calculus integration question

We have a weekly assignment and the teacher posts solution but doesn't EXPLAIN how she got the answer. just gives you the answer.

So I got this question wrong and I need help on how the answer was found..

The length of a rectangle is increasing at 2 m/s and with is increasing 1 m/s. when the length is 5m and width is 3 m how fast is the area increasing?

a ladder 10 meters long is leaning against a wall, with the foot of the ladder 8 meters away from the wall. if the foot of the ladder is being pulled away from the wall at 3 meters per second, how fast is the top of the ladder sliding down the wall?

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It looks to me you need differentials here, not integrals... seeing we're all about rates of change in this question... – J. M. Nov 29 '11 at 10:34

I'll solve your second problem. Which is better stated as:

A ladder 10 meters long is sliding against a wall. If the foot of the ladder is being pulled away from the wall at 3 meters per second, how fast is the top of the ladder sliding down the wall when the foot of the ladder is 8 meters away from the wall. ?

1. Identify the variables in the problem. What is changing? It is very important to introduce and name the variables here.

The height $h$ from the top of the ladder to the floor and the length $l$ from the bottom of the ladder to the wall are changing.

2. Ask yourself: "What rates of change do I know?" and "What rate of change is being asked for?".

You know $l$ is increasing at a rate of 3, so ${dl\over dt}=3$.

You need to find the rate of change of $h$ when $l=8$. So, you want to find ${dh\over dt } \Bigl |_{l=8}$.

Ok, we need to find a rate of change of $h$ and we have these variables $l$ and $h$...

3. Write an equation relating the variables.

By the Pythagorean Theorem: $$\tag{2}l^2+h^2=100.$$

But, we want to find $h'$. How to get that?
4. Implicitly differentiate (2) with respect to time $t$ to obtain: $$\tag{3}2l{dl\over dt}+2h{dh\over dt} =0.$$
5. Now substitute what you know into (3) and solve for what you don't:

You are given ${dl\over dt}=3$ and $l=8$ and you can calculate $h=\sqrt{100-64}=6$.

${dh\over dt}\Bigl |_{l=8}$ is what we are trying to find.

Now substitute this information into (3): $$2\cdot8\cdot 3+2\cdot6\cdot{dh\over dt}\Biggl |_{l=8} =0,$$ and solve for ${dh\over dt}\Bigl |_{l=8}$: $${dh\over dt} \Biggl|_{l=8}=-{ 48\over 12}=-4.$$

I should have used units throughtout, but was to lazy to... Note that the answer should be negative since the top of the ladder would be moving down.

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Thank you for all your help I understand it now. You should be the teacher ^_^ – Raynos Nov 29 '11 at 14:50

Let's look at your first question: let $w(t)$ be the width, $l(t)$ the height, and $A(t)$ the area of your rectangle, as a function of time.

We know from geometry that $A(t) = l(t)w(t)$.

We want to find the rate at which the area is changing: in other words, the derivative of area with respect to time. Using the product rule, $$A'(t) = l'(t)w(t)+l(t)w'(t).$$ To evaluate the formula on the right, we need to know (from left to right): the rate of change of the length, the current width, the current length, and the rate of change of the width. These were given in the problem, so $$A'(t) = 2\ \textrm{m/s} \cdot 3\ \textrm{m} + 5\ \textrm{m} \cdot 1\ \textrm{m/s} = 11\ \textrm{m}^2\textrm{/s}.$$ (Sanity check: the units of the answer, meters squared per second, make sense for measuring the rate of change of area.)

Can you now do the ladder problem on your own?

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Thank you so much it makes a lot of sense now – Raynos Nov 29 '11 at 14:50