/*! 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 199 The co-ordinates of a moving par... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 199

The co-ordinates of a moving particle at any time \(t\) are given by \(x=\alpha t^{3}\) and \(y=\beta t^{3}\). The speed of the particle at time \(t\) is given by (A) \(3 t \sqrt{\alpha^{2}+\beta^{2}}\) (B) \(3 t^{2} \sqrt{\alpha^{2}+\beta^{2}}\) (C) \(t^{2} \sqrt{\alpha^{2}+\beta^{2}}\) (D) \(\sqrt{\alpha^{2}+\beta^{2}}\)

Short Answer

Expert verified
The speed of the particle at time \(t\) is given by (B) \(3 t^{2} \sqrt{\alpha^{2}+\beta^{2}}\).

Step by step solution

01

Differentiate the given functions with respect to \(t\)

To find the velocity components, we need to differentiate the given functions \(x = \alpha t^3\) and \(y = \beta t^3\) with respect to \(t\) to get \(v_x = \frac{dx}{dt}\) and \(v_y = \frac{dy}{dt}\): \[ \frac{dx}{dt} = \frac{d(\alpha t^{3})}{dt} = 3 \alpha t^{2} \] \[ \frac{dy}{dt} = \frac{d(\beta t^{3})}{dt} = 3 \beta t^{2} \]
02

Find the speed of the particle

The speed of the particle can be found by determining the magnitude of the velocity vector with components \(v_x\) and \(v_y\). We can use the Pythagorean theorem: \[ v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(3 \alpha t^{2})^{2} + (3 \beta t^{2})^{2}} \]
03

Simplify the expression for the speed

To simplify the expression for the speed \(v\), we can factor out common terms from the terms inside the square root: \[ v = \sqrt{9 \alpha^{2} t^{4} + 9 \beta^{2} t^{4}} = \sqrt{9t^{4}(\alpha^{2} + \beta^{2})} = 3t^{2}\sqrt{\alpha^{2} + \beta^{2}} \]
04

Compare the solution to the given options

Comparing the simplified expression for the speed \(v = 3t^{2}\sqrt{\alpha^{2} + \beta^{2}}\) to the given options, we can see that the correct answer is: (B) \(3 t^{2} \sqrt{\alpha^{2}+\beta^{2}}\)

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Most popular questions from this chapter

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