Math Is Fun Forum

explore

Member

Registered: 2010-01-27

Posts: 5

Application in the derivative ..?

I hope that someone help me in solving these application. (plz, I want to solution step by step)..

Q1:

A Cassna plane takes off from an airport at sea level and it's altitude (in feet) at time t (in minutes) is
given by h= 2000 ln (t+1) .
find the rate of climb at time t = 3min .

Q2:

If the variable sound power W is given by W = T^2 + t + 1 , Find the rate of change of the sound
pressure P , at time t= 3 s .

Q3:

A computer is programmed to inscribe a series of rectangle in the first quadrant under the curve of y=e^-x
What is the greatest area of the largest rectangle that can be inscribed ? .

Q4:

A point moves along the x axis so that it's position x at time t is specificed by the function x(t)= t^3-8t+8 .
determine the following :
a- the acceleration at times when the velocity is zero .
b- the average velocity over the time interval (0,6) .

I'm sorry for all this quastion ..

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Application in the derivative ..?

Hi explore;

Q3 is the most interesting problem to me so:

If you draw the graph of y = e^(-x) here:

http://www.mathsisfun.com/graph/functio … 4163972222

You see that I am going to have to make an assumption to simplify the problem. I am going to assume that the rectangle starts at the y axis and is drawn horizontally from there. If we choose any point on the y axis and draw a horizontal line to the the curve exp(-x) and then drop the end to the x axis we will have a rectangle.

We can now try to find the one with the largest area of those.

Choosing a point (0, p )  yields a corresponding point (- log(p),p) on the curve e^(-x) so the area of that rectangle would just be p * - log(p). For instance if I chose (0,.2) and drew a horizontal line to the curve e^(-x) it would intersect it at (-log(,2) , .2). This rectangle is .2 high and - log(.2) long, so the area is .2 * - log(.2)

For the general point p we must maximize A = p * - log(p). Differentiate p * - log(p)

Set the derivative to 0 and solve for p:

p = 1 / e So the maximum area occurs when we choose p = 1 / e and the maximum Area is also 1 / e which is approximately .367879

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Application in the derivative ..?

Hi;

For Q4

My mechanics is a little rusty so I hope I get this right.

The equation of motion is:

x(t)= t^3-8t+8

The velocity v is the first derivative:

x '(t)= 3t^2 - 8

Set this to 0 and solve you get:

t = 2 √(2 / 3) and t = - 2 √(2 / 3)

You can't have a negative time so we drop the second root.

t = 2 √(2 / 3) is when the velocity v is 0. The acceleration is the second derivative.

x ''(t)= 6 t

So for a) the acceleration when v = 0 is 6 * 2 √(2 / 3) = 12 √(2 / 3)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

explore

Member

Registered: 2010-01-27

Posts: 5

Re: Application in the derivative ..?

Hi bobbym ..

Now I feel that these applications were easy. I really really thank you my friend for this solution .

Regards ..^_^

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Application in the derivative ..?

Thanks, glad to help:

For Q1)

given by h= 2000 ln (t+1) .

The rate of climb is given by the derivative:

So at T= 3 we get:  2000 / 4 = 500 ft/min

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

explore

Member

Registered: 2010-01-27

Posts: 5

Re: Application in the derivative ..?

Nice, but why you make the first derivative for the rate of climb ??

regards .

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Application in the derivative ..?

Hi explore;

The rate of change of a function is the derivative. Climbing is a change in height.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

explore

Member

Registered: 2010-01-27

Posts: 5

Re: Application in the derivative ..?

Thank's alot dear bobbym ..

this forum now in my favorites ..

regards..

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Application in the derivative ..?

Hi explore;

Thank's alot dear bobbym ..

this forum now in my favorites ..

Your welcome and I hope some of the above work is actually correct. My physics is worse than my german or my math.

Glad you like the forum but that has nothing to do with me. MathsisFun is the creator of it. I just hang out here.

You will agree that this forum is friendlier than the other one you posted these problems on. (I am a member there also)

For Q2) I am having a problem, how do you determine P?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

explore

Member

Registered: 2010-01-27

Posts: 5

Re: Application in the derivative ..?

The questions from this book "Mathematics HL - International Baccalaureate Diploma Programme"

I don't have any information about how they determine P ..

but I think this is the solution :

W = t^2 + t + 1.
dW/dt = 2t + 1.
Plug in t = 3 s to get
dW/dt = 2(3) + 1 = 7 (presumably 7 Pa/s).

regards ^_^ ..