Use differentials to find the approximate change in the surface area when the radius of the sphere is increased from 10cm to 10.2cm.

I know A=4pir^2 and A' is 8pir but I'm not sure how to set it up from here. Can someone please show me all the steps involved? (This isn't HW, just a textbook question that I can't figure out, thanks!)

Ha! I figured it out.

dA=8pir dr
r=10 , dr=10.2

dA=8(10)(.2)pi
dA=16pi (approx)

check:
A(10)=4(10)^2pi
A(10)=400pi
A(10.2)=4(10.2)^2pi
A(10.2)=416.16

A(10.2)-A(10)=16.16pi (exact)

Well, well, well, it seems like you've stumbled onto a mathematical conundrum! Don't worry, I'm here to help you through it with my clowny wisdom.

Let's start by finding the differential equation for the surface area of a sphere. You correctly identified that A = 4πr^2, and A' represents the derivative of A with respect to r.

So, to find the approximate change in the surface area, we'll use the differentials:

dA = A' * dr

To approximate the change, we need the value of A' (the derivative of A), which you already provided as 8πr. Now, we need to evaluate it at r = 10 cm (the initial radius).

A' = 8π(10) = 80π cm²

Now, substitute this value in the differential equation:

dA = (80π cm²) * (dr)

Since we are only interested in the approximate change in surface area when the radius increases from 10 cm to 10.2 cm, we can use the small change in radius, which is 10.2 cm - 10 cm = 0.2 cm.

Substitute the change in radius into the equation:

dA ≈ (80π cm²) * (0.2 cm)

Now, solving this equation will give us the approximate change in the surface area of the sphere when the radius increases from 10 cm to 10.2 cm.

dA ≈ (16π cm²)

So, the approximate change in the surface area is approximately 16π cm². Voila!

Keep on juggling those math problems, my friend!

To find the approximate change in the surface area (ΔA), we can use differentials. The differential of the surface area dA is given by:

dA = 8πr dr,

where dr is the small change in the radius.

Now we can substitute the values into the equation:

dA = 8π(10 cm)(0.2 cm) = 16π cm^2.

This gives us the approximate change in the surface area when the radius is increased by 0.2 cm.

Therefore, the approximate change in the surface area is 16π cm^2.

To find the approximate change in the surface area when the radius of the sphere is increased from 10 cm to 10.2 cm, we can use differentials.

First, we need to find the derivative of the surface area formula with respect to the radius. Since A = 4πr^2, we have:

dA/dr = d(4πr^2)/dr

To simplify, we differentiate each term with respect to r:

dA/dr = 4π(d r^2/dr)

Using the power rule of differentiation, we get:

dA/dr = 4π(2r)

Simplifying further, we have:

dA/dr = 8πr

Now, to find the approximate change in surface area (∆A), we can use the differentials:

∆A ≈ dA = (dA/dr)∆r

where ∆r is the change in radius (10.2 cm - 10 cm = 0.2 cm) and dA/dr is the derivative we found earlier.

Substituting the values, we have:

∆A ≈ (8πr)∆r

∆A ≈ (8π)(10 cm)(0.2 cm)

∆A ≈ 16π cm^2

Therefore, the approximate change in surface area is approximately 16π cm^2 when the radius of the sphere is increased from 10 cm to 10.2 cm.