Related Rates: Part II

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Hckyplayer8

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Thank you all who have been helped me out in the previous thread. Hopefully this one goes a bit smoother for me.

The length of a rectangle is increasing at the rate of 5 m/min while the width is decreasing at the rate of 3 m/min. At a certain instant, the length is 20m and the width is 10m. At that instant determine the rate of change of the area and the rate of change of the perimeter.

First, the rate of of change of the area.

A = L times W. So of L=x and W=y, then dA/dt = (x) dy/dt + (y) dx/dt.

dx/dt= 5 m/min
dy/dt= -3 m/min
X=20m
Y=10m

dA/dt= x (-3) + y (5) = 20 (-3) + 10 (5) = -10m
 
Looks good except for the change to uppercase X and Y and the units of the time rate of change of area, which should be in [MATH]\frac{\text{m}^2}{\text{min}}[/MATH].
 
MarkFL said:
Looks good except for the change to uppercase X and Y and the units of the time rate of change of area, which should be in [MATH]\frac{\text{m}^2}{\text{min}}[/MATH].
Click to expand...

Thanks. Are x,X and y,Y interpreted as different variables, which makes what I wrote incorrect?
 
Yes, changing the case means a different variable. It's a common issue, just wanted you to be aware of it. It is obvious what you mean, but at least some professors will mark against doing that. It's good practice to retain the same case for a variable for the duration of the problem.
 
MarkFL said:
Yes, changing the case means a different variable. It's a common issue, just wanted you to be aware of it. It is obvious what you mean, but at least some professors will mark against doing that. It's good practice to retain the same case for a variable for the duration of the problem.
Click to expand...

Makes sense. Thank you.
 
Hckyplayer8 said:
Thank you all who have been helped me out in the previous thread. Hopefully this one goes a bit smoother for me.

The length of a rectangle is increasing at the rate of 5 m/min while the width is decreasing at the rate of 3 m/min. At a certain instant, the length is 20m and the width is 10m. At that instant determine the rate of change of the area and the rate of change of the perimeter.

First, the rate of of change of the area.

A = L times W. So of L=x and W=y, then dA/dt = (x) dy/dt + (y) dx/dt.

dx/dt= 5 m/min
dy/dt= -3 m/min
X=20m
Y=10m

dA/dt= x (-3) + y (5) = 20 (-3) + 10 (5) = -10m
Click to expand...

Okay, now the perimeter. P = 2(x + y) which means

dP/dt = 2 (dx/dt+dy/dt) = 2 [ 5 + (-3)] = 4m/min
 
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