A particle is moving in the plane according to the law x = 162 - t^3, y = 59 - t^2 (0...

Question:

A particle is moving in the plane according to the law {eq}x = 162 - t^3, y = 59 - t^2 (0 \leq t \leq 3) {/eq}Starting from the time {eq}t = 3 {/eq}, the particle is moving with constant velocity - the same velocity it had at the time {eq}t = 3 {/eq}. At what time and at what point will it hit a coordinate axis for the first time?

Calculus:

A mathematical quantity that represents the analyse of the continuous variation of the function with respect to other variable is known as calculus. The differentiation and integration are sub-parts to help in analyses of the equation.

Answer and Explanation:


Given Data:

  • The x coordinate is: {eq}x = 162 - {t^3} {/eq}
  • The y coordinate is: {eq}y = 59 - {t^2} {/eq}
  • {eq}0 \le t \le 3 {/eq}


The rearrange the y coordinates

{eq}\begin{align*} y &= 59 - {t^2}\\ {t^2} &= 59 - y\\ t &= \sqrt {59 - y} \end{align*} {/eq}


Substitute the value and solve the x coordinate

{eq}\begin{align*} x &= 162 - {\left( {\sqrt {59 - y} } \right)^3}\\ {\left( {\sqrt {59 - y} } \right)^3} &= 162 - x\\ 59 - y &= {\left( {162 - x} \right)^{\dfrac{2}{3}}}\\ y &= 59 - {\left( {162 - x} \right)^{\dfrac{2}{3}}} \end{align*} {/eq}


Differentiate the above expression with respect to {eq}t {/eq}

{eq}\begin{align*} \dfrac{{dy}}{{dt}} &= \dfrac{{d\left( {59 - {{\left( {162 - x} \right)}^{\dfrac{2}{3}}}} \right)}}{{dt}}\\ &= - \dfrac{2}{3}{\left( {162 - x} \right)^{ - \dfrac{1}{3}}} \end{align*} {/eq}


Equate the differentiation of y to zero to get critical points

{eq}\begin{align*} \dfrac{{dy}}{{dt}} &= 0\\ - \dfrac{2}{3}{\left( {162 - x} \right)^{ - \dfrac{1}{3}}} &= 0\\ 162 - x &= 0\\ x &= 162 \end{align*} {/eq}


Substitute the value {eq}x = 162 {/eq} in x coordinates

{eq}\begin{align*} 162 &= 162 - {t^3}\\ t &= 0 \end{align*} {/eq}


Substitute the value {eq}t = 0 {/eq} in y coordinates

{eq}\begin{align*} y &= 59 - {\left( 0 \right)^2}\\ &= 59 \end{align*} {/eq}


Thus the at time {eq}t = 0 {/eq} it hit a coordinates {eq}\left( {162,59} \right) {/eq}


Learn more about this topic:

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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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