How Do You Calculate the Rate of Area Increase in an Expanding Rectangle?

[ace123]

it would be 10 if they were moving towards each other. If i am moving 30 ft/sec in one direction and you move 40 ft/sec in the other direction. The distance between us in 1 second would be what?

 

[bondgirl007]

When I put it in 1 triangle, here's what I get.

a = 210 km
da/dt = 10km

b = 50 km
db/dt = 0

c = square root (46600)
dc/dt = 9.73 after solving it but that is way off!

What am I doing wrong? :crying:

 

[bondgirl007]

it would be 10 if they were moving towards each other. If i am moving 30 ft/sec in one direction and you move 40 ft/sec in the other direction. The distance between us in 1 second would be what?

It worked after I put 70 in for da/dt!

Thank you so much, everyone!

 

[ace123]

I think only one of us should continue because all of us are saying the samething but in a different way and we end up confusing her. It shouldnt' be me because I haven't touched this topic in years

 

[rocomath]

It worked after I put 70 in for da/dt!

Thank you so much, everyone!

you're given dA\dt and dB\dt

let x be the distance in the x-axis

let x+y be the distance in the y-axis

$$x^{2}+(x+y)^{2}=z^{2}$$

you got your answer by luck, you're given dA\dt was not 70. i encourage you to keep working this problem!

 

[ace123]

He is correct the dA/dt was not 70. If you look at coomast and his posts the reason he got his equation is by a^2(t ^2)+ b^2. Then plug in the numbers you have. L(t)=sqrt(4900t^2+2500)

 

[bondgirl007]

For my answer, I got 14700/sqrt(46600) and the back of my textbook has 1470/sqrt(466), which are both equivalent. I think I have it right.

 

[arildno]

For my answer, I got 14700/sqrt(46600) and the back of my textbook has 1470/sqrt(466), which are both equivalent. I think I have it right.

Indeed you do; problem solved