Question

Related Rates Problem::Perimeter and Area of a rectangle...

The length of a rectangle is increasing at 5 meters per second while the width is decreasing at 3 meters per second. At a certain instant, the length is 2 meters and the width is 10 meters. At that time, determine the rate of change of the area and perimeter.

Answer 1

DA/dt = W(dW/dt)(L) + L(dL/dt)(W) is this equation right?

Answer 2

dP/dt = 2W(dW/dt) + 2L(dL/dt) is this right?

Answer 3

actually, it appears as I was using the chain rule instead of the product rule, I get 44 for the area rate, is this right?

Answer 4

I get 4 m/s for the perimeter rate...is this right?

Answer 5

HI!!

Answer 6

what variables are you using?

Answer 7

looks like \(W\) and \(L\) so
\[A=WL\\
A'=W'L+L'W\] by the product rule

Answer 8

you are given all the numbers, go ahead and plug them in

Answer 9

@misty1212 do those answers look correct to you?

Answer 10

hmm not the first one

Answer 11

it is not \(WW'L\) just \(W'L\) etc

Answer 12

it is a product, the product rule says \[A=WL\\
A'=W'L+L'W\]

Answer 13

sorry, yes, corrected that before plugging in, the answer is adjusted for that formula

Answer 14

answers are just below there

Answer 15

oh i didn't plug in the numbers i can check though

Answer 16

\[L'=5,W'=-3, L-2,W=10\] right ?

Answer 17

\[-3\times 2+5\times 10=44\]

Answer 18

\[P=2L+2W\\
P'=2W'+2L'\]

Answer 19

that one is even easier

Answer 20

thanks for assisting

Answer 21

\[2\times5+2\times -3\] right?

Answer 22

yw \(\color\magenta\heartsuit\)