Question

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Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tables, or other objects that you use. Justifications require that you give mathematical reasons, and that you verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit.

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A particle moves along the l-axis so that its position at time t>0 is given by x(t)= (t^2 - 9)/(3t^2 + 8)

Show that the velocity of the particle at time t is given by v(t)=70t/(30t^2 + 8)^2

Answer 1

Answer:

We showed that if we have the position of the particle $x(t)= \frac{t^2 - 9}{3t^2 + 8}$ the velocity of the particle at time t is given by $v(t)=\frac{70t}{\left(3t^2+8\right)^2}$.

Step-by-step explanation:

Velocity is defined as the rate of change of position with respect to time.

                                                  $v(x)=\frac{dx}{dt}$

To find velocity, we take the derivative of the position function $x(t)= \frac{t^2 - 9}{3t^2 + 8}$

$\mathrm{Apply\:the\:Quotient\:Rule}:\quad \left(\frac{f}{g}\right)'=\frac{f\:'\cdot g-g'\cdot f}{g^2}$

$\frac{d}{dt}\left(\frac{t^2-9}{3t^2+8}\right)=\frac{\frac{d}{dt}\left(t^2-9\right)\left(3t^2+8\right)-\frac{d}{dt}\left(3t^2+8\right)\left(t^2-9\right)}{\left(3t^2+8\right)^2}$

Next, we find the values of $\frac{d}{dt}\left(t^2-9\right)$ and $\frac{d}{dt}\left(3t^2+8\right)$

$\frac{d}{dt}\left(t^2-9\right)=\frac{d}{dt}\left(t^2\right)-\frac{d}{dt}\left(9\right)=2t-0=2t$

$\frac{d}{dt}\left(3t^2+8\right)=\frac{d}{dt}\left(3t^2\right)+\frac{d}{dt}\left(8\right)=6t+0=6t$

So,

$\frac{d}{dt}\left(\frac{t^2-9}{3t^2+8}\right)=\frac{2t\left(3t^2+8\right)-6t\left(t^2-9\right)}{\left(3t^2+8\right)^2}$

Next, we expand $2t\left(3t^2+8\right)-6t\left(t^2-9\right)$

$2t\left(3t^2+8\right)-6t\left(t^2-9\right)=6t^3+16t-6t\left(t^2-9\right)=6t^3+16t-6t^3+54t=70t$

Therefore,

$v(t)=\frac{d}{dt}\left(\frac{t^2-9}{3t^2+8}\right)=\frac{70t}{\left(3t^2+8\right)^2}$