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Applications of differentiation - Assistance required!!

After washing the kitchen floor, Alex put his favorite mop flat against the wall and left it there. A few minutes later, the mop starts to slide down the wall. Let h be the height of the top end of the mop above the floor and let y be the horizontal distance of the bottom end of the mop from the wall at any time, t.

a) If the mop is 1.2m tall, express y in terms of x.

b) If the top end of the mop slides down with a constant speed of 5cm/s, find the speed (in terms of h) with which the bottom end of the mop moves away from the wall.

c) Find the speed with which the bottom end of the mop moves away from the wall when the top of the mop:

    i) is 0.8m from the floor
   ii) has slid down by 20cm

d) Find the speed with which the bottom end of the mop moves away from the wall 6 seconds after the top of the mop started sliding down.

I attempted this question but am not sure of my answers, any attemps would be appreciated.

Show us what you've done so far. StuRat 14:27, 2 October 2006 (UTC)

Yes you have to show us what you've done. In the future, sign your naem by typing ```` at the end (or clicking sign your name). 203.122.88.79 04:23, 3 October 2006 (UTC)

Actually, you enter your signature with --~~~~ (two dashes or minus signs and four tildes), not ````. You can insert them by clicking this button: on the toolbar, above the editing area. --CiaPan 06:00, 3 October 2006 (UTC)

That's not entirely accurate either. The signature itself (plus timestamp) is just 4 tildes, ~~~~. It is in some places recommended to put 2 dashes (--) before it. I personally have added -- to my raw signature, and myself only type the ~~~~. -- Meni Rosenfeld (talk) 06:28, 3 October 2006 (UTC)

sorry i didnt sign, here it is: --J p xyz 10:20, 3 October 2006 (UTC)

Ah, the classic example of the falling ladder, except a mop in a kitchen? LOL. Anyways, like all of these sorts of problems, first draw a diagram of what you know, then write down what you DO know (in terms of constants), and then write down what you want to know (in terms of variables). And then try to express what you want to know in terms of what you do know.

Similar example problem:

Given: Definition of square

Given: Definition of right triangle

Given: Pythagorean theorem

Given: A square with a side 1 meter long.

Find: The ratio between the length of the diagonal squared to the area of the square.

• Step One: Draw a diagram.

A square with its diagonal should be drawn.

• Step two: Assign the constants to measurements on the diagram.

Only a single constant, 1, is given: the length of the side of the square.

• Step three: Assign variables to measurements on the diagram.

The length of the diagonal can be called Δ, and the area of the square can be called z.

• Step four: Assign a variable to the answer.

The ratio of the diagonal squared to the area of the square can be called R.

• Step five: Write an equation that relates the answer to the variables.

The ratio of the diagonal squared to the area of the square is our answer. So then, expressed in an equation, it would be:

R = Δ²/z

• Step six: Write equations that relate the constants to the variables, and solve for the variables in step five.

Part One: This is the part where information learned is used in order to solve the problem. This is where knowledge gained is applied to the problem. If nothing was known about the properties of squares, this step could stil be reached. But in order to complete this step, something about the properties of squares would have to be known (which is why it is one of the givens). In this case, two things need to be known: the relationship between the side of a square to the area, and the relationship between the side of a square to its diagonal.

Part two:The relationship between the side of a square to the area is this formula (information given, part of the definition of a square):

A = s² (The area of a square is the square of its side)

The known variables and known constants can be substituted for the variables in this formula. In the square, the side is of length 1 m. The area is z. So that means:

z = 1² = 1 z = 1

So the area of the square is 1 meter squared. In this case (and in yours), it is important to keep track of units.

Part three: Now, the numerator of the right side of the equation in step five, Δ², needs to be found.

To do this requires a second application of knowledge about the properties of squares.

And we now apply the Pythagorean theorem(information given):

c² = a² + b²

The known variables and known constants can be substituted for the variables in this formula. In the square, the side is of length 1 m. The diagonal is Δ. So that means:

Δ² = 1² + 1²

Δ² = 2 Δ² = 2 meters squared

• Step Seven: Solve the equation in step five using information gained from step six.

R = Δ²/z R = (2 meters squared)/(1 meter squared) R = 2/1

And that is our final answer.

• Step Eight: The answer.

The ratio of the length of the diagonal squared to the area of the square is 2/1.

• End

Now, we can use these same problem-solving steps to find the answer to YOUR question.

So let's see the question again:

After washing the kitchen floor, Alex put his favorite mop flat against the wall and left it there. A few minutes later, the mop starts to slide down the wall. Let h be the height of the top end of the mop above the floor and let y be the horizontal distance of the bottom end of the mop from the wall at any time, t.

a) If the mop is 1.2m tall, express y in terms of x.

b) If the top end of the mop slides down with a constant speed of 5cm/s, find the speed (in terms of h) with which the bottom end of the mop moves away from the wall.

c) Find the speed with which the bottom end of the mop moves away from the wall when the top of the mop:

    i) is 0.8m from the floor
   ii) has slid down by 20cm

d) Find the speed with which the bottom end of the mop moves away from the wall 6 seconds after the top of the mop started sliding down.

So let's begin:

Let's do the preliminary first. In my example, it was already done. But now we need to do the preliminary for your problem. Looking at the problem, I'm only going to deal with d, since the rest of those, besides c, have to be found in order to find d. So let's begin:

Preliminary: Rewrite the problem into given/find format.

For the example problem I wrote, I explicity defined what knowledge we had. But unless the problem says otherwise, your first given should always be:

Given: Everything you know about mathematics

An example of a problem where we have to limit ourselves is when we have to prove something using only a certain method. If, for instance, a problem states to use the synthetic division to manipulate a polynomial, but you use the quadratic formula, that is an incorrect solution, because the quadratic formula was not given. (But of course, when you want to check your answer, you can use whatever the heck method you wish.)

