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Mik

Guest

Related Rates problem

Lately I've been practicing my calculus but have stumbled upon two questions that I just can't figure out the answers to. The problem statement:

A ladder 50 feet long is leaning against a wall.  If the foot of the ladder is being pulled away from the wall at a certain rate, derive the formula for how fast the top of the ladder is descending. Let x be the distance of the foot of the ladder from the wall, and let h be the distance of the top from the ground.

From the pythagorean theorem I have:

Differentiating h with respect to t:

Then,

If the foot is being pulled away at the rate of 3 ft/min: how fast is the top descending when the foot is 14 feet from the
wall?

I solve this by plugging in the rate and distance x and solving:

Now comes the questions I can't figure out the answers to:

When is the top descending at the rate of 4 ft/min?

When will the top and bottom move at the same rate?

I know that the answer to the first question is 40 feet, and 25*sqrt(2) for the second, but how to get there?!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,163

Re: Related Rates problem

hi Mik,

Welcome to the forum.

Answer up to -7/8 is correct.

Last part:

Square both sides and you'll get that answer easily.

I'm a bit confused by the other question as it asks 'when?' and the answer is a distance not a time.  Also my method involves integration and I cannot see how to find the constant of integration.  Here's what I did:

so we can rearrange as

The right hand side will integrate fairly easily but with a +C.  Then what?

I'll give it some more thought and post again if I have any inspiration. 

LATER EDIT:

If you put in that answer (40) and dh/dt = -4 you get dx/dt = 3

So maybe that's what the questioner intended.  Put dx/dt = 4 and dh/dt = -4 and solve for x.

Bob

Last edited by Bob (2014-09-28 05:41:30)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Mik

Member

Registered: 2014-09-30

Posts: 1

Re: Related Rates problem

Bob bundy - thankyouthankyouthankyou!

I've been banging my head into the wall for far too many times the last couple of days, trying to figure out the solution, but after reading (and re-reading) your post several times it finally dawned on me. It felt like something just *clicked* in place and I "leveled up". Gotta love math!

For posterity's sake I'll post the solutions:

When is the top descending at the rate of 4 ft/min?

Like bob bundy wrote, the trick is to use the derived formula and set the descending rate to -4 (since the height h will be decreasing), and then solve for x. We have:

When will the top and bottom move at the same rate?

The solution is to realize that when the top and bottom are moving at the same rate, then dh/dt will be equal to dx/dt. From previously, we have the formula:

If we solve for x, dx/dt goes to the other side as the reciprocal:

Then substituting h:

Once again, thanks for the help!