture, the displacement vector of the plane is found by subtracting initial coordinates from final coordinates. This means the displacement vector is $\bold{x} = 6\bold{i}-12\bold{j}+12\bold{k}$ .

So the distance travelled in the 1 minute is $|\bold{x}| = \sqrt{6^2 + (-12)^2 + 12^2} = 18 \mathrm{ km}$ . So the velocity of the plane is 18 km/min (i.e. $18\times 60 = 1080 \mathrm{ km/h}$ ) in the direction of $6\bold{i}-12\bold{j}+12\bold{k}$ . (I assume you're asking for the velocity of the plane.)
Edit: just realised the second image contains the questions

braintic · Mar 28, 2015

What are they getting at in part c? Surely time itself would be used as the parameter. Are they saying m is the time?

InteGrand · Mar 28, 2015

bubb said:
View attachment 31967

View attachment 31968

c) The fighter's position vector r is given by $\bold{r} = \bold{r_0} + t \bold{v}$ , where $\bold{v}$ is the velocity vector and $\bold{r_0}$ is the initial position of the fighter, and t corresponds to the time elapsed (you could use m instead of t as the parameter, but t seems more natural to me).

The velocity vector is found by dividing the displacement vector in my previous post by the time taken (which is $\frac{1}{60}\mathrm{ h}$ ).

Hence $\bold{v}=360\bold{i}-720\bold{j}+720\bold{k}$ , where the units for the components of this vector are km/h. (Units of the scalar t are hours, so $t\bold{v}$ has units of km.)

So the fighter's position vector is $\bold{r} = \left< 2\bold{i}+8\bold{j}+\bold{k}\right> + t \left<360\bold{i}-720\bold{j}+720\bold{k}\right>$ for $t\in \mathbb{R}:t\geq 0$

d) The position vector of the fighter is $\left<2+360t,8-720t,1+720t\right>$ . Hence the squared distance of the fighter from the station (the origin) at any time is $s=(2+360t)^2 + (8-720t)^2 + (1+720t)^2$ .

To minimise distance, we can minimise squared distance.

$\frac{\mathrm{d} s}{\mathrm{d} t} = 720(2+360t)-1440(8-720t)+1440(1+720t)$ .

Set this to 0 to find minimum distance (if the solution is a positive t, this corresponds to a positive time, hence a feasible solution).

$720(2+360t)-1440(8-720t)+1440(1+720t)=0 \Rightarrow t = \frac{1}{270}$ .

Hence the fighter plane is closest when $\frac{1}{270}$ hours have passed.

Plugging in this value of t to the fighter's position vector, we get the required point to be $\left(\frac{10}{3}, \frac{16}{3}, \frac{1}{3}\right)$ .

e) required distance is $\sqrt{\left( \frac{10}{3}\right)^2 + \left( \frac{16}{3}\right)^2 + \left( \frac{1}{3}\right)^2 }} \approx 6.30\mathrm{ km}$ .

f) Done in my previous post. 1080 km/h

g) The fighter's rate of ascent or descent is given by the third component of the velocity vector, which is +720. Hence the plane is ascending at a rate of 720 km/h, or, in more natural units, 200 m/s.

Trebla · Mar 28, 2015

please post questions outside the HSC course in the Extracurricular Topics forum

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What is a parameter and how to solve? (1 Viewer)

bubb

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InteGrand

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braintic

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InteGrand

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Trebla

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