# A) One airplane is approaching an airport from the north at 184 kilometer per hour. A second...

## Question:

A) One airplane is approaching an airport from the north at {eq}184 {/eq} kilometer per hour. A second airplane approaches from the east at {eq}210 {/eq} kilometer per hour. Find the rate at which the distance between the planes changes when the southbound plane is {eq}33 km {/eq} away from the airport and the westbound plane is {eq}22 km {/eq} from the airport.

B) A man {eq}6 ft {/eq} tall walks at a rate of {eq}3 {/eq} feet per second away from a lamppost that is {eq}21 ft {/eq} high. At what rate is the length of his shadow changing when he is {eq}30 ft {/eq} away from the lamppost?

C) A piece of land is shaped like a right triangle. Two people start at the right angle of the triangle at the same time, and walk at the same speed along different legs of the triangle. If the area formed by the positions of the two people and their starting point (the right angle) is changing at {eq}2 m^2 {/eq} per second, then how fast are the people moving when they are {eq}3 m {/eq} from the right angle?

## Rate Of Change:

The rate of change of a function is the ratio of the infinitesimal change in the value of the function to the infinitesimal change in the value of the dependent variable.

This rate of change is also obtained by differentiating the function with respect to the dependent variable.

Consider two variables related by {eq}y= f \left( x \right) {/eq} such that they both are dependent on another variable {eq}t {/eq}, then their rate of change with respect to {eq}t {/eq} are also related and can be calculated as

{eq}\dfrac{{dy}}{{dt}} = \dfrac{{df}}{{dx}}\left( {\dfrac{{dx}}{{dt}}} \right) {/eq}

## Answer and Explanation:

A)

The distance between the plane is given by;

{eq}D = \sqrt {{x^2} + {y^2}} {/eq}

Where, {eq}x {/eq} and {eq}y {/eq} are the distances of the planes from the airport.

Differentiating it with respect to time, we get;

{eq}\dfrac{d}{{dt}}\left( D \right) = \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}\left( {\dfrac{{dx}}{{dt}}} \right) + \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}\left( {\dfrac{{dy}}{{dt}}} \right) {/eq}

Given,

{eq}\eqalign{ & x = 33km \cr & y = 22km \cr & \dfrac{{dx}}{{dt}} = - 184km/hr \cr & \dfrac{{dy}}{{dt}} = - 210km/hr \cr} {/eq}

(The -ve sign indicates that the distance is decreasing with time.)

Substituting, we get;

{eq}\dfrac{d}{{dt}}\left( D \right) = \dfrac{{33}}{{\sqrt {{{\left( {33} \right)}^2} + {{\left( {22} \right)}^2}} }}\left( { - 184} \right) + \dfrac{{22}}{{\sqrt {{{\left( {33} \right)}^2} + {{\left( {22} \right)}^2}} }}\left( { - 210} \right) = - \dfrac{{998}}{{\sqrt {13} }} \approx - 276.795km/hr {/eq}

B)

The lamppost, the man and the tip of the shadow form two right-angled triangle, one projected over the other.

Let the distance of the man from the lamppost at any instant be {eq}x {/eq} and the distance of the man from the tip of the shadow at any instant be {eq}y {/eq}.

Therefore, using the similarity of triangles, we can say that;

{eq}\eqalign{ & \dfrac{{21}}{{x + y}} = \dfrac{6}{y} \cr & 21y = 6x + 6y \cr & y = \dfrac{2}{5}x \cr} {/eq}

Differentiating it with respect to time, we get;

{eq}\dfrac{{dy}}{{dt}} = \dfrac{2}{5}\left( {\dfrac{{dx}}{{dt}}} \right) {/eq}

Given,

{eq}\dfrac{{dx}}{{dt}} = 3 {/eq}

Substituting it in the above equation, we get,

{eq}\dfrac{{dy}}{{dt}} = \dfrac{2}{5}\left( 3 \right) = \dfrac{6}{5} = 1.2ft{s^{ - 1}} {/eq}

(Note: The rate of change of the length is independent of the instantaneous position of the man.)

C)

Let the length of one leg of the triangle be {eq}x {/eq} and the other be {eq}y {/eq}.

Now, since both the people are walking at the same speed, we can say;

{eq}\dfrac{{dx}}{{dt}} = \dfrac{{dy}}{{dt}} {/eq}

The area of the triangle will be given by;

{eq}A=x \times y {/eq}

Differentiating it with respect to time, we get;

{eq}\eqalign{ & \dfrac{{dA}}{{dt}} = y\left( {\dfrac{{dx}}{{dt}}} \right) + x\left( {\dfrac{{dy}}{{dt}}} \right) \cr & \dfrac{{dA}}{{dt}} = y\left( {\dfrac{{dx}}{{dt}}} \right) + x\left( {\dfrac{{dx}}{{dt}}} \right) \cr & \dfrac{{dA}}{{dt}} = \left( {x + y} \right)\left( {\dfrac{{dx}}{{dt}}} \right) \cr} {/eq}

Given,

{eq}\eqalign{ & \dfrac{{dA}}{{dt}} = 2{m^2}/\sec \cr & y = 3m \cr & x = 3m \cr} {/eq}

Therefore, substituting it in the above relation, we get;

{eq}\eqalign{ & 2 = \left( {3 + 3} \right)\dfrac{{dx}}{{dt}} \cr & \dfrac{{dx}}{{dt}} = \dfrac{1}{3} \approx 0.33m/\sec \cr} {/eq}

#### Learn more about this topic:

from Common Core Math Grade 8 - Functions: Standards

Chapter 4 / Lesson 4