MHB How Much Should the Radius Increase to Enlarge the Circle's Area by b Units?

  • Thread starter Thread starter mathdad
  • Start date Start date
  • Tags Tags
    Increasing Radius
mathdad
Messages
1,280
Reaction score
0
The radius of a circle is r units. By how many units should the radius be increased so that the area increases by b units?

The information in this problem tells me that using the area of a circle formula is needed.

A = pi•r^2

I think b should be added to r and squared.

A = pi(r + b)^2

Can someone provide a one or two hints?
 
Mathematics news on Phys.org
RTCNTC said:
The radius of a circle is r units. By how many units should the radius be increased so that the area increases by b units?

The information in this problem tells me that using the area of a circle formula is needed.

A = pi•r^2

I think b should be added to r and squared.

A = pi(r + b)^2

Can someone provide a one or two hints?
You certainly should not add $b$ to $r$, because $b$ is an area and $r$ is a distance. Adding a two-dimensional quantity to a one-dimensional quantity does not make sense.

What you want to know is how much to increase $r$ so that $\pi r^2$ becomes $\pi r^2 + b$. Suppose that $x$ is the amount that has to be added to $r$. Then the equation is $\pi(r+x)^2 = \pi r^2 + b$. So you need to solve that equation for $x$.
 
Opalg said:
You certainly should not add $b$ to $r$, because $b$ is an area and $r$ is a distance. Adding a two-dimensional quantity to a one-dimensional quantity does not make sense.

What you want to know is how much to increase $r$ so that $\pi r^2$ becomes $\pi r^2 + b$. Suppose that $x$ is the amount that has to be added to $r$. Then the equation is $\pi(r+x)^2 = \pi r^2 + b$. So you need to solve that equation for $x$.

I understand what you are saying.

π(r + x)^2 = A + b

π(r + x)^2 = πr^2 + b

(r + x)^2 = (πr^2 + b)/π

sqrt{(r + x)^2} = sqrt{(πr^2 + b)/π}

r + x = sqrt{(πr^2 + b)/π}

x = sqrt{(πr^2 + b)/π} - r

Correct?
 
Last edited:
RTCNTC said:
π(r + x)^2 = A + b

π(r + x)^2 = πr^2 + b

(r + x)^2 = (πr^2 + b)/π

sqrt{(r + x)^2} = sqrt{(πr^2 + b)π}

r + x = sqrt{(πr^2 + b)π}

x = sqrt{(πr^2 + b)π} - r

Correct?
The method is correct, but in the last three lines you should have $(\pi r^2 + b)/\pi$ instead of $(\pi r^2 + b)\pi$.
 
Opalg said:
The method is correct, but in the last three lines you should have $(\pi r^2 + b)/\pi$ instead of $(\pi r^2 + b)\pi$.

I just forgot to include the slash symbol in the last 3 lines. It has now been edited.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top