

Related Rates Explained
I will now give a sample problem for us to work with. Problem  You have a spherical balloon, and the volume is
increasing by 4 cm^{3}/s. At this point in time, the radius has reached
8 cm. How fast is the radius changing at this point in time? [The equation for a
sphere is V = (4/3)∏r^{3}, see reference
tables] The method for solving for a variable's rate of change
with respect to time is as follows: Step 1  Write down every variable that changes with
respect to t (time). In this case, that is both V and r. Step 2  Write down an equation that relates all
these variables. I gave it here, in case you don't know it. V = (4/3)∏r^{3}
Step 3  Take the derivative with respect to t. In
order to do this you must know how to do implicit
differentiation. I am not going to repeat that in its entirety over here.
Basically, you multiply the normal derivative of a variable with its prime. x^{2}
becomes 2x(x'). In this case, our equation is differentiated to (1)(V’) =
(4/3)(∏)(3r^{2})(r’). You see that both V and r were
differentiated with respect to t, not each other. That is why we have the V' and
r' in the equation. Step 4  Replace each variable with its number. V’
= 4, as stated in the problem. r = 8, also stated. That leaves one unknown,
r’. This is the rate of change of r with respect to t, and you can now solve
for it. 4 = 4(∏)(64)(r’) r’ = 1/(64∏) Important Cautionary Note: My notation in these problems may be a bit confusing to you, so let me clarify something. The proper way to denote "derivative of V with respect to t" is dV/dt. With this notation, the last equation (1)(V’) = (4/3)(∏)(3r^{2})(r’) should have read (1)(dV/dt) = (4/3)(∏)(3r^{2})(dr/dt). Writing it in this way is actually more detailed than the way that I do it, but I feel I like to keep things simple. Just don't be confused when you come across a problem with x and y, and think that y' means dy/dx. These steps can be used for many questions on related
rates. Here is another problem I have on the topic. Problem – A 10 foot ladder is leaning on a wall, slowly sliding down. If the ladder is sliding away from the wall at 1 foot/second, how fast is the top sliding down the wall, when the bottom is 6 feet away? Solution: Step 1  The distance from the bottom of the ladder to the wall is x, and the height of the top is y. These are both changing with respect to t. Step 2 – x^{2} + y^{2} = 10^{2} = 100 This works because the Pythagorean theorem must hold at any point while the ladder slides down the wall. The length of the ladder will be the hypotenuse, and it is a constant 10. Step 3 – 2x(x’) + 2y(y’) = 0 This equation relates our given variables to time. Step 4 – 2(6)(1) + 2y(y’) = 0 You still have two unknowns, so a bit of ingenuity is
required. Y can be solved, through x^{2} + y^{2} = 100, because
at this time x is 6. Solve for y and get y = 8.
(It should occur to you that if you ever have a variable you do not know, the
answer will be somewhere in the question, or in the original equation that you
created at Step 2 of the problem.) Now plug this into the equation with two unknowns:
2(6)(1) + 2(8)(y’) = 0 y’
= 3/4 Note in this last step how we solved what y was at the point where x is 6, and plugged that into our final formula. I just want to make were that you understand that it is at this point that you plug it in, and not earlier. Y is a variable changing with respect to time, and if we entered its value at that point before the derivative was taken, it would be a constant! We had to keep its identity throughout the problem before narrowing down its value to the point at which we were planning on checking it.
Problem  You are given a watch, with the minutes hand 5 mm long, and the seconds hand 8 mm long. At 2:10, how fast is the distance between the tips of the hands changing? Solution: The length of the minutes hand is 5, the length of the seconds hand is 8. These do not change with respect to time. X is the distance between the tips, and A is the angle theta between the two hands. X and A are changing with respect to time. The equation that relates these is apparently the law of
cosines. This equation is X^{2} = 5^{2} + 8^{2}  2(5)(8)(cos A) The derivative of this equation is: 2X(X') = 80(sinA)(A') To find these values to plug in, we must use the equations
given. At 2:10, We also do not know the speed that A is changing, or A'. To do that, use commons sense. The minutes hand is moving away from the seconds hand at 2∏/60 per minute, and the seconds hand is approaching at 2∏ per minute (one revolution per minute). At 2:10, the angle is shrinking, and the speed is at 2∏  2∏/60 which is 59∏/30. 2X(X') = 80(sinT)(T') ==> 2(7)(X')
= 80(√3/2)(59∏/30) That's not a very clean answer, is it? I finally found a couple of decent problems to put in the problem set for this section. (Link is below) If you have any suggestions for others, don't be shy! Prev Lesson  Problem Sets  Next Lesson  Index of Topics  Reference 
