How a Changing Length Changes an Area

Start with a square you can measure

Begin with two copies of x as the side lengths. At x equal to two, each side is two and the area is four. The original rectangle gives us a reference. We will change the input and watch how the area responds.

Increase the input by three tenths. Both sides grow from two to two point three. The growing strips and corner fill the space between the original square and the new boundary.

The new area is five point two nine. Subtract the original four to find an increase of one point two nine. The input grew by three tenths, each side grew by three tenths, and the area grew by one point two nine. These are different quantities.

Shrink the input change to one tenth while keeping the starting input at two. The new sides become two point one, and their area becomes four point four one. A smaller input change produces smaller added regions.

One input can change two lengths differently

Now keep the first side equal to x but choose x squared for the second. At x equal to two, the old lengths are two and four. Their product is eight. One input determines both lengths through two different functions.

Grow an input change of three tenths. The first length becomes two point three. The second becomes two point three squared, or five point two nine. Its change is one point two nine, which is larger than the input change itself.

Look at the first added strip. Its width is the change in the first factor, three tenths. Its height is the old second factor, four. Their product is one point two. The original side supplies the unchanged dimension of this strip.

Now look at the other strip. Its width is the old first factor, two, and its height is the second factor change, one point two nine. Its area is two point five eight. The two strips are unequal because the two functions respond differently.

Where both changes meet

Where the two new lengths meet, there is one more region. Its dimensions are three tenths by one point two nine. The corner contributes zero point three eight seven. It is separate from both strips.

View the full changed rectangle again. One point two, two point five eight, and zero point three eight seven total four point one six seven. Add the original eight and obtain twelve point one six seven, the new product.

Reduce the input change to one tenth. The first length changes by one tenth; the squared length changes by zero point four one. Watch all three added regions shrink while the original two-by-four rectangle stays fixed.

The new area is two point one times four point four one, or nine point two six one. The increase above eight is one point two six one. The corner is smaller now, but it is still part of the exact changed area.

Change the starting input instead

Return to a three-tenths input step, but start at one point five. The old lengths are one point five and two point two five. Their area is three point three seven five. Changing the starting input changes the reference rectangle itself.

Grow the same three-tenths step from this new starting point. The lengths become one point eight and three point two four. Their product is five point eight three two. The area increase is two point four five seven.

Compare the earlier starting input, two. The same step now gives an area increase of four point one six seven. The input step is unchanged, but the starting rectangle and the second factor change are larger.

These are two separate experiments: changing the starting input, and changing the size of the step from it. Keep one fixed while comparing the other. The controls let you make that distinction directly.

Swap the factors while preserving the product

Return to x times x squared at input two, with an input step of three tenths. The old side lengths are two and four, and the old area is eight. One side changes by three tenths and the other by one point two nine. Keep those unequal responses in mind before swapping the functions.

Exchange the functions: the first is now x squared and the second x. The old rectangle is four by two, still area eight. Its changed lengths are five point two nine and two point three, giving the same new product twelve point one six seven. Swapping factors changes the orientation of the calculation, not the resulting area.

Emphasize the strip associated with the first factor change. Its width is now one point two nine and its unchanged height two, giving area two point five eight. This is the contribution that previously belonged to the other strip. The function roles exchanged, so their strip contributions exchanged too.

Return to the full partition. The other strip contributes one point two and the corner zero point three eight seven. Their total remains four point one six seven. The corner multiplies both changes, so reversing their order leaves its area unchanged. The complete finite identity respects the symmetry of multiplication.

Let both side lengths respond quadratically

Choose x squared for both side lengths, starting at input two. Each old side is four, giving area sixteen. The input is still two; it is not equal to either side length. This distinction becomes especially useful when both functions differ from the input itself.

Grow an input step of one tenth. Each side becomes two point one squared, or four point four one. Each side change is therefore zero point four one, not one tenth. The completed area is four point four one squared, or nineteen point four four eight one.

Look at one strip. Its unchanged dimension is four, and its changed dimension is zero point four one. The area is one point six four. The other strip has the same area because the two selected functions are equal. Equality of the strip areas comes from equality of the actual side responses.

The corner has both changed dimensions, zero point four one by zero point four one, giving zero point one six eight one. Two strips of one point six four plus that corner total three point four four eight one. Adding the original sixteen recovers the complete new area.

Use the input step as the rate denominator

Keep the two squared functions and the one-tenth input step. The area increase is three point four four eight one. Dividing by the input change one tenth gives an average rate of thirty-four point four eight one. Dividing by the side change zero point four one would answer a different question, an area change per unit side change.

Compare x times x squared starting at one point five, with a one-tenth step. The old product is three point three seven five and the new product four point zero nine six. The increase is zero point seven two one, giving an average rate of seven point two one. The same input step does not produce the same area rate when the functions or their starting values differ.

Return to x times x squared, now starting at input two point five. The old sides are two point five and six point two five, with product fifteen point six two five. We will grow the input by two tenths. The reference area has changed because the starting input changed.

Grow that two-tenths step. The new input is two point seven, giving lengths two point seven and seven point two nine. Their product is nineteen point six eight three. The area increase is four point zero five eight, so the average rate across the input step is twenty point two nine.

Shrink the step while holding the starting input two point five. The original rectangle stays fixed, and the added regions contract. The quotient approaches the local rate eighteen point seven five. The changing display still uses a positive step; its purpose is to compare finite responses, while the exact limiting value comes from the product relationship.

From an area change to a rate

Use the two-by-four reference and a step of three tenths again. The area increase is four point one six seven. Dividing by the input change gives thirteen point eight nine units of area change per unit input. This is an average rate across a finite step.

Reduce the step to one tenth. The area increase is one point two six one, so dividing by one tenth gives twelve point six one. The average rate changed even though the starting input stayed two.

Shrink the step toward the smallest positive setting. The added regions become thin. The displayed quotient approaches twelve, although the finite setting is still above twelve. A local rate asks what these quotients approach as the step approaches zero.

The World stops at a small positive step, not at zero. We can compare its finite regions and readouts; explaining the exact limiting rate will require connecting those regions to the rates of the two side lengths. That is the next idea in this sequence.