The shortest distance ...

Level pending

Given the function f ( t ) f(t)

f ( t ) = 1105 t 2 − 356 t + 29 f(t)=\sqrt{1105t^{2}-356t+29}

which describes the distance between the two vectors P t P_{t} and Q t Q_{t} - we would like to know:

What is approximately (rounded to two decimal figures) the minimum distance between vector P t P_{t} and vector Q t Q_{t} ?

Additional information (not really necessary for solving this problem):

P t Q t ⃗ = ( 5 − 32 t 2 − 9 t ) \vec{P_{t}Q_{t}}={5 - 32t \choose 2 - 9t}

0.57 1.39 5.47 0.46

Lukaa Ro by Lukaa Ro , 25, Germany

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Lukaa Ro
May 29, 2014
  • 1st: derive equation: => 2210 t − 356 2 ∗ 1105 t 2 − 356 t + 29 \frac {2210t - 356}{2* \sqrt{1105t^{2}-356t+29} }
  • 2nd: equate the derivative to zero => 2210 t − 356 2 ∗ 1105 t 2 − 356 t + 29 = 0 \frac {2210t - 356}{2* \sqrt {1105t^{2}-356t+29} }=0
  • 3rd: We solve for t. By knowing that 2210 t − 356 2210t - 356 must be equal to zero, we can easily solve 2210 t − 356 = 0 2210t - 356=0 , which gives us: t = 178 1105 t=\frac {178}{1105}
  • 4th: Now, we plug in t = 178 1105 t=\frac {178}{1105} into our original function: 1105 ∗ ( 178 1105 ) 2 − 356 ∗ 178 1105 + 29 \sqrt {1105*(\frac {178}{1105})^{2}-356*\frac {178}{1105}+29}
  • 5th: We calculate this and we get approximately: 0.57 0.57
0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...