Modeling with function composition | Functions and their graphs | Algebra II | Khan Academy
Articles Blog

Modeling with function composition | Functions and their graphs | Algebra II | Khan Academy

October 19, 2019


Voiceover:Fuel is being
pumped into a storage tank. The volume, V, of the fuel in the tank depends on the depth, d,
according to the formula. V of d is equal to four
times three d squared plus five to the third power, where d is in measured in meters. Suppose that the depth, d, of the fuel depends on time measured in hours according to the formula. d of t is equal to one over
the square root of three times the square root of t minus five. That makes sense, that if I’m filling, if I’m pumping something
into a storage tank the depth of what I’m filling, of the fuel I guess is what I’m filling is going to depend on time and then the volume could
be dependent on the depth. Use function composition
to write the volume of the fuel of the tank
as a function of time. Simplify your answer as much as possible. Let me get my scratch pad out, so I’ve already written
what they told us here. This is volume as a function of depth and now this is depth
as a function of time. If we want volume as a function of time, what we really want to
do is figure out what is, so we could take V of d where we could take V of d of t. Now this would give me
volume as a function of time. Everywhere where I see a d, I should replace it with this description of how does d vary with respect to time. Let’s do that. V of d of t is going to be equal to four times three, times d squared. Instead of writing d squared, I’m going to write all
of this business squared, so one over square root of three, times the square root of t minus five, all of that squared plus five to the third power. I really hope this simplifies
in a reasonable way. Let’s see, let me actually
copy and paste this so I can have a little
bit more real estate. Let me just paste it down
here so I could work it out. All right, so then we get
this is going to be equal to, this is going to be equal to, let’s see if we square
this right over here and I could do it all in. This is going to be four times three, times this whole thing squared. One over square root of three squared is going to be 1/3 and then square root
of t minus five squared is going to be t minus five and then of course we have
plus five to the third power. Let’s see three times 1/3, well that’s just going to be equal to one. We are left with, this is equal to four times t minus five plus five, this does simplify nicely. Plus five, we’re going to be left with four times t to the third power. Let’s input that into our V of t and let me just make it clear. This is now V, this is giving our volume as a function of time. We can now say that this is V of t. V of t is four t squared. Once again this is the
volume as a function of time. You give me a time, I’ll
tell you what the volume is. Four t to the third. Four t to the third power. V of t, did I say four t squared? I don’t know, sometimes
I do strange things. I said four t to the third power, I did format it correctly and then there’s a second
part to this question. How many cubic meters
of fuel are in the tank after two hours? Round your answer to the nearest 10th. Well two hours, that means t is equal two. They say t is measured in hours. When t is equal to two, the volume is going to be two
to the third power is eight. Eight times four is 32. Let’s check our answer
and we got it right.

Only registered users can comment.

Leave a Reply

Your email address will not be published. Required fields are marked *