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Chapter 2: Problem 28

Find a number \(t\) such that the distance between (-2,1) and \((3 t, 2 t)\) is as small as possible.

Short Answer

Expert verified
The value of t that minimizes the distance between the points (-2, 1) and (3t, 2t) is \(t = -\frac{4}{13}\).

Step by step solution

01

Write down the distance formula

The distance formula for two points (x1, y1) and (x2, y2) in a plane is given by: \(D = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}\)
02

Calculate the distance between the two points in terms of t

Using the distance formula, we can calculate the distance between the points (-2, 1) and (3t, 2t) as follows: \(D(t) = \sqrt{((3t) - (-2))^2 + ((2t) - 1)^2}\)
03

Simplify the expression

Let's simplify the expression to find the distance in terms of t: \(D(t) = \sqrt{(3t + 2)^2 + (2t - 1)^2}\)
04

Minimize the distance

To minimize the distance, we want to find the value of t that makes D(t) as small as possible. Since the square root function is monotonic (i.e., if a > b 鈮� 0, then 鈭歛 > 鈭歜 ), we can minimize the distance by minimizing the squared distance: \(D(t)^2 = (3t + 2)^2 + (2t - 1)^2\) Now, let's differentiate D(t)^2 with respect to t and set the result equal to zero to find the minimum value: \(F(t) = (3t + 2)^2 + (2t - 1)^2\)
05

Differentiate F(t) with respect to t

Differentiating F(t) with respect to t: \(F'(t) = 2(3t + 2)(3) + 2(2t - 1)(2)\)
06

Set F'(t) equal to zero and solve for t

Now we set the derivative equal to zero to find the value of t that minimizes the distance: \(0 = 2(3t + 2)(3) + 2(2t - 1)(2)\) Divide both sides by 2: \(0 = 3(3t + 2) + 2(2t - 1)\) Expanding the above equation gives: \(0 = 9t + 6 + 4t - 2\) Simplifying the equation: \(0 = 13t + 4\) Now, solving for t: \(t = -\frac{4}{13}\)
07

Conclusion

The value of t that minimizes the distance between the points (-2, 1) and (3t, 2t) is \(t = -\frac{4}{13}\).

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