Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 14

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At what rate is the distance between te ships changing at 4:00 PM?

a.) State the quantities given in the problem.

b.) State the unknown.

c.) Illustrate a picture of the situation for any time $t$.

d.) Write an expression that relates the quantities.

e.) Solve the problem

a.) Given:

$\qquad $ distance of ship A to ship B = $150 km$

$\qquad $ speed of ship A due east = $35 km/h$

$\qquad $ speed of ship B due north = $25 km/h$

b.) Required: Distance between the ships when $t = 4$

c.)

d.) By using Pythagorean Theorem,

$ \begin{equation} \begin{aligned} z^2 = x^2 + y^2 \qquad \text{Equation 1} \end{aligned} \end{equation} $

e.) By getting the derivative with respect to time,

$ \begin{equation} \begin{aligned} \cancel{2} z \frac{dz}{dt} =& \cancel{2}x \frac{dx}{dt} + \cancel{2}y \frac{dy}{dt} \\ \\ \frac{dz}{dt} =& \frac{\displaystyle x \frac{dx}{dt} + y \frac{ydy}{dt}}{z} \qquad \text{Equation 2} \end{aligned} \end{equation} $

To get the value of $x, y$ and $z$, we substitute the instance $t = 4$ to their corresponding speed.

$ \begin{equation} \begin{aligned} x =& (35 km/h) (4h) = 140 km \\ \\ y =& (25 km/h) (4h) = 100 km \end{aligned} \end{equation} $

We have the initial distance of $150 km$ between the ship and we got $140 km$ at time = $4$ and we know that ship A is approaching ship B so the equivalent distance $x = 150 - 140 = 10 km$, Also $\displaystyle \frac{d}{dx}$ is negative because the rate is decreasing.

$z = \sqrt{x^2 + y^2}$ (from equation 1) = $\sqrt{10^2 + 100^2} = 10 \sqrt{101} km$

$\displaystyle \frac{dz}{dt} = \frac{10(-35) + 100(25)}{10 \sqrt{101}}$

$ \begin{equation} \begin{aligned} & \boxed{\displaystyle \frac{dz}{dt} = 21.40 km/h} \end{aligned} \end{equation} $