Video Transcript
The length of a rectangle is
increasing at a rate of 15 centimeters per second and its width at a rate of 13
centimeters per second. Determine the rate at which the
area of the rectangle increases when the length of the rectangle is 25
centimeters and its width is 12 centimeters.
We’re given quite a lot of
information about the area and information about the length and width of a
rectangle. Let’s begin by defining the
length of our rectangle to be equal to 𝑙 centimeters and its width to be equal
to 𝑤. And so it’s area 𝐴 will be
equal to its length times its width 𝑙𝑤. That’s not quite enough,
though. We’re looking to find
information about the rate of change of the area, in other words, the derivative
of area 𝐴 with respect to 𝑡. And of course, the information
we have about the area of the rectangle is in terms of its length and its
width.
Luckily, we can use a
combination of implicit differentiation and the product rule to find an
expression for d𝐴 by d𝑡 in terms of 𝑙, 𝑤, and their respective
derivatives. Let’s begin with the product
rule. 𝐴 itself is the product of
what we assume are two differentiable functions 𝑙 and 𝑤. Now we can make that assumption
because in the question we’re given the rate of change of 𝑙 and the rate of
change of 𝑤. So the derivative with 𝐴 with
respect to 𝑡 according to the product rule is 𝑙 times d𝑤 d𝑡 plus 𝑤 times
d𝑙 by d𝑡.
And this is really useful
because we’re told the length of the rectangle increases at a rate of 15
centimeters per second. And so the derivative of 𝑙
with respect to 𝑡 must be positive since it’s increasing and it must be equal
to 15. Similarly, we can say that the
derivative of 𝑤 with respect to 𝑡, the rate of change of its width, is 13. Substituting these values into
our earlier expression for d𝐴 by d𝑡, and we get 13𝑙 plus 15𝑤. And actually that’s everything
we need because we’re told that the length of the rectangle at a given point is
25 centimeters and its width is 12.
And so we’re going to
substitute these values into our expression for d𝐴 by d𝑡. That gives us 13 times 25 plus
15 times 12, which is 505. This is the rate of change of
its area. And area will be in square
centimeters. And we’re also told that the
time is given in seconds. And so we see the rate at which
the area of the rectangle is increasing is 505 square centimeters per
second. Note that even if we hadn’t
been told that it was increasing, we could infer that from the sign of the
derivative; it’s positive.
