Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

When driving in England, Brian drove through several speed cameras. On one drive, Brian left his hotel and after 7 minutes he was 2.5 km away and at that moment was measured going 0.5 kpm (kilometers per minute). Later, 18 minutes after he started driving, he was 14.8 km from the hotel and at that moment was measured going 0.8 kpm.
A. Find the linearization at 7 minutes.
B. Find the linearization at 18 minutes.
C. Compute two approximations for Brian's distance from his hotel at 14 minutes, using the linearizations from parts a and b.
D. Which approximation do you think is better? Why?

Sagot :

Solution :

The formula for the linearization of a function [tex]$f(x)$[/tex] at a point [tex]$x$[/tex] = a is given as

[tex]$L(x)=f(a)+(x-a)f'(a)$[/tex]

Assuming the time is t and the distance travelled is [tex]$f(t)$[/tex], that makes the speed as [tex]$f'(t)$[/tex].

So substituting them in the linearization formula,

A. At t = 7 minutes

   f(7) = 2.5 km

    f'(7) = 0.5 kpm

  ∴ [tex]$L_7(t)=f(7)+(t-7)f'(7)$[/tex]

             [tex]$=2.5+(t-7)0.5$[/tex]

              [tex]$=2.5+0.5t-3.5$[/tex]

              [tex]$=0.5t-1$[/tex]

B. At t = 18 minutes

      f(18) = 14.8 km

    f'(18) = 0.8 kpm

  ∴ [tex]$L_{18}(t)=f(18)+(t-18)f'(18)$[/tex]

               [tex]$=14.8+(t-18)0.8$[/tex]

              [tex]$=14.8+0.8t-14.4$[/tex]

              [tex]$=0.8t-0.4$[/tex]

C. Substituting the value of t as 14 in both the linearization to determine the position at 14 minutes, we get

[tex]$L_7(14)=0.5(14)-1.0$[/tex]

          = 7 - 1

          = 6 km

[tex]$L_{18}(14)=0.8(14)+0.4$[/tex]

            = 11.2 + 0.4

            = 11.6 km

D. According to the linearization at 7, the distance travelled between the 7 minutes and 14 minutes is = 6 km - 2.5 km

                                           = 3.5 km

And between the 14 minutes and 18 minutes is = 14.8 km - 6 km

                                                                              = 8.8 km

This is an average speed of 0.5 kpm in the first interval and an average speed of 2.2 kpm.

Now, according to the linearization of 18, the distance travelled between the 7 minutes and the 14 minutes is = 11.6 km - 2.5 km

                                                           = 9.1 km

And between 14 minutes and 18 minutes is = 14.8 km - 11.6 km

                                                                       = 3.2 km

This gives an average speed of 1.3 kpm in the first interval and 0.8 kpm in the second interval.

Therefore, the second approximation is the better one since the average speed are closer to the actual readings in the second linearization.  

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.