/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Find a number \(t\) such that th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 13

Find a number \(t\) such that the distance between (2,3) and \((t, 2 t)\) is as small as possible.

Short Answer

Expert verified
The value of \(t\) such that the distance between (2,3) and \((t, 2t)\) is minimized is \(t = \frac{8}{5}\).

Step by step solution

01

Write down the distance formula

The distance between two points (x1, y1) and (x2, y2) is given by the formula: \[D = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}\]
02

Substitute the given points into the distance formula

In this case, (x1, y1) = (2, 3) and (x2, y2) = (t, 2t). Substituting these values into the distance formula, we get: \[D = \sqrt{(t - 2)^2 + (2t - 3)^2}\]
03

Find the derivative of the distance function with respect to t

To minimize the distance, we need to find the value of t at which the derivative of the distance function is 0. So let's differentiate the function D with respect to t. Our main goal should be to find the derivative of \(D^2\) with respect to t, as minimizing D and D^2 will give the same result, and it will be much easier. \[D^2 = (t - 2)^2 + (2t - 3)^2\]Now differentiate D^2 with respect to t: \[\frac{d(D^2)}{dt} = 2(t - 2) + 4(2t - 3)\]
04

Set the derivative equal to 0 and solve for t

Now, set the derivative equal to 0 and solve for t: \[2(t - 2) + 4(2t - 3) = 0\]Expanding and simplifying the equation: \[2t - 4 + 8t - 12 = 0\]Combine like terms: \[10t - 16 = 0\]Now, isolate t: \[t = \frac{16}{10}\]Simplifying the fraction, we get: \[t = \frac{8}{5}\]
05

Substitute the value of t back into the point (t, 2t)

Finally, substitute the value of t back into the point (t, 2t) to find the coordinates of the point with the minimum distance between (2,3) and (t, 2t): \[(t, 2t) = \left(\frac{8}{5}, 2\left(\frac{8}{5}\right)\right) = \left(\frac{8}{5}, \frac{16}{5}\right)\] The value of t such that the distance between (2,3) and (t, 2t) is minimized is \(t = \frac{8}{5}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that \(\sinh x \approx x\) if \(x\) is close to 0 [The definition of sinh was given before Exercise 52 in Section \(4.3 .\)

About how many years does it take for \(\$ 300\) to become \(\$ 2,400\) when compounded continuously at \(5 \%\) per year?

Suppose a colony of bacteria has a continuous growth rate of \(35 \%\) per hour. How long does it take the colony to triple in size?

Suppose a colony of bacteria has doubled in two hours. What is the approximate continuous growth rate of this colony of bacteria?

(a) Using a calculator or computer, verify that $$ \left(1+\frac{\ln 10}{x}\right)^{x} \approx 10 $$ for large values of \(x\) (for example, try \(x=1000\) and then larger values of \(x)\) (b) Explain why the approximation above follows from the approximation \(\left(1+\frac{r}{x}\right)^{x} \approx\) \(e^{r}\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.