You notice a bug flying around the room. The distance s(t) (in feet) that the bug has traveled...

Question:

You notice a bug flying around the room. The distance s(t) (in feet) that the bug has traveled after t seconds is given by

{eq}s\left ( t \right ) = \int_{0}^{3t^2}\sqrt{2x^2 + cos^2\left ( \pi x \right )}dx {/eq}

a. Find the speed {eq}\frac{\mathrm{d} s}{\mathrm{d} t} {/eq} of the bug at time {eq}t = \frac{1}{2} {/eq} seconds.

b. After t = 1 second, the bug has traveled a distance.

{eq}s\left ( 1 \right ) = \int_{0}^{3}\sqrt{2x^2 + cos^2\left ( \pi x \right )}dx {/eq}

Fundamental Theorem of Calculus - Part I


Part I of the Fundamental Theorem of Calculus states that

{eq}\displaystyle \frac{d}{dx}\left(\int_a^x f(t)\ dt \right)=f(x), \text{ where } a \text{ is constant}. {/eq}

Or the Chain rule form {eq}\displaystyle \frac{d}{dx}\left(\int_a^{u(x)} f(t)\ dt\right) =f(u(x))\cdot u'(x), \text{ where } '=\frac{d}{dx}. {/eq}

It is important to have a constant as a lower limit of integration and as the upper limit of integration,

to have a function of {eq}x {/eq}, the variable with respect to which the derivative is taken.

If not, we need to use the properties of the definite integrals to bring it to the required form.

Answer and Explanation:


a. The speed of the bug whose position function is given as {eq}\displaystyle s\left ( t \right ) = \int_{0}^{3t^2}\sqrt{2x^2 + cos^2\left ( \pi x \right )}dx {/eq}

is obtained by taken the derivative with respect to {eq}\displaystyle t, {/eq} as below.

{eq}\displaystyle \begin{align} \text{speed }&=s'(t)=\frac{d}{dt}\left(\int_{0}^{3t^2}\sqrt{2x^2 + cos^2\left ( \pi x \right )}dx\right)\\ &= \sqrt{2(3t^2)^2 + cos^2\left ( \pi \cdot 3t^2 \right )}\cdot \frac{d}{dt}\left(3t^2\right), &\left[\text{ using the Fundamental theorem of calculus part I}\right]\\ &= 6t\sqrt{18t^4 + \cos^2\left (3 \pi t^2 \right )}\\ \implies s'\left(\frac{1}{2}\right)&=3\sqrt{18\cdot \frac{1}{4} + \cos^2\left (3 \pi \frac{1}{4} \right )}\\ &=3\sqrt{\frac{9}{2} + \frac{1}{2} }\\ &=\boxed{3\sqrt{5}}. \end{align} {/eq}


b. The distance traveled in the first one second is {eq}\displaystyle s\left ( 1 \right ) = \int_{0}^{3}\sqrt{2x^2 + cos^2\left ( \pi x \right )}dx\boxed{\approx 6.88} {/eq} using numerical integration, Riemann sum, because otherwise is challenging to solve it analytically.


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The Fundamental Theorem of Calculus

from Math 104: Calculus

Chapter 12 / Lesson 10
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