Video Transcript
A particle of unit mass is moving
under the action of a force 𝐅, which is equal to negative two 𝐢 plus four 𝐣. Its displacement as a function of
time is given by 𝐒 of 𝑡 equals two 𝑡𝐢 plus seven 𝑡 squared plus two 𝑡𝐣. Find the value of the derivative of
the dot product of 𝐅 and 𝐒 with respect to 𝑡 when 𝑡 is equal to four.
In this question, we’ve been given
two vectors. One describes the force and one
describes the displacement of the particle. We’re asked to find the derivative
of the dot product of these with respect to time. So let’s just begin by finding the
dot product of 𝐅 and 𝐒. To achieve this, we multiply the
components for 𝐢. So that’s negative two times two
𝑡. And then we add the products of the
𝐣 components. So that’s four times seven 𝑡
squared plus two 𝑡. We distribute our parentheses and
simplify it. And we find that the dot product of
𝐅 and 𝐒 is negative four 𝑡 plus 28𝑡 squared plus eight 𝑡. Now in this question, we’re looking
to find the derivative of this expression with respect to time. These are simple power terms. So we can differentiate term by
term.
The derivative of negative four 𝑡
with respect to 𝑡 is negative four. When we differentiate 28𝑡 squared,
we multiply the entire term by the exponent and then reduce the exponent by one. So that’s two times 28𝑡, which is
56𝑡. Then the derivative of eight 𝑡 is
eight. And so we simplify it, and we find
that the derivative of the dot product of 𝐅 and 𝐒 with respect to 𝑡 is four plus
56𝑡. We’re told to evaluate this when 𝑡
is equal to four. So that’s four plus 56 times four
which is equal to 228. Now actually, what we’ve calculated
is the power. The dot product of 𝐅 and 𝐒 is the
work. Then when we differentiate the work
with respect to time, we get the power. And so we can say that this will be
measured in power units. The derivative of the dot product
of 𝐅 and 𝐒 with respect to 𝑡 when 𝑡 equals four is 228 power units.
