Help on Year 11 Volume & Surface Area of Cones Question

[Nimbot96]

Homework Statement

I am in Year 11 at the moment and just learned about rates of change. We have an assignment which dwelves into volume and surface area of cones. I just want to know if I am going on the right track for these questions. (because I am new to this forum, i might have not put this in the right place... move if necessary)

Q = Radius
θ = Angle of sector area (the part of the circle which becomes the cone without the base)

All angles HAVE to be in radians

Homework Equations

1) Find a formula in terms of θ and Q for the sector area in the domain 0≤θ≤2∏. This area is equal to the cone's surface area but without the base. Using excel or similar graphing package, calculate the values of the surface area for every 1/4 of a radian for the domain with Q = 1 unit. Plot the graph of Surface Area versus θ. Comment on the rate of change for this graph and identify the relationship between the variables.

The Attempt at a Solution

This is my attempt at the question -

The formula found to calculate the sector area in the domain of 0 ≤ θ ≤ 2(pi) will be
½ × (θ × π/180) × r2

excel graph is attached -

In this graph, we can see the relationship between the three variables, surface area, the radius and the angle (in degrees). Looking at the line, it goes up periodically in a straight line. The radius is the distance between the middle to the edge of a circle (in this case, the circle used to make the cone). So as the radius is increased, the sector area would also increase. Because this is a straight line graph, it also shows that the relationship between the surface area and θ is constant. This means that as the angle of the arc in the circle increases, the more surface area would be exposed on the cone (without the base).

The rate of change can be defined as the speed at which a variable changes over time. In this case, the value which defines the rate at which the angle increases along with the surface area of the cone without the inclusion of the base (or the sector area of the circle). The rate of change is calculated by deriving the gradient of the graph. To calculate the gradient, the formula: rise/run is applied.

2-0.25/1-0.125 = 1.625 (seen on the graph)

So 1.625 is the average rate of change.

As this is a straight graph, the rate of change will remain constant. This means that as the circle’s radius and angle is increased in the sector, the sector area will increase at the rate of 1.625 every period.

Am i on the right track here? Do i need to explain anything else? Have I calculated the rate of change correctly?

 

[Nimbot96]

Dont Worry guys... i got it... nver mind...

 

[Mark44]

Some tips if you post again.
1. Spend some time on coming up with a meaningful thread title. "Help in a question" doesn't give readers any idea about the problem you're trying to solve. A better title might be "Volume and surface area of a cone".
2. If you say there is an attachment, be sure to provide it.
3. When you write expressions with fractions (such as 2-0.25/1-0.125), use parentheses around the entire numerator and the entire denominator.
Due to the order of operations, what you wrote means 2 - (.25/1) - .125 = 1.625, which is what you got. Since you were calculating the slope, it probably should have been (2 - .25)/(1 -.125) = 1.75/.875 = 2.