Question #180468

1. An observer stands 600 meters from the launch pad of a rocket, watches it       ascend vertically at 100 m/s.


a.  Find the rate of change of the distance between the rocket and the observer. (Hint: Use Pythagorean Theorem)


 b.    Find the distance between the observer and the rocket when the rocket is 400 meters high.

  


 c.   What is the rate of change of the distance between the rocket and the observer when the rocket is 400 meters high?


1
Expert's answer
2021-04-13T11:10:06-0400

To be given in question

Observer and rocket between distance  =600meter

Vertically velocity of rocket =100m/sec

To be asked in question

Rate of change distance between observer and rocket dhdt=?\frac {dh}{dt}=?

Solution (a)


x2+y2=h2x^2+y^2=h^2

Take differentiate both side


2ydydt=2hdhdt2y \frac{dy}{dt}=2h\frac{dh}{dt}


ydydt=hdhdty \frac{dy}{dt}=h\frac{dh}{dt}

dhdt=yhdydt\frac{dh}{dt}=\frac{y}{h}\frac{dy}{dt}

dhdt\frac {dh}{dt} =5005002+6002×100=\frac {500}{\sqrt {500^2+600^2}}\times100

dhdt=64.02meter/sec\frac {dh}{dt}=64.02meter/sec

Solution (b)

When Y=400meter/sec


Rocket and observer between distance pithagorse theorem

Use

h2=x2+y2h^2=x^2+y^2

h2=6002+4002h^2=600^2+400^2

h=5200h=√5200

h=72.11meter

Solution (c)

Pauthagorse theorem

x2+y2=h2x^2+y^2=h^2

Take differenciate 

ydydt=hdhdty \frac{dy}{dt}=h\frac{dh}{dt}

2ydydt=2hdhdt2y \frac{dy}{dt}=2h\frac{dh}{dt}


dhdt=yhdydt\frac{dh}{dt}=\frac{y}{h}\frac{dy}{dt}

dhdt=4006002+4002×100\frac{dh}{dt}=\frac{400}{\sqrt{600^2+400^2}}\times100

dhdt=55.44meter/sec\frac{dh}{dt}=55.44meter/sec




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