a particle moves along the curve y = -2x² - 4x - 9 in such a way that as it moves, dx/dt = 8 units/minute…

Question 13

Explanation:

Step1: Differentiate $y$ with respect to $t$

Given $y = - 2x^{2}-4x - 9$, using the chain - rule $\frac{dy}{dt}=( - 4x - 4)\frac{dx}{dt}$. Substitute $x = 6$ and $\frac{dx}{dt}=8$ into the equation. So $\frac{dy}{dt}=(-4\times6 - 4)\times8=(-24 - 4)\times8=-28\times8=-224$ units/minute. The distance $s$ from the point $(x,y)$ to the origin is $s=\sqrt{x^{2}+y^{2}}$. Differentiating both sides with respect to $t$ using the chain - rule gives $\frac{ds}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{\sqrt{x^{2}+y^{2}}}$. Substitute $x = 6$, $y=-105$, $\frac{dx}{dt}=8$, and $\frac{dy}{dt}=-224$ into the formula: [ \begin{align*} \frac{ds}{dt}&=\frac{6\times8+( - 105)\times(-224)}{\sqrt{6^{2}+(-105)^{2}}}\ &=\frac{48 + 105\times224}{\sqrt{36 + 11025}}\ &=\frac{48+23520}{\sqrt{11061}}\ &=\frac{23568}{\sqrt{11061}}\approx\frac{23568}{105.171}\approx224.096 \end{align*} ]

Answer:

$224.096$

Question 14

Explanation:

Step1: Find the angular velocity $\omega$

The lighthouse beam makes a full - turn (i.e., $2\pi$ radians) every $5$ seconds. So the angular velocity $\omega=\frac{2\pi}{5}$ rad/s. Let $x$ be the distance along the shoreline from the point on the shore closest to the lighthouse and $d = 450$ be the distance from the lighthouse to the shore. Let $\theta$ be the angle between the beam and the line perpendicular to the shore. We know that $\tan\theta=\frac{x}{450}$, so $x = 450\tan\theta$.

Step2: Differentiate $x$ with respect to $t$

Differentiating $x = 450\tan\theta$ with respect to $t$ using the chain - rule gives $\frac{dx}{dt}=450\sec^{2}\theta\frac{d\theta}{dt}$. When the beam meets the shoreline at a point $1200$ meters from the lighthouse, $\sec\theta=\frac{\sqrt{450^{2}+1200^{2}}}{450}=\frac{\sqrt{202500 + 1440000}}{450}=\frac{\sqrt{1642500}}{450}=\frac{1281.601}{450}$. And $\frac{d\theta}{dt}=\omega=\frac{2\pi}{5}$ rad/s. [ \begin{align*} \frac{dx}{dt}&=450\times\left(\frac{\sqrt{450^{2}+1200^{2}}}{450}\right)^{2}\times\frac{2\pi}{5}\ &=450\times\frac{450^{2}+1200^{2}}{450^{2}}\times\frac{2\pi}{5}\ &=(450 + \frac{1200^{2}}{450})\times\frac{2\pi}{5}\ &=(450+\frac{1440000}{450})\times\frac{2\pi}{5}\ &=(450 + 3200)\times\frac{2\pi}{5}\ &=3650\times\frac{2\pi}{5}\ &=1460\pi\approx4581.459 \end{align*} ]

Answer:

$4581.459$

Question 15

Explanation:

Step1: Define the variables and their rates of change

Let $y$ be the position of the first car (north - south direction), $x$ be the position of the second car (east - west direction), and $s$ be the distance between the two cars. So $s=\sqrt{x^{2}+y^{2}}$. We know that $y = 200-15t$ and $x = 130 + 18t$. After $t = 3$ seconds, $y=200-15\times3=200 - 45 = 155$ meters and $x=130+18\times3=130 + 54 = 184$ meters. Differentiating $s=\sqrt{x^{2}+y^{2}}$ with respect to $t$ using the chain - rule gives $\frac{ds}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{\sqrt{x^{2}+y^{2}}}$. We have $\frac{dx}{dt}=18$ m/s and $\frac{dy}{dt}=-15$ m/s.

Step2: Calculate $\frac{ds}{dt}$

Substitute $x = 184$, $y = 155$, $\frac{dx}{dt}=18$, and $\frac{dy}{dt}=-15$ into the formula: [ \begin{align*} \frac{ds}{dt}&=\frac{184\times18+155\times(-15)}{\sqrt{184^{2}+155^{2}}}\ &=\frac{3312-2325}{\sqrt{33856 + 24025}}\ &=\frac{987}{\sqrt{57881}}\ &=\frac{987}{240.585}\approx4.095 \end{align*} ]

Answer:

$4.095$