Related rates word problems- poliece car& spot light

[wind]

Hi, I am having trouble with the following questions, can some one pleas cheack over the first one and help me with the second? Thanks

1. A police car is travelling west toward the inersection with a north/south road. A car is traveling south on the intersecting road. the police radar detects that the other car is speeding. At the time of the reading.The police car is 2kn east of the intersection, travelling 80km/h, the other car is 1.5 km south of the intersection, traviling south. the speed limit of the north/south road is 90km/h. the distance between the 2 cars is decresing at 1km/h. Is the other car seeding? if so, by how mutch?

Let P re the poilece car
Let c rep the other car
Let x rep the distance between the two cars

dc/dt=?
dp/dt=80km/h
dx/dt=1km/h

c=1.5 km
p=2 km
x= ?

Find X
c^2+ p^2 = x^2
2.5=x

Sub in
d/dt c^2+d/dt p^2 =d/dt x^2
d/dt 2c + d/dt 2p = 0
d/dt 3 + 320 = 0
d/dt = -320/3
d/dt = 106.7

therefor the car is speeding by 16.7km/h

Is this right?

2. A spotlight on the ground shines on the outside wall of a parking garade 12m away. if a 2m tall man walks toward the garage at a speed of 0.75m/s, how fast is the height of the man's shadow on the garage wall decresing when he is 4m from the building?

Want
dh/dt

Let x rep the distance form the light to the wall
Let h rep the hight of the shadow on the wall

Small tri
h= 2m
b= x

Big tri
h= h
b= 4+x

Similar triangle's side lenghts are proportionate

2/h = x/4+x
2(4 + x) = xh
8 + 2x/x = h

...now what?

Thanks.

 

[galactus]

Hi, I am having trouble with the following questions, can some one pleas cheack over the first one and help me with the second? Thanks

1. A police car is travelling west toward the inersection with a north/south road. A car is traveling south on the intersecting road. the police radar detects that the other car is speeding. At the time of the reading.The police car is 2kn east of the intersection, travelling 80km/h, the other car is 1.5 km south of the intersection, traviling south. the speed limit of the north/south road is 90km/h. the distance between the 2 cars is decresing at 1km/h. Is the other car seeding? if so, by how mutch?

Let P re the poilece car
Let c rep the other car
Let x rep the distance between the two cars

dc/dt=?
dp/dt=80km/h
dx/dt=1km/h

c=1.5 km
p=2 km
x= ?

Find X
c^2+ p^2 = x^2
2.5=x

Sub in
d/dt c^2+d/dt p^2 =d/dt x^2
d/dt 2c + d/dt 2p = 0
d/dt 3 + 320 = 0
d/dt = -320/3
d/dt = 106.7

therefor the car is speeding by 16.7km/h

Is this right?

I got something slightly different. Check me out.

x=2; y=1.5; z=2.5; dz/dt=-1; dx/dt=-80

We want dy/dt. If it's > 90, then the car is speeding.

As you done, Pythagoras:

\(\displaystyle \L\\z^{2}=x^{2}+y^{2}\)

Differentiate:

\(\displaystyle \L\\z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\)

\(\displaystyle \L\\(2.5)(-1)=2(-80)+(1.5)\frac{dy}{dt}\)

\(\displaystyle \L\\\frac{dy}{dt}=105\)

The car is speeding by 15 km/hr.

2. A spotlight on the ground shines on the outside wall of a parking garade 12m away. if a 2m tall man walks toward the garage at a speed of 0.75m/s, how fast is the height of the man's shadow on the garage wall decresing when he is 4m from the building?

Want
dh/dt

Let x rep the distance form the light to the wall
Let h rep the hight of the shadow on the wall

Small tri
h= 2m
b= x

Big tri
h= h
b= 4+x

Similar triangle's side lenghts are proportionate

2/h = x/4+x
2(4 + x) = xh
8 + 2x/x = h

...now what?

Thanks.

This ones a little trickier. I think anyway. Make sure you check out my 'gazintas' or get a second opinion.

We'll use similar triangles, of course. x and h are the independent variables. That is, they're the ones changing; x as the man walks toward the wall and h the height of the shadow.

\(\displaystyle \L\\\frac{x}{2}=\frac{12}{h}\).....[1]

\(\displaystyle \L\\h=\frac{24}{x}\).........[2]

Differentiate [1]:

\(\displaystyle \L\\\frac{1}{2}\frac{dx}{dt}=\frac{-12}{h^{2}}\frac{dh}{dt}\)

Solve for \(\displaystyle \L\\\frac{dh}{dt}\)

\(\displaystyle \L\\\frac{dh}{dt}=\frac{1}{2}\frac{dx}{dt}\cdot\frac{-h^{2}}{12}\)

dx/dt=0.75, x=8 when he is 4 feet from the wall.

\(\displaystyle \L\\\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{-h^{2}}{12}=\frac{dh}{dt}\)

We can use [2] to find h, 24/8=3=h

\(\displaystyle \L\\\frac{dh}{dt}=\frac{-3}{8}\cdot\frac{9}{12}=\frac{-27}{96}=\frac{-9}{32} \;\ m/s\)

Whatcha think?. Look OK?. Seems OK.

 

[wind]

Thanks galactus

your anwser is right, I just relized this question is in my textbook. The only thing I am confused about is- when you solve for dh/dt why did the -12/h^2 flip to -h^2/12?

 

[galactus]

Basic algebra. I divided by -12/h^2, so I multiplied by -h^2/12.

 

[wind]

oh, ok