Engineering Mathematics Exam > Engineering Mathematics Videos > Understanding function notation (example 3)
Understanding function notation (example 3) Video Lecture - Engineering Mathematics
FAQs on Understanding function notation (example 3) Video Lecture - Engineering Mathematics
| 1. What is function notation in engineering mathematics? |  |
| 2. How is function notation useful in engineering mathematics? | |
Ans. Function notation is useful in engineering mathematics as it simplifies the representation and analysis of mathematical functions. It allows engineers to express complex relationships between variables and perform operations on functions more efficiently.
| 3. Can you provide an example of function notation in engineering mathematics? | |
Ans. Certainly! Consider the function f(x) = 2x + 3. In this example, "f(x)" represents the dependent variable, and "x" represents the independent variable. By substituting different values for "x," we can determine corresponding values for "f(x)."
| 4. What are the benefits of using function notation in engineering mathematics? | |
Ans. Using function notation in engineering mathematics offers several benefits. It provides a concise and standardized way to represent functions, making it easier to communicate and understand mathematical concepts. It also allows engineers to apply various mathematical operations, such as differentiation and integration, to analyze functions effectively.
| 5. How does function notation relate to real-life engineering applications? | |
Ans. Function notation is extensively used in real-life engineering applications. It helps engineers model and analyze various physical phenomena, such as electrical circuits, fluid flow, and structural behavior. By representing these phenomena as functions, engineers can make predictions, optimize designs, and solve complex engineering problems more effectively.
Text Transcript from Video
Frank wants to fill
up a spherical water
balloon with as much
water as possible.
The balloons he
bought can stretch
to a radius of 3
inches-- not too big.
If the volume of a sphere
is-- and this is volume
as a function of radius--
is equal to 4/3 pi r cubed,
what volume of water
in cubic inches
can Frank put into the balloon?
So this function
definition is going--
if you give it a
radius in inches,
it's going to produce a
volume in cubic inches.
So let's rewrite it.
Volume as a function of radius
is equal to 4/3 pi r cubed.
Now, they say the
balloons he bought
can stretch to a
radius of 3 inches.
So let's think about, if the
radius gets to 3 inches, what
the volume of that
balloon is going to be.
So we essentially would
just input 3 inches
into our function definition.
So everywhere where we see an
r, we would replace it with a 3.
So we could write--
and just to be clear,
let me rewrite it
in the same color.
V of-- that's not
the same color.
We do it in that brownish
color right over here.
So V of 3 is equal to 4/3
pi-- and instead of r cubed,
I would write 3
cubed-- 4/3 pi 3 cubed.
This is how the function
definition works.
Whatever we input
here, it will replace
the r in the expression.
So V of 3 is going to
be equal to-- so this
is going to be equal to 4/3 pi
times-- 3 to the third power
is 27.
27 divided by 3 is
9, so this is 9.
9 times 4 is 36 pi.
So this is equal to 36 pi.
And since this was
in inches, our volume
is going to be in inches
cubed or cubic inches.
So that's the volume
of water that Frank
can put in the balloon--
36 pi cubic inches.
up a spherical water
balloon with as much
water as possible.
The balloons he
bought can stretch
to a radius of 3
inches-- not too big.
If the volume of a sphere
is-- and this is volume
as a function of radius--
is equal to 4/3 pi r cubed,
what volume of water
in cubic inches
can Frank put into the balloon?
So this function
definition is going--
if you give it a
radius in inches,
it's going to produce a
volume in cubic inches.
So let's rewrite it.
Volume as a function of radius
is equal to 4/3 pi r cubed.
Now, they say the
balloons he bought
can stretch to a
radius of 3 inches.
So let's think about, if the
radius gets to 3 inches, what
the volume of that
balloon is going to be.
So we essentially would
just input 3 inches
into our function definition.
So everywhere where we see an
r, we would replace it with a 3.
So we could write--
and just to be clear,
let me rewrite it
in the same color.
V of-- that's not
the same color.
We do it in that brownish
color right over here.
So V of 3 is equal to 4/3
pi-- and instead of r cubed,
I would write 3
cubed-- 4/3 pi 3 cubed.
This is how the function
definition works.
Whatever we input
here, it will replace
the r in the expression.
So V of 3 is going to
be equal to-- so this
is going to be equal to 4/3 pi
times-- 3 to the third power
is 27.
27 divided by 3 is
9, so this is 9.
9 times 4 is 36 pi.
So this is equal to 36 pi.
And since this was
in inches, our volume
is going to be in inches
cubed or cubic inches.
So that's the volume
of water that Frank
can put in the balloon--
36 pi cubic inches.
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Apr 15, 2025
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