Question

7. (1 point) A fighter plane, which can shoot a laser beam straight ahead, travels along the path ~r(t) = 5 − t, 21 − t 2 , 3 − t 3 27 Show that there is precisely one time t at which the pilot can hit a target located at the origin.

Answer 1

Answer:

At t = 3, pilot can hit the target located at the origin.

Step-by-step explanation:

We have been given displacement vector:

r(t) = 5 - t, 21 - t², 3-t³/27

so, in this case r(t) and r'(t) should be in the parallel but opposite direction.

In order to get r'(t) we need to differentiate r(t).

r(t) = 5 - t, 21 - t², 3-t³/27

r'(t) = -1, -2t, -t²/9

Now perform the cross product among these two r(t) and r'(t).

r'(t) x r(t) = $\left[\begin{array}{ccc}i&j&k\\-1&-2t&\frac{-t^{2} }{9} \\5-t&21-t^{2} &3-\frac{t^{3} }{27} \end{array}\right]$

            = i ((-6t + $\frac{2t^{4} }{27}$ )+ ( $\frac{21t^{2} }{9}$ - $\frac{t^{2} }{9}$ )) -j ((-3 + $\frac{t^{3} }{27}$ + $\frac{5t^{2} }{9}$ - $\frac{t^{2} }{9}$ )) + k ((-21 + $t^{2}$  + 10t -2$t^{2}$))

           = ($\frac{-t^{4} }{27}$ + $\frac{7t^{2} }{3}$ - 6t )i + ($\frac{2t^{3} }{27}$ - $\frac{5t^{2} }{9}$ +3) j + (-t² + 10t -21)k

In order to find the value of t, we need to put

$\frac{-t^{4} }{27}$ + $\frac{7t^{2} }{3}$ - 6t = 0

-t² + 10t -21 = 0

So, after solving for t, we will get

- ( t-3) (t-7) = 0

t = 3 or t = 7

In this case, only t = 3 satisfies the other two equations as well. t=7 is not satisfying. So take t =3 as the time. and for further assurance, we need to check are our vectors r(t) and r'(t) opposite at t = 3 or not. Let's check it out.

r(3) = 5-3, 21 - 3²,  3 - 3³/27

r (3) = 2, 12, 2

r'(3) = -1, -2(3) , -3²/9

r'(3) = -1, -6 -1

Here, we can easily see that, r(3) = -2 r'(3) which is opposite and hence it is proved that, at exactly t = 3, pilot can hit a target located at the origin.