Answer
Verified
390.3k+ views
Hint: In this problem, the volume of a sphere is increasing at a rate of proportional to be reciprocal of its radius. We have to find the radius of the sphere as a function of time
Using the function, we will find the relation between that radius and time and some constants. Then, using the condition we can find the values of the constants.
Complete step-by-step answer:
It is given that; at time \[t > 0\] the volume of a sphere is increasing at a rate of proportional to be reciprocal of its radius. At \[t = 0\] the radius of the sphere is \[1\] unit and \[t = 15\] the radius is \[2\] units.
We have to find the radius of the sphere.
Let us consider, \[V\] is the volume of the sphere and \[r\] is the radius of the sphere.
Since, the volume of a sphere is increasing at a rate of proportional to be reciprocal of its radius, the function can be written as,
\[\dfrac{{dV}}{{dt}} \propto \dfrac{1}{r}\]
It can be written as,
\[\dfrac{{dV}}{{dt}} = \dfrac{k}{r},k\] be any constant. … (1)
We know that, the volume of a sphere of radius \[r\] is \[V = \dfrac{4}{3}\pi {r^3}\]
Let us substitute the volume in equation (1) we get,
\[\dfrac{{d(\dfrac{4}{3}\pi {r^3})}}{{dt}} = \dfrac{k}{r}\]
Differentiate with respect to \[t\] we get,
\[\dfrac{4}{3}\pi .3{r^2}.\dfrac{{dr}}{{dt}} = \dfrac{k}{r}\]
Let us simplify the above equation, we get,
\[\pi{4} {r^3}dr = kdt\]
Since, \[k\] is constant.
Now let us integrate both sides we get,
\[\pi{4} \int {{r^3}} dr = k\int {dt} \]
Solving we get,
\[\pi{4} \dfrac{{{r^4}}}{4} = kt + C\]
So, we have,
\[\pi {r^4} = kt + C\]… (2)
We know that, when \[t = 0,r = 1\]
Substitute the values in equation (2) we get,
\[C = \pi \]
Then equation (2) becomes,
\[\pi {r^4} = kt + \pi \]… (3)
Again, we know that, when \[t = 15,r = 2\]
On substituting the values in (3) we get,
\[\pi {2^4} = 15k + \pi \]
On solving we get,
\[15k = 16\pi - \pi = 15\pi \]
So, \[k = \pi \]
Therefore, equation (3) becomes,
\[\pi {r^4} = \pi t + \pi \]
Now let us simplify the above equation, we get,
\[\pi {r^4} = \pi (t + 1)\]
Cancel the like terms that are in common and taking fourth root on both sides we get,
So, \[r = {(t + 1)^{\dfrac{1}{4}}}\]
Hence, the radius is \[r = {(t + 1)^{\dfrac{1}{4}}}\]
Note: We know that, the volume of a sphere of radius \[r\]is \[V = \dfrac{4}{3}\pi {r^3}\]while finding the radius we substitute the given values of t and r it is an important step which helps in finding the value of K and C.
Using the function, we will find the relation between that radius and time and some constants. Then, using the condition we can find the values of the constants.
Complete step-by-step answer:
It is given that; at time \[t > 0\] the volume of a sphere is increasing at a rate of proportional to be reciprocal of its radius. At \[t = 0\] the radius of the sphere is \[1\] unit and \[t = 15\] the radius is \[2\] units.
We have to find the radius of the sphere.
Let us consider, \[V\] is the volume of the sphere and \[r\] is the radius of the sphere.
Since, the volume of a sphere is increasing at a rate of proportional to be reciprocal of its radius, the function can be written as,
\[\dfrac{{dV}}{{dt}} \propto \dfrac{1}{r}\]
It can be written as,
\[\dfrac{{dV}}{{dt}} = \dfrac{k}{r},k\] be any constant. … (1)
We know that, the volume of a sphere of radius \[r\] is \[V = \dfrac{4}{3}\pi {r^3}\]
Let us substitute the volume in equation (1) we get,
\[\dfrac{{d(\dfrac{4}{3}\pi {r^3})}}{{dt}} = \dfrac{k}{r}\]
Differentiate with respect to \[t\] we get,
\[\dfrac{4}{3}\pi .3{r^2}.\dfrac{{dr}}{{dt}} = \dfrac{k}{r}\]
Let us simplify the above equation, we get,
\[\pi{4} {r^3}dr = kdt\]
Since, \[k\] is constant.
Now let us integrate both sides we get,
\[\pi{4} \int {{r^3}} dr = k\int {dt} \]
Solving we get,
\[\pi{4} \dfrac{{{r^4}}}{4} = kt + C\]
So, we have,
\[\pi {r^4} = kt + C\]… (2)
We know that, when \[t = 0,r = 1\]
Substitute the values in equation (2) we get,
\[C = \pi \]
Then equation (2) becomes,
\[\pi {r^4} = kt + \pi \]… (3)
Again, we know that, when \[t = 15,r = 2\]
On substituting the values in (3) we get,
\[\pi {2^4} = 15k + \pi \]
On solving we get,
\[15k = 16\pi - \pi = 15\pi \]
So, \[k = \pi \]
Therefore, equation (3) becomes,
\[\pi {r^4} = \pi t + \pi \]
Now let us simplify the above equation, we get,
\[\pi {r^4} = \pi (t + 1)\]
Cancel the like terms that are in common and taking fourth root on both sides we get,
So, \[r = {(t + 1)^{\dfrac{1}{4}}}\]
Hence, the radius is \[r = {(t + 1)^{\dfrac{1}{4}}}\]
Note: We know that, the volume of a sphere of radius \[r\]is \[V = \dfrac{4}{3}\pi {r^3}\]while finding the radius we substitute the given values of t and r it is an important step which helps in finding the value of K and C.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE