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Chapter 2: Problem 27

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 \(t = \frac{8}{5}\).

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