Find the requested function and sketch the curve

[BrendanM]

help me please!

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..

 

[Tide]

Write r as a function of time then plot:
[tex]r = 1 + t[/tex]

 

[M98Ranger]

The requested function in this case is the surface area of the sphere as a function of time, which can be represented as A(t). The curve that would be sketched would be a linear curve, as the rate of change of the surface area is constant (8*pi*r). The x-axis would represent time (t) and the y-axis would represent surface area (A).

To sketch the curve, we can plot the points you have calculated: (0, 8*pi), (1, 16*pi), (2, 24*pi), etc. These points represent the surface area at different points in time. We can then connect these points with a straight line to create the linear curve.

However, as you mentioned, the radius at t=0 is 1, which means the surface area at t=0 would be 4*pi*1^2 = 4*pi. This means the first point on the graph should be (0, 4*pi) instead of (0, 8*pi). This will result in a curve that starts at (0, 4*pi) and increases linearly from there.

In summary, the requested function and curve for this problem would be:
A(t) = 4*pi + 8*pi*t
Curve: a linear curve starting at (0, 4*pi) and increasing at a constant rate of 8*pi.