In the previous video, we introduced Leibniz's notation which expresses the derivative in the form of a ratio of differentials. In this video, we'll discuss some contrasting applications by interpreting differentials as small changes in the respective variables. This is closely related to the tangent line approximation of the curve near any particular point of interest. In our first example, we take a square and consider the increase in area as we change the width slightly. Here we've drawn a square of width a 100 meters and colored in the area in blue. Let A denote the the area of a square of side length x, so A equals x squared. The value of A when x equals a 100 is 10,000. So, this square has area 10,000 square meters. How does the area change if we increase the width by say 10 meters, or 5 meters, or 1 meter, or 1 centimeter? In the first case, we expand the width by 10 meters to become 110 meters. We've shaded in pink, the new extra area, which is the area of the larger square minus the area of the original square, and you can check this is exactly 2,100 square meters. In the second case, we expand the width now only by 5 meters to become 105 meters, and we can calculate this extra pink area to be exactly 1,025 square meters. In the third case, we expand the width by just 1 meter to become 101 meters, and the extra area becomes exactly 201 square meters. In the fourth case, we expand the width by miniscule of one centimeter that is 0.01 meters, and the extra area turns out to be 2.0001 square meters. Though the exact calculations are not difficult in this example, I want to show you how you can exploit Leibniz's notation of the derivative to get fast, and perhaps surprisingly accurate estimates for these extra areas. We have this function A equals x squared. The derivative of A with respect to x is just 2x. The trick is to treat the differentials dA and dx like numbers, and dA/dx like an ordinary fraction. So, the multiplying through by dx, we get dA equals 2x dx which we think of as an equation involving differentials. This is an idealized form of an approximation namely, Delta A is approximately equal to 2x Delta x. This gives us a good approximation formula for the change in area Delta A in terms of x and Delta x, the change in the width x. This says that the change in the area of a square is approximately two times the width times the change in the width. In our example with the blue square X equals 100 meters, in case one, the change in x is 10 meters, so the formula for the change in area gives approximately 2,000 square meters. In case two, the change in x is 5 meters and the formula gives the change in area approximately 1,000 square meters. In case three, Delta x is 1, and Delta A is approximately 200 square meters. In case four Delta x is 0.01 and Delta A is approximately 2 square meters. We can compare those estimates with a true increases in area in each of the four cases. One can see that the estimates are quite close to the true increases, and in fact the quality of the estimates improves as the change in x gets smaller and smaller. The changes in area themselves get smaller but the proportional error in the estimates gets vanishingly small as Delta x tends towards zero. This is related to the underlying idea of tangent lines used to approximate curves which motivates the definition of the derivative. In this particular example, there's also a clear visual explanation of why the estimate should be so good. Take a general square of side length x and add a small change in x to the width. To form a slightly larger square of width x plus Delta x, color the original square blue and the oblong areas pink. There's a new much small square colored beige in the diagram that makes up the corner. The area of the largest squares x plus Delta x all squared which we can expand out. The part of this expansion 2x times Delta x is in fact the formula for the estimate of increase that we obtained by considering differentials before, which is the result of combining the two pink colored areas. The last part of the expansion Delta x all squared is the area of the tiny beige square and it's exactly the error in our estimates. As Delta x tends towards zero, this beige square vanishes out of sight becoming insignificant compared to the pink areas, and the quality of the approximation improves towards perfect agreement. The next example is an application of differentials to estimate changes in the volume of a sphere. Suppose you're manufacturing metal ball bearings in the shape of hopefully close to perfect spheres of radius five millimeters. However, your manufacturing process is not perfect, and you expect up to about two percent error in the true radius for any particular ball bearing. Our problem is to estimate the percentage error in the volume of metal required to manufacture these ball bearings. Let V be the volume of a sphere of radius r. It's a fact and also follows from methods we learned in the final module on integral calculus, that v is four-thirds Pi r cubed. Let's differentiate regarded as a function of r. The four-thirds Pi is a constant that can come out the front of the derivative of r cubed. Remember, the derivative of x cubed with respect to x is 3x squared so using r instead of x, the derivative of r cubed with respect to r is 3r squared. Hence dv dr simplifies quickly to four Pi r squared. Multiplying through by the differential dr gives dv equals four Pi r squared dr which we can interpret as an equation involving differentials, which is really an idealization of an approximation involving changes in v and r. Delta v is approximately four Pi r square Delta