Surface area of expanding sphere as function of time

[BrendanM]

Im very rusty here's my problem, please help thanks!

The radius of a sphere is expanding at the rate of 1 meter/min. At t=0 the radius is 1. In earch case, find the requested function and sketch the curve.

a)The surface area of the sphere as a function of the time;
For this i took: dR/dt = 1 m/min
A=4*Pi*r^2
dA/dt=8*pi*r*dR/dt
dA/dt = 8*pi*r <--- if this is right how do i sketch this graph as a function of time??
my graph points that i ploted were (0,8pi),(1,16pi),(2,24pi) etc... but this is confusing me because at t=0 radius is 1... is my x-axis supposed to be time or radius someone help me I am confused..

 

[phoenixthoth]

Im very rusty here's my problem, please help thanks!

The radius of a sphere is expanding at the rate of 1 meter/min. At t=0 the radius is 1. In earch case, find the requested function and sketch the curve.

a)The surface area of the sphere as a function of the time;
For this i took: dR/dt = 1 m/min
A=4*Pi*r^2
dA/dt=8*pi*r*dR/dt
dA/dt = 8*pi*r <--- if this is right how do i sketch this graph as a function of time??
my graph points that i ploted were (0,8pi),(1,16pi),(2,24pi) etc... but this is confusing me because at t=0 radius is 1... is my x-axis supposed to be time or radius someone help me I am confused..

You are plotting the rate of change of area, not area. So based of the graph of the derivative of area, reconstruct what the graph of area looks like knowing that what you have are slopes. Start at (0,1) and draw a tangent line with slope 8pi. Then at (1,?) where ? > 1, the tangent line will have slope 16pi. Or if you've done integration, integrate dA/dt and to solve for the "C", plug in the value (0,1). I think your area is parabolic.

 

[M98Ranger]

Firstly, your approach to finding the function for the surface area is correct. The formula for the surface area of a sphere is A=4πr², where r is the radius. Since the radius is expanding at a rate of 1 m/min, we can write this as dR/dt=1 m/min. Using the chain rule, we can find the rate of change of the surface area with respect to time, which is dA/dt=8πr*dR/dt. As you correctly calculated, the function for the surface area as a function of time is dA/dt=8πr.

To sketch this graph, we need to consider the relationship between the surface area and the radius. As the radius increases, the surface area also increases. However, the rate of increase of the surface area is not constant. It is proportional to the radius, which means that as the radius increases, the rate of increase of the surface area also increases. This is why the graph of dA/dt vs. time will not be a straight line, but rather a curve.

To plot this graph, we can use the values you calculated. However, instead of using the radius as the x-axis, we can use time as the x-axis. This is because we want to see how the surface area changes over time. So, at t=0, the radius is 1 and the surface area is 4π. This gives us our first point (0, 4π). Then, at t=1, the radius is 2 and the surface area is 16π. This gives us our second point (1, 16π). Similarly, at t=2, the radius is 3 and the surface area is 36π. This gives us our third point (2, 36π). We can continue this pattern to plot more points and then connect them to get the curve. The resulting graph will be an increasing curve, with a steeper slope as time increases.

In summary, your approach to finding the function for the surface area was correct. However, when sketching the graph, it is important to consider the relationship between the variables and choose the appropriate axis for the independent variable. In this case, since we want to see how the surface area changes over time, time should be plotted on the x-axis and the surface area on the y-axis.