Group+E+Period+6

Welcome Group E Period 6! You must complete problems 5, 11, and 17 in the space provided. Feel free to use the discussion board to bounce ideas off one another. Each member of the group must contribute.

By: Bobby Dorigo Jones, Kevin Hodges, Dean Sansovich

 * 5. Diagonals: If //x,y//, and //z// are lengths of the edges of a rectangular box, the common length of the box's diagonals is //s=(x^2 + y^3 + z^2)^1/2//. how is //ds/dt// related to //dx/dt//, and //dz/dt//?**
 * remember, it is a box, so it is 3D.


 * Take the derivative. s' = [1/2//(x^2 + y^3 + z^2)^-1/2] x (2x dx/dt + 2y dy/dt + 2z dz/dt)//

//*// Then you get: //ds/dt=// //(2x dx/dt + 2y dy/dt + 2z dz/dt)// / //(x^2 + y^3 + z^2)^1/2//

//* And thats the answer.//

**(a) How fast is the balloon's radius increasing at the instant the radius is 5 ft?**
When I have to use related rates, I like to think of an acronym I learned from a YouTube video, DREDS. Here is a link to the video I learned it from:

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If you don't want to watch the video, I'll sum it up for you. DREDS stands for Diagram, Rates, Equation, Derivative, and Substitute. These are the steps to using related rates successfully. So let's get started.

DIAGRAM: I drew a sphere (a circle that I labeled sphere) and labeled the radius 5, because that's the radius at the instant described in the question.

RATES: Find the rate of change in the equation. The rate of change for the volume in this problem (dV/dt) is 100pi ft^3/min, but we don't need to worry about the ft^3/min.

EQUATION: Now set up an equation. Remember that our goal is to find the rate of change of the volume (dr/dt). We write the equation for the volume of a sphere: (4/3)(pi)(r^3) = V

DERIVE: Derive the top equation to get dr/dt. (4)(pi)(r^2)(dr/dt) = (dV/dt)

SUBSTITUTE: We can substitute in all the information we have. We know the radius and the rate or change of the volume of the sphere (dV/dt): (4)(pi)(5^2)(dr/dt) = 100(pi) After a little algebra... 100(pi)(dr/dt) = 100(pi) After a little more algebra... (dr/dt) = 1

And that's your answer! The rate of change of the radius is 1 ft/min/ On to part 2.

**(b) How fast is the surface area increasing at that instant?**
Here we can skip ahead to EQUATION. Write the equation for the surface area of a sphere. We'll label surface area "S" 4(pi)(r^2) = S

DERIVE:

8(pi)(r)(dr/dt) = (dS/dt)

SUBSTITUTE:

8(pi)(5)(1) = (dS/dt)

After a little algebra...

40(pi) = (dS/dt)

And there's your answer! I hope this helped you understand how to do number 11. Credit to PatrickJMT on YouTube for DREDS.

=17) Water is flowing at the rate of 50 m^3/min from a concrete conical reservoir (vertex down) of base radius 45m and height 6m.= =A) How fast is the water level falling when the water is 5m deep?=

Because it is a cone, its equation for the area of a cone is A = (1/3)(pi)(r^2)(h). Take the derivative of the equation. This turns out to be (da/dt) = (1/3)(pi)(2r(dr/dt)(h)+(r^2)(dh/dt)) Our goal is to find what (dh/dt) is. (da/dt) = 50 r = 37.5 (dr/dt) = (15/2)(dh/dt) h = 5 You now substitute your new values into the equation of the derivative to find what (dh/dt) 50 = (1/3)(pi)(2(37.5)((15/2) *(dh/dt))(5)+(37.5^2)(dh/dt)) Do a little bit of algebra and (dh/dt) equals 32/(9(pi) The answer is 32/(9(pi)

=**B) How fast is the radius of the water's surface changing at the moment**= As seen in the step above, (dr/dt) = (15/2)(dh/dt) We can use this formula to find the rate that the radius is changing. With the terms plugged into the formula, it looks like this (dr/dt) = (15/2)(32/(9*pi)) We knew what (dh/dt) was because of part A of the problem. The answer is = -80/(3*pi) This means that the radius is **decreasing** at a rate of 80/(3*pi)