r, where Delta v is the change in v propagated by Delta r, a small change in r. In our application, v represents the volume of a ball bearing of radius r equals five millimeters, and we're told to expect up to two percent error in the radius. Here's all the information so far. Delta r may be positive or negative depending on whether the actual radius is a bit more or a bit less than five millimeters. We can ignore the sign by taking the absolute value and then the information about the percentage error is really saying that the proportion, the magnitude of Delta r divided by r is less than or equal to 0.02. We're working towards estimating the percentage error in the volume. So, we really want the proportion of the volume v represented by the magnitude of the change in volume Delta v. Thus, we want to estimate the fraction the magnitude of Delta v divided by v. Here's a summary of what we now have, and what we're looking for. We just have to put the pieces together. The magnitude of Delta v over v is approximately this expression, and the numerator can be rewritten as four Pi r squared outside of the magnitude of Delta r because r squared is positive, and then there are some cancellations, and everything simplifies to three times the magnitude of Delta r over r which must be less than or equal to three times 0.02 which is 0.06. Thus, we have the fraction, the magnitude of Delta v over v is approximately less than or equal to 0.06, and we therefore expect up to about six percent variation in the volume of the ball bearings. It's worth noting that the mathematics shows that this estimate is independent of the size of the ball bearings. This method only makes an approximate prediction when translating differentials into actual changes in the variables. Because the percentage error two percent was small, we expect the approximation to be quite good. In the next example, we apply differentials to estimate cube roots, and also interpret our answers in terms of a tangent line to the cube root curve. Let's estimate some cube roots close to the cube root of 64 which is 4. Say the cube roots of 70, 65, and 63. These numbers are all in the vicinity of the 64 whose cube root is exactly 4. So, expect to get cube roots in the vicinity of four. Note that cube roots are fractional powers with exponents one-third. So, we consider the function y equals f of x equals x to the third. Then the derivative we saw last time is a third x to the minus two-thirds. Multiplying through by dx this becomes an equation involving differentials. Which may be interpreted as the following approximation involving changes in y and x. Delta y is approximately equal to a third x to the minus two-thirds times Delta x. In our particular problem, all the values of x are in the vicinity of 64. So, in forming this approximation, we take x equal to 64 so that y equals its cube root which is 4, and the approximation for Delta y simplifies in a few steps to Delta x divided by 48. Now, 48 is close to 50, so we can approximate Delta y even more coursely by Delta x over 50 if we want to. Of course, Delta over 48 is more accurate, but Delta x on 50 may produce simpler rounded numbers if we're only looking for rough approximations. We have all the information we need to start making approximations. Let's estimate the cube root of 70. To move from 64 to 70, Delta x, the change in x is 6. So, Delta y should be approximately 6 divided by 50 which is equal to 0.12. The cube root of 70 which is y plus the change in y Delta y, will be approximately 4 plus 0.12 which is 4.12. The cube root of 70 in fact has this decimal expansion. So, a rough and ready approximation is correct to two decimal places. We estimate the cube root of 65 using the same manipulations, but Delta x now becomes one, and Delta y is approximately one divided by 50 which is equal to 0.02. So, the cube root of 65 which is y plus Delta y now becomes 4 plus 0.02 which is 4.02. The true value again agrees with the approximation to two decimal places. If we use the more accurate estimate for Delta y, that is, Delta x over 48, then our approximation for the cube root of 65 becomes more refined, which astonishingly agrees to the true value to almost four decimal places. Finally, estimating the cube root of 63 this time Delta x is negative 1 because the change from 64 to 63, we must take away one unit, and now Delta y is approximately negative 1 over 50 which is negative 0.02. So, the cube root of 63 is approximately 4 minus 0.02 which is 3.98. Again, agreeing to the true value to two decimal places. Using the more accurate approximation with 48 instead of 50 in the denominator, we get a more refined estimate which again agrees to the true value to four decimal places. To interpret what's happening visually, here's the graph of y equals x to the third. We're focusing on behavior near x equals 64 giving this point on the graph with y equals four. The underlying idea is to take the tangent line to the curve at this point which is a very good approximation to the curve nearby. The value of the derivative at x equals 4 is in fact one of the 48 following from our earlier calculations, and it's not hard to find the equation of the tangent line which is y equals x on 48 plus 8 over 3, and we can use this equation to estimate y values on the actual curve. If you plug in x equals 65 and x equals 63, the equation of the line produces the same approximations we found earlier correct to four decimal places implied by the differentials. In fact, using differentials and tangent lines to approximate points on the curve are entirely equivalent processes, just expressed with different notation.
