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
 

MarkFL

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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 \(\displaystyle \frac{\text{m}^2}{\text{min}}\).
 

Hckyplayer8

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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 \(\displaystyle \frac{\text{m}^2}{\text{min}}\).
Thanks. Are x,X and y,Y interpreted as different variables, which makes what I wrote incorrect?
 

MarkFL

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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.
 

Harry_the_cat

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The units attached to dA/dt is the A unit per the t unit , ie m2 /min.
 

Hckyplayer8

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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.
 

Jomo

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I would add that you should use L and W OR use x and y but NOT both.
 

Hckyplayer8

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235

Hckyplayer8

Junior Member
Joined
Jun 9, 2019
Messages
235
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
 
Last edited:

Harry_the_cat

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Your value is correct, but dP/dt units are P units per t units which is not m2/min. Can you fix that up?
 

Hckyplayer8

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Your value is correct, but dP/dt units are P units per t units which is not m2/min. Can you fix that up?
Duh. Area is squared. Perimeter is just the values be measured by a certain scale. In this case, meters.

So 4m/min.

Thank you.
 

Harry_the_cat

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YAY!! Always check your units by looking at the units of the numerator per units of denominator.
 
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