Related Rates (rectangular prism)

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SUMMARY

The discussion centers on calculating the rate of change of the volume of a rectangular prism with dimensions increasing at specified rates. The length increases by 12 cm/min, width by 4 cm/min, and height by 2 cm/min. The volume formula V = LWH is used, leading to the derivative dV/dt. The correct rate of volume change at the given dimensions (L=200 cm, W=50 cm, H=30 cm) is determined to be 62,000 cm³/min, confirming that the calculations align with the principles of related rates in calculus.

PREREQUISITES
  • Understanding of related rates in calculus
  • Familiarity with the volume formula for a rectangular prism (V = LWH)
  • Knowledge of differentiation and the product rule
  • Ability to solve polynomial equations
NEXT STEPS
  • Study the product rule for differentiation in calculus
  • Learn how to apply related rates problems in real-world scenarios
  • Explore the concept of implicit differentiation
  • Practice solving polynomial equations to improve accuracy in calculations
USEFUL FOR

Students in calculus courses, educators teaching related rates, and anyone seeking to understand the dynamics of changing geometric dimensions in mathematical contexts.

bdraycott
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Homework Statement



A rectangular prism has its length increasing by 12 cm/min, its width increasing by 4 cm/min and its height increasing by 2 cm/min. How fast is it's volume changing when the dimensions are 200 cm in length, 50 cm in width and 30 cm in height?


Homework Equations





The Attempt at a Solution



I am reworking a few things for an online calculus course that I am about to finish. This questions seems to have me baffled though. I have worked out the equations to arrive at dv/dt, however when I try to complete ( by attempting to arrive at (t) or use the quadratic equations the values I arrive at are always negative?

Here we go,

dL/dt= 12cm/min dW/dt=4cm/min dH/dt=3cm/min

How fast is the change occurring when∶ L=200 W=50 H=30

V=LWH

V=(200+12t)(50+4t)(30+3t)

dv/dt=10,000+800t+600t+48t^2 (30+3t)

=300000+20000t+24000t+1600t+18000t+1200t^2+1440t^2+96t^3

= 96t^3+ 2648t^2+63600t+300000

=2〖(96t)〗^2+2(2648t)+63600

=192t^2+5296t+63600

∴at t=2cm/min

This is where I get lost, I have tried using v(t)=x(t)y(t)z(t) and keep running into (-) values.Hopefully someone can shed a little light on this one for me.
 
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Aren't you trying to find the volume when t = 0? So wouldn't that mean the volume is changing at a rate of 63600 cm^3/min? (Assuming your calculations are correct)
 
bdraycott said:

Homework Statement



A rectangular prism has its length increasing by 12 cm/min, its width increasing by 4 cm/min and its height increasing by 2 cm/min. How fast is it's volume changing when the dimensions are 200 cm in length, 50 cm in width and 30 cm in height?


Homework Equations





The Attempt at a Solution



I am reworking a few things for an online calculus course that I am about to finish. This questions seems to have me baffled though. I have worked out the equations to arrive at dv/dt, however when I try to complete ( by attempting to arrive at (t) or use the quadratic equations the values I arrive at are always negative?

Here we go,

dL/dt= 12cm/min dW/dt=4cm/min dH/dt=3cm/min

How fast is the change occurring when∶ L=200 W=50 H=30

V=LWH

V=(200+12t)(50+4t)(30+3t)

dv/dt=10,000+800t+600t+48t^2 (30+3t)

=300000+20000t+24000t+1600t+18000t+1200t^2+1440t^2+96t^3

= 96t^3+ 2648t^2+63600t+300000

=2〖(96t)〗^2+2(2648t)+63600

=192t^2+5296t+63600

∴at t=2cm/min

This is where I get lost, I have tried using v(t)=x(t)y(t)z(t) and keep running into (-) values.Hopefully someone can shed a little light on this one for me.

Relevant Equations:

V (prism) = Bh = LWH

How to Tackle The Problem:

Whenever starting any related rates problem, always 1. )STATE your givens.

You already did it--but here's to summarize:

dL/dt = 12 cm/min
dw/dt = 4 cm/min
dh/dt = 2 cm/min

W = 50
L = 200
H = 30

Next 2.) Find out what it's ASKING.


"How fast is it's volume changing"

Ding ding ding! Basically, in English, it's asking you to solve for dV/dt.

Since V = lwh

Find dV/dt ... don't do any substitutions yet. After you've solved for dV/dt THEN 3.) substitute your givens in the equation.
I got an answer of 62,000 cm^3 / min . Not too far from moemoney's answer.
 
carlodelmundo's is correct and is far simpler.

You just have to remember the multiplication identity for taking derivatives of the product of 3 variables. This will save you time rather than having to expand everything out.

Y = abc ==> Y' = a'bc + ab'c + abc' if I remember correctly.
 
Yep that's correct, just an extension of the product rule.
 

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