Differentials and paint needed problem

In summary, to estimate the amount of paint needed to apply a coat of paint 0.05cm thick to a hemispherical dome with diameter 50m, we can use differentials to calculate the corresponding change in volume, ∆V, for a small change in radius, ∆r. This can be done by taking the derivative of the volume formula for a hemisphere, which is (2/3)pi.r^3, and multiplying it by ∆r. The resulting expression, 2pi.r^2.∆r, will give us an estimate of the amount of paint needed in terms of volume.
  • #1
synergix
178
0

Homework Statement



Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05cm thick to a hemispherical dome with diameter 50m

Homework Equations



A= 2(pi)r2

The Attempt at a Solution



dr=0.0005 m
r=25m
dA=?

(A=2(pi)r2)'

dA= 4(pi)r*dr

I won't go any further because the number is very small and i think incorrect.
I am pretty sure I'm missing a step but I can't figure out what or why.
Then again I could be way off.
 
Physics news on Phys.org
  • #2


the amount of paint will be a volume, not area, think of a thin spherical shell
 
  • #3


What is the formula for volume of a sphere?

([itex]\pi r^2[/itex] is the area of circle and not relevant here.)
 
  • #4


volume of a sphere= (4/3)pi*r^2

volume of hemisphere= (4/6)pi*r^2

dv= (4/3)pi*r*dr

dv= (4/3)*pi*25*.0005

am I wrong still?
 
  • #5


synergix said:
volume of a sphere= (4/3)pi*r^2

volume of hemisphere= (4/6)pi*r^2

dv= (4/3)pi*r*dr

dv= (4/3)*pi*25*.0005

am I wrong still?

yes...

volume of a sphere is (4/3)pi*r^3

and when you differntiate a power, you multiply by the orgiginal power, not divide
 
  • #6


that looks like the whole sphere, how about the hemisphere part?
 
  • #7


dv=2pi*r^2
 
  • #8


almost... you just need to add a dr in there

so to sumamrise
V(r) = (1/2)(4/3)pi.r^3 = (2/3)pi.r^3

then the derivative is
dV/dr = 2pi*r^2

so for a small change in r, ∆r the approximate corresponding change in ∆V volume will be
∆V= (dV/dr).∆r = 2pi.r^2.∆r
 

Related to Differentials and paint needed problem

1. How do I calculate the number of paint cans needed for a room?

To calculate the number of paint cans needed for a room, you will need to measure the length, width, and height of the walls. Then, multiply the length by the width to get the total square footage. Divide this number by the coverage area listed on the paint can to determine the number of cans needed. Don't forget to factor in multiple coats and subtract for windows and doors.

2. What is the difference between differential and integral calculus?

Differential calculus involves the study of rates of change, slopes, and derivatives, while integral calculus focuses on the accumulation of quantities and finding areas under curves. In simpler terms, differential calculus is about finding the rate at which something is changing, while integral calculus is about finding the total amount of something.

3. How do I find the average rate of change using differentials?

To find the average rate of change using differentials, you will need to calculate the difference between the final and initial values of a function and divide it by the change in the independent variable. This will give you the average rate of change or the slope of the secant line between the two points.

4. What is the paint differential equation?

The paint differential equation is a mathematical equation that models the rate of change of the amount of paint in a paint can over time. It takes into account the rate of paint being used and the rate of paint being added, as well as any other factors that may impact the amount of paint in the can.

5. How can I use differentials to optimize my paint usage?

You can use differentials to optimize your paint usage by finding the minimum or maximum of a function that represents the cost of paint. By taking the derivative of the cost function and setting it equal to zero, you can find the critical points and use them to determine the most efficient amount of paint to use for a given project.

Similar threads

Replies
2
Views
4K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
754
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
5K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
Replies
4
Views
31K
  • Calculus and Beyond Homework Help
Replies
6
Views
11K
Back
Top