Related Rates: Part II

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].
 
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].

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.
 
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.

Makes sense. Thank you.
 
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

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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