Consider a closed container in the shape of a cylinder of radius 10 cm and height 15 cm with a h

Consider a closed container in the shape of a cylinder of...

Consider a closed container in the shape of a cylinder of radius 10 $\mathrm{cm}$ and height 15 $\mathrm{cm}$ with a hemisphere on each end, as shown in the accompanying figure. The container is coated with a layer of ice 1$/ 2 \mathrm{cm}$ thick. Use a differential to estimate the total volume of ice. (Hint: Assume $r$ is radius with $d r=1 / 2$ and $h$ is height with $d h=0 .$ )

Consider a closed container in the shape of a cylinder of radius 10 $\mathrm{cm}$ and height 15 $\mathrm{cm}$ with a hemisphere on each end, as shown in the accompanying figure.
The container is coated with a layer of ice 1$/ 2 \mathrm{cm}$ thick. Use a differential to estimate the total volume of ice. (Hint: Assume $r$ is radius with $d r=1 / 2$ and $h$ is height with $d h=0 .$ )

Key Concepts

Differential Calculus

Differential calculus is used to approximate how a small change in one variable of a function can affect the function as a whole. In the context of volume estimation, differentials allow us to estimate the additional volume resulting from a small increase in the dimensions of a shape by using the derivative of the volume function.

Linear Approximation

Linear approximation involves using the tangent line at a point to estimate the value of a function near that point. When a dimension such as the radius is increased by a small amount, the change in volume can be approximated by multiplying the derivative of the volume with respect to that dimension by the small change.

Volume of Composite Solids

Understanding the volume of composite solids involves breaking down a complex shape into simpler components (e.g., cylinders, hemispheres) and calculating the volume for each part. The total volume is then determined by summing the individual volumes. This approach is essential when the object consists of different geometric shapes.

Transcript

00:01 Okay, we have this cylinder of radius 10 centimeters, and then it has a half of the sphere on the top and a half a sphere on the bottom.

00:10 And it's covered with ice half a centimeter thick, and we want to know what's the volume of the ice.

00:16 Okay, so my first thought was, i'm going to find the surface area, but that doesn't work.

00:22 So what i'm going to do is i'm going to find the volume and and use that to find the volume of the whole thing, and then use that to find the volume of the ice.

00:36 Okay, so the volume of the thing is the volume of the cylinder, which is pi r squared h plus the volume of the two half spheres, which is the volume of a whole sphere, which is four thirds pi r cubed.

00:59 Okay, so now what i'm going to do is i'm going to take the differential of both sides, and then i'm just going to plug in the stuff that i know.

01:11 I know the radius is 10 and the height is 15, and here's the important part.

01:17 The ice is one half centimeter thick, and they tell you to let that be dr, the change in the radius, and then the height, dh, is going to be zero here, change in the height.

01:32 Okay, because the height of the cylinder part is not going to get any bigger, but the height of the spheres is going to be bigger, but it's also the change in the radius.

01:45 Okay, because the radius of the sphere is the height of the sphere right here.

01:51 Okay, so letting dh equal zero means we're not changing the height of this thing in here...

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