4.3 Related Ratesap Calculus



This video lesson explores the concept of Related Rates, which is the study of what is happening over time.

Ratesap4.3 Related Ratesap Calculus

Setting up Related-Rates Problems. In many real-world applications, related quantities are changing with respect to time. For example, if we consider the balloon example again, we can say that the rate of change in the volume, (V ), is related to the rate of change in the radius, (r ). AP Calculus LHopitals Rule and Related rates. AP Calculus Learning Objectives Explored in this Section. Calculate derivatives; Interpret the meaning of a derivative.

To solve problems with Related Rates, we will need to know how to differentiate implicitly, as most problems will be formulas of one or more variables.

But this time we are going to take the derivative with respect to time, t, so this means we will multiply by a differential for the derivative of every variable!

But don’t let this confused you, as there are only four basic steps when problem-solving with Related Rates:

  1. Write down every given information or quantity, even what you’re looking for, and make a sketch.
  2. Determine the equation needed to solve for your unknown quantity.
  3. Take the derivative implicitly, not forgetting to multiply by your differential for every single variable!
  4. Plug in all of your given information or quantities and solve.

And here’s a big hint…

…most of the time all we have to do is find an equation that relates the rate we’re looking for to a rate that we already know, as Paul’s Online Notes so nicely states.

This means we need to look for geometric shapes, known formulas, and ratios.

Ladder Sliding Down a Wall

Together we will solve these 8 classic questions:

  • Pebble Dropped In Pond
  • Air Being Pumped Into A Balloon
  • Edges Of Cube Expanding
  • Radar Tracking Station & Airplane Flight Path
  • Sand Falling onto a Conical Pile
  • Water Pouring into a Conical Tank
  • Boat Pulled into a Dock
  • Ladder Sliding Down a Wall

How to Solve Related Rates – Video

1 hr 35 min

  • Overview of Related Rates + Tips to Solve Them
  • 00:02:58 – Increasing Area of a Circle
  • 00:12:30 – Expanding Volume of a Sphere
  • 00:21:15 – Expanding Volume of a Cube
  • 00:26:32 – Calculate the Speed of an Airplane
  • 00:39:13 – Conical Sand Pile
  • 00:51:19 – Conical Water Tank
  • 00:59:59 – Boat & Winch
  • 01:09:13 – Ladder Sliding Down A Wall

Related Rates


Related Rates – 2 Examples

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The radius of a sphere is increasing at a rate of 2 meters per second. At what rate is the volume increasing when the radius is equal to 4 meters?

4.3 related ratesap calculus problems

This type of problem is known as a 'related rate' problem. In this sort of problem, we know the rate of change of one variable (in this case, the radius) and need to find the rate of change of another variable (in this case, the volume), at a certain point in time (in this case, when r = 4). The reason why such a problem can be solved is that the variables themselves have a certain relation between them that can be used to find the relation between the known rate of change and the unknown rate of change.

Related rate problems can be solved through the following steps:
Step one: Separate 'general' and 'particular' information.
General information is information contained in the problem that is true at all times. Particular information is information that is true only at the particular instant that the problem is asking about.

For example, in this problem, the general information is that the radius is growing at a rate of 2 meters per second. Thus, if we let the radius of the sphere be represented by r, we can say that r'(t)=2 m/s.

Calculus

The particular information is that at the moment of interest, r(t) = 4.

It is important to separate out these two types of information because information about the particular case must be set aside and must not be used until the very last step of solving the problem. One mistake that students frequently make is to use the particular information too early. This invariably leads to incorrect answers.

4.3


Step two: Draw a sketch of the situation using the general information only.This often helps to illuminate the relationship between the variables.
In this case, the general information leads to the following sketch:

Although we know the rate of change of r, we cannot depict a rate in the sketch.

Note here that it would be a mistake to write 'r = 4' in the sketch, since this is only true at one particular instant.


Step three: Identify the known rate and the desired rate

In this case, we know , and we want to find , the rate of change of the volume.


Step four: Use geometry or trigonometry to relate the variable with the known rate of change to the variable with the unknown rate of change.

The appropriate relation will often involve one of the following:

  1. Spheres: Radius vs. Surface area (4Πr2)vs. Volume (Πr3)
  2. Circles: Radius vs. Area (Πr2) vs. Circumference (2Πr)
  3. Cylinders: Radius and Height vs. Volume (Πr2h)
  4. Cones: Radius and Height vs. Volume (Πr2h) Also see similar triangles below.
  5. Right Triangles: Pythagorean Theorem (a2 + b2 = c2)
  6. Similar Triangles: = =
  7. Angles: Problems with angles will typically involve a trigonometric relation. Recall the 'SOHCAHTOA' relations (sin = , cos = , and tan = )
      In this case, the appropriate relation is
      Recall that both V and r are variables with change with time. Therefore, it is best to write the relation in a way that indicates this fact:
      V(t) = Πr(t)


      Step Five: Differentiate both sides of the equation with respect to timeNow that we have found a relationship between the variables, we must differentiate both sides of the equation to find a relationship between their rates of change (i.e., their derivatives). It is crucial to remember to use the chain rule here, since we are taking derivatives with respect to t:




      Step 6: Plug in the particular information to solve for the desired rate.Now, we can finally use the particular fact that r = 4 to solve for the unknown rate:


      = 4Πr(t)
      = 4Π42
      = 128Π

      So, when the radius is 4 meters long, the volume is increasing at a rate of 128Π cubic meters per second.