Video Transcript
As the upper end of a ladder slides
down a wall at 𝑘 centimeters per second, its base on the ground slides away from
the wall. How fast is the base sliding at the
moment when the angle 𝜃 between the ladder and wall satisfies csc 𝜃 equals five
over four?
Let us first start by drawing a
diagram to help us understand the problem. We can imagine a right triangle
representing the side-on view of the situation, where the left side is the wall, the
bottom side is the ground, and the hypotenuse is the ladder. As described in the problem, the
upper end is sliding down the wall, while the base on the ground is sliding away
from the wall. Furthermore, the angle between the
ladder and the wall is denoted by 𝜃.
It will also be helpful for us to
label the axes of this graph using Cartesian coordinates. Note that we will choose to
orientate 𝑦 as going down the wall, since this means we can think of motion in this
direction as positive.
Now, in this question, that we want
to find the speed of the base sliding to the right, which is the change of 𝑥 with
respect to time, or d𝑥 by d𝑡. However, we have not been
explicitly given what this actually is. Instead, we have been given the
speed of the ladder sliding down the wall, which is d𝑦 by d𝑡, and this is given to
be 𝑘 centimeters per second.
We have also been told that csc 𝜃
equals five over four. Now, by definition, csc 𝜃 equals
one over sin 𝜃, so we have one over sin 𝜃 equals five over four. And by taking the reciprocal of
both sides, we have sin 𝜃 equals four over five, which, by basic trigonometry, is
the ratio between the opposite side to the angle and the hypotenuse.
So, assuming that our example does
in fact form a right triangle, we can mark these sides, five and four, on the
diagram. We can also figure out that the
missing side is three, since these numbers form a Pythagorean triplet. Note that these are just the ratios
of the side lengths and not the actual lengths.
Now, how can this information be
used to solve our current problem? Well, since we want d𝑥 by d𝑡 and
have d𝑦 by d𝑡 and a relation between 𝑥 and 𝑦 in terms of the triangle, we can
use what we know about related rates of change to help us. Specifically, due to the chain
rule, we have that d𝑥 by d𝑡 is equal to d𝑥 by d𝑦 times d𝑦 by d𝑡. Incidentally, while these
derivatives are not fractions, we can verify that the rule we have written is
correct by considering what would happen if we treated them so, giving us d𝑥 by
d𝑡.
Now, we already have d𝑦 by d𝑡,
but what about d𝑥 by d𝑦? We want to find d𝑥 by d𝑦 at the
instant that csc 𝜃 equals five over four, so let us consider the diagram which
tells us the lengths of 𝑥 and 𝑦 at that instant. To get a relation between the two
variables, we can start by writing them in terms of some dummy variable, 𝑎. We then rearrange each equation in
terms of 𝑎 and then set the left-hand sides equal to each other, which we can then
rearrange in terms of 𝑥, giving us 𝑥 equals four 𝑦 over three. It is also possible to figure this
equation out just via inspection. Finally, if we differentiate this
with respect to 𝑦, we get d𝑥 by d𝑦 equals four over three.
Now, let us use the chain rule to
get d𝑥 by d𝑡. So, on the right-hand side, we have
d𝑥 by d𝑦 times d𝑦 by d𝑡, which is 𝑘 times four over three. And by rearranging and adding in
the correct units, we get four over three 𝑘 centimeters per second, which is our
final answer.
