Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 72

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The equation $x = \sqrt{b^2 + c^2 t^2}, t \geq 0$ represents a particle that moves along a horizontal line where $b$ and $c$ are positive constants.

a.) Determine the velocity and acceleration functions.

$ \begin{equation} \begin{aligned} \text{velocity } = \frac{dx}{dt} =& \frac{d}{dt} \sqrt{b^2 + c^2 t^2} = \frac{d}{dt} (b^2 + c^2t^2)^{\frac{1}{2}} \\ \\ \frac{dx}{dt} =& \frac{1}{2} (b^2 + c^2 t^2)^{\frac{-1}{2}} \cdot c^2 (2t) \\ \\ \frac{dx}{dt} =& \frac{\cancel{2} c^2 t}{\cancel{2} \sqrt{b^2 + c^2 t^2}} \\ \\ \frac{dx}{dt} =& \frac{c^2 t}{\sqrt{b^2 + c^2 t^2}} \\ \\ \text{acceleration } = \frac{d^2 x}{dt^2} =& \frac{d}{dt} \frac{c^2 t}{\sqrt{b^2 + c^2 t^2}} \\ \\ \frac{d^2 x}{dt^2} =& \frac{\displaystyle \sqrt{b^2 + c^2 t^2} \cdot c^2 \frac{d}{dt} (t) - (c^2t) \cdot \frac{d}{dt} \sqrt{b^2 + c^2t^2}}{(\sqrt{b^2 + c^2t^2})^2} \\ \\ \frac{d^2 x}{dt^2} =& \frac{\displaystyle \sqrt{b^2 + c^2t^2} \cdot c^2(1) - (c^2 t) \cdot \frac{\cancel{2} c^2 t}{\cancel{2} \sqrt{b^2 + c^2t^2}} }{b^2 + c^2 t^2} \\ \\ \frac{d^2 x}{dt^2} =& c^2 \frac{\displaystyle \left[ \sqrt{b^2 + c^2 t^2} - \frac{c^2 t^2}{\sqrt{b^2 + c^2 t^2}} \right]}{b^2 c^2 t^2} \\ \\ \frac{d^2 x}{dt^2} =& c^2 \frac{\displaystyle \frac{b^2 + c^2t^2 - c^2 t^2}{\sqrt{b^2 + c^2 t^2}}}{b^2 + c^2 t^2} \\ \\ \frac{d^2 x}{dt^2} =& \frac{b^2 c^2}{(b^2 + c^2t^2)^{\frac{3}{2}}} \end{aligned} \end{equation} $

b.) Show that the particle always moves in positive direction.

The particle is moving in a positive direction if $v(t) > 0$ that is $\displaystyle \frac{c^2 t}{\sqrt{b^2 + c^2 t^2}} > 0$. Sinc the denominator is always positive, recall that the square root function is defined only for positive solution. Therefore, the sign of the velocity function is affected only by the sign of the numerator. But we know that the function is defined only for $t \geq 0$ and we have $c$ as positive constant. Hence, $v(t)$ is always positive and the particle always moves in the positive direction.

$ \begin{equation} \begin{aligned} h'(x) =& \frac{1}{2} \left[ \frac{f9x)}{g(x)} \right] ^{\frac{-1}{2}} \cdot \left( \frac{g(x) f'(x) - f(x) g'(x)}{[g(x)]^2} \right] \\ \\ h'(x) =& \frac{g(x) f'(x) - f(x) g'(x)}{\displaystyle 2 \sqrt{\frac{f(x)}{g(x)}} [g(x)]^2 } \\ \\ h' =& \frac{gf' - fg'}{2 \sqrt{fg^3}} \end{aligned} \end{equation} $