[Math] calculus integration question

calculusderivativesintegration

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?

Best Answer

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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