Surface area of expanding sphere as a 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..

 

[Sirus]

Please do not post the same thing in multiple places. This should really be posted only once, and in the Calculus & Analysis forum.

 

[M98Ranger]

Hi there,

It looks like you're on the right track for finding the surface area of the sphere as a function of time. Your formula for dA/dt is correct, and you can use that to find the surface area at any given time by plugging in the value of r. Your graph points are correct as well, and they show the relationship between the surface area and time. However, since the question is asking for a graph of surface area as a function of time, your x-axis should represent time. This means that at t=0, the surface area is 0, since the radius is also 0 at that point. As time increases, the surface area also increases, as shown by your plotted points.

To sketch the curve, you can plot your points and then connect them with a smooth curve. Since we know that the surface area increases as time increases, the curve should be an upward slope. It may also help to label your axes as "time" and "surface area" to make it clear what the graph represents.

I hope this helps and good luck with your homework!