1. Calculus: Applications of Derivatives?

Having trouble with my calculus homework... The questions are:

1. The area of a rectangle is increasing at a rate of 5 m^2/s, while the length is increasing at a rate of 10 m/s. Determine how fast the width is changing when the length is 20m and the width is 16m.

2. A road crosses over a canal at a right angle. A boat traveling in the canal at 36 km/h passes under the road just as a car on the overhead road crosses over the canal travelling at 48 km/h. Determine how fast the boat and the car are separating 10 minutes after they cross.

3. Sawdust is falling onto a pile at a rate of 1/2 m^3/min. If the pile maintains the shape of a cone with height equal to the diameter of the base, then determine how fast the height of the pile is increasing when the pile is 3m high.

I don't even know where to start. =/

2. Originally Posted by Cougar22
Having trouble with my calculus homework... The questions are:

1. The area of a rectangle is increasing at a rate of 5 m^2/s, while the length is increasing at a rate of 10 m/s. Determine how fast the width is changing when the length is 20m and the width is 16m.

2. A road crosses over a canal at a right angle. A boat traveling in the canal at 36 km/h passes under the road just as a car on the overhead road crosses over the canal travelling at 48 km/h. Determine how fast the boat and the car are separating 10 minutes after they cross.

3. Sawdust is falling onto a pile at a rate of 1/2 m^3/min. If the pile maintains the shape of a cone with height equal to the diameter of the base, then determine how fast the height of the pile is increasing when the pile is 3m high.

I don't even know where to start. =/
I understand the feeling. Related rates can be quite troublesome when you're first doing them.

I have the first two solved. I'll display them here step by step, and while you ponder them over, I'll try to figure out #3.

#1

You're given a rectangle with a length of 20 m and a width of 16 m. You're also given two rates:

$\frac{dA}{dt}=5m^2/sec$
$\frac{dl}{dt}=10 m/s$

What you're asked to find is find the rate of the width: $\frac{dw}{dt}$.

First, find a formula for a rectangle that includes area, width, and length. That formula is A = lw. Of course, we can't plug in the rates to just that, so take the derivative of the formula. Notice that we have to use the product rule on the right side.

$\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

Now plug in everything you know, and solve for the missing variable.

$5 = 20\frac{dw}{dt} + 16(10)$

Simplify, and you'll get.

$\frac{dw}{dt} = -7.75 m/s$

Be sure to include the second unit.

#2

Ok, this is a right triangle. x = 36 km/h and y = 48 km/h. There is no rate here except the one we need to find (the hypotenuse, which I'm going to call w). However, notice that the problem is asking for the rate after ten minutes have passed, so what you can do, is take x and y, and divide each of those by 60, in order to convert it to minutes. That'll give you your dx and dy.

Here's what we know now.

x = 36 km/h

y = 48 km/h

$\frac{dx}{dt} = .6 km/min$

$\frac{dy}{dt} = .8 km/min$

Now, we need to find a formula for this triangle. To find the missing side, we will need to use the Pythagorean Theorm. Before we get into anything, lets go ahead and find w (trust me on this).

$36^2 + 48^2 = w^2$
$3600 = w^2$
$60 = w$

Ok, we found w. Lets go ahead and take the derivative of the Pythagorean Theorem.

$x^2 + y^2 = w^2$
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2w\frac{dw}{dt}$

Notice that we now need w to solve the equation. At this point, plug everything you know in.

$2(36)(.6) + 2(48)(.8) = 2(60)\frac{dw}{dt}$

Solve for the missing variable.

$43.2 + 76.8 = 120\frac{dw}{dt}$
$120 = 120\frac{dw}{dt}$
$1 = \frac{dw}{dt}$

Keep in mind that this is in kilometers/minute after one minute has passed. To find it at the 10 minute mark, we have to multiply by 10 to find the total distance.

1(10) = 10 km

3. Originally Posted by Cougar22
Having trouble with my calculus homework... The questions are:

3. Sawdust is falling onto a pile at a rate of 1/2 m^3/min. If the pile maintains the shape of a cone with height equal to the diameter of the base, then determine how fast the height of the pile is increasing when the pile is 3m high.
This one is actually the easiest of the three you posted, because the only extra thing you have to find is the radius.

Here's what we know right now.

$\frac{dV}{dt}= \frac{1}{2}m^3/min$

We also know height, but only at the instant the pile is 3m high; in other words, we do not know the original height.

Find a formula that has volume, and height in it. The formula we will use is $V = \frac{1}{3}\pi r^2h$.

Ok, lets take the derivative of the equation, keeping in mind that 1/3 and pie, and r are constants (r is a constant because there is absolutely no way to find $\frac{dr}{dt}$ if we were to take the derivative of it).

$\frac{dV}{dt} = \frac{1}{3}\pi r^2 \frac{dh}{dt}$

Ok, we need an r, but we're only given height. Keep in mind that the height will always be the same as the diameter. If the height is 3m, then the diameter is 3m. The radius of a circle is half of the diameter, so the radius is $\frac{3}{2}$.

Plug in what we know into the derivative (the height was only there to enable us to find the radius; it is essentially useless from this point on).

$\frac{1}{2}= \frac{1}{3}\pi(\frac{3}{2})^2\frac{dh}{dt}$

Simplify.

$\frac{1}{2}=\frac{1}{3}\pi\frac{9}{4}\frac{dh}{dt}$

$\frac{1}{2}=\frac{3}{4}\pi\frac{dh}{dt}$

$\frac{2}{3} = \pi\frac{dh}{dt}$

$\frac{2}{3\pi} = \frac{dh}{dt}$

The last part was found by dividing by pie, then simplifying the complex fraction by mulitplying both numerator and denominator by $\frac{1}{\pi}$.

Be sure to include the unit meters/minute.

Hope the last couple of posts helped you in some way

4. Thank you very much. You explained it better than my teacher, lol.