Question

I'm doing related rates problems and I don't have the hang of it yet. Could someone guide me through this problem?

Answer 1

So far I've drawn a circle, written down the rate the snowball is melting, and written down the equation for the surface area of a sphere.

Answer 2

\[SA: 4\pi r^{2}\]

Answer 3

I know it's dA/dt = something, but I'm not sure how to get to the next step.

Answer 4

put r=d/2 and diff SA.

Answer 5

4 pi (dd/dt) where first d for diff. and second for diameter

Answer 6

How did you do that? I'm lost on how you got the derivative of (d/2)^2

Answer 7

SA=4*pi*(d/2)^2
=pi*d^2
now diff it

Answer 8

2pi*dd/dt?

Answer 9

2*pi*d*(dd/dt)

Answer 10

Why d(dd/dt)? Are we acting as if it is y and the deriv of y is yy'?

Answer 11

derivative of y is y' and y^2 is 2 y y' so of d^2 is 2 d d'

Answer 12

Ohh. I understand now, okay

Answer 13

Okay, so now we want to get dd/dt alone right?
dA/dt/(2pi*d) = dd/dt and then substitute what we know -> 3/(2pi*10) = dd/dt?

Answer 14

correct

Answer 15

dA/dt =-3

Answer 16

since its decrease.

Answer 17

Oh, I keep forgetting that. Sorry for the wait, I was writing down the steps

Answer 18

I got -0.048 cm/min.

Answer 19

I think I did something wrong bc webassign (where I submit the answers) says my answer is wrong

Answer 20

correct

Answer 21

Should I try -0.15pi?

Answer 22

Try +0.048

they ask for how much it decreases... they may want a positive value

Answer 23

Ahh. That was right, thanks!

Answer 24

Thank you Er.Mohd.Amir!