And now we rewrite the rest:

Given: h, the height of the top end of the mop above the floor at any time t

Given: y, the horizontal distance of the bottom end of the mop from the wall at any time t

Given: t, time

Given: 1.2 meters, the length of the mop

Given: 5 centimeters per second, the rate of change of the height of the top end of the mop above the floor with respect to time

Given: 6 seconds, the amount of time lapsed.

Find: The rate of change of the distance between the bottom end of the mop and the wall with respect to time.

• Step One: Draw a diagram.

It would look like a right triangle, with the hypotenuse being the mop, the vertical leg being h, and the horizontal leg being y.

• Step two: Assign the constants to measurements on the diagram.

The length of the mop is the only constant given, so in our diagram we can label the length of the hypotenuse as 1.2 meters.

• Step three: Assign variables to measurements on the diagram.

This has already been done for us.

• Step four: Assign a variable to the answer.

First, let's look at the description of the answer:

The rate of change of the distance between the bottom end of the mop and the wall with respect to time.

The distance between the bottom end of the mop and the wall is y. And the unit is time. So we can write the answer in the form dy/dt.

So our variable for the answer will be dy/dt.

• Step five: Write an equation that relates the answer to the variables.

• Step six: Write equations that relate the constants to the variables, and solve for the variables in step five.

• Step Seven: Solve the equation in step five using information gained from step six.

Because this is differentiation, we have to combine these steps because our answer, dy/dt, is already related to the variables, since dy/dt is by definition related to both y and t.

Since our diagram is a triangle, we use the Pythagorean Theorem to solve for h in terms of y. We do this because we want to find dy/dt and not dh/dt (we already have dh/dt, do you know how?), so it makes sense to eliminate the variable we have no use for:

Using the Theorem:

(1.2)² = y² + h²

1.44 = y² + h²

Solve for h:

h² = 1.44 - y² h = sqrt(1.44 - y²) The square root of the quantity 1.44 minus y squared.

So now we save this and return back to our Pythagorean Theorem:

1.44 = y² + h²

Differentiate with respect to time:

0/dt = 2y(dy/dt) + 2h(dh/dt)

Solve for dy/dt:

0 = 2(y(dy/dt) + h(dh/dt))

0 = y(dy/dt) + h(dh/dt)

y(dy/dt) = -h(dh/dt)

dy/dt = (-h(dh/dt))/y

Congratulations, we have found an equation that relates our answer to our variables!

Now, we simply substitute for the variables we already know.

Let's recap:

h = sqrt(1.44 - y²)

dh/dt = 5 centimeters per second

So now we subsitute:

dy/dt = (-sqrt(1.44 - y²))(5))/y

Now we need to solve for y. I'll leave it up to you. I hope all of this helped you. It took me an hour to write all of that. --Ķĩřβȳ♥ŤįɱéØ 15:30, 3 October 2006 (UTC)

Trying to identify a distribution

I have a distribution with pdf

$f_{Q}(q;\lambda ,k,l)={\frac {q^{k-1}\lambda ^{l}}{\left(\lambda +q\right)^{k+l}B\left(k,l\right)}}$

It arises as the quotient of two independent gamma-distributed RVs: $Q\sim X/Y\,$ where $X\sim \Gamma \left(\alpha ,k\right)$, $Y\sim \Gamma \left(\beta ,l\right)$ and $\lambda =\alpha /\beta \,$. It's reasonably well behaved; moments are ${\frac {\lambda ^{n}\Gamma (k+n)\Gamma (l-n)}{\Gamma (k)\Gamma (l)}}$.

However, I can't seem to find it in any of the standard lists of distributions. Does it have a name, or is it not considered interesting enough? EdC 21:27, 2 October 2006 (UTC)

Incircle of a pentagon

Could someone tell me how I find the radius of the incircle of a perfect pentagon? (i.e. a formula or a ratio etc.) Thanks in advance. - ^Рэд_хот 21:30, 2 October 2006 (UTC)

See apothem ☢ Ҡi∊ff⌇↯ 23:48, 2 October 2006 (UTC)

Since a regular pentagon is constructed of 10 identical right angled triangles (of angles 36, 54 and 90), where the adjecent edge to the centre (edge 36,90), is the radius of an incircle.

Given that the length of one side of the regular pentagon is $x$

$sin^{-1}54({\frac {sin36}{0.5x}})$

is the radius, i dont really know what information you've got though. So maybe this equation is not of that much help either. Philc _T^E_C^I 22:57, 2 October 2006 (UTC)

The radius $r$ should be proportional to the length $x$ of a side. Would you believe $r={\frac {x}{2}}\cot 36^{\circ }$?

BTW, $\cot 36^{\circ }={\frac {\sqrt {25+10{\sqrt {5}}}}{5}}$.  --Lambiam^Talk 23:46, 2 October 2006 (UTC)

We are all assuming that by “perfect pentagon” you mean a regular convex pentagon, with all sides the same length. What we do not know is the dimension(s) you already have. Is it the length of a side? Is it the distance from center to vertex (the radius of the circumscribing circle)? As others have indicated, the reasoning is simple trigonometry. A radial segment from the center to the midpoint of an edge produces two right triangles.

However, there is another regular five-sided figure which is also sometimes called a pentagon. It is not convex, and is more often called a pentagram. If this is what you mean, please let us know. --KSmrq^T 00:10, 3 October 2006 (UTC)

No, I had the height of the overall pentagon, and it's overall width (didn't have length of indiviual sides) but I've figured a way around it, but thanks for the answers anyway. - ^Рэд_хот 15:48, 3 October 2006 (UTC)