### Video Transcript

A particle is moving in a straight
line such that its velocity π£ after π‘ seconds is given by π£ equals negative π‘
squared minus 68π‘ plus 63 meters per second when π‘ is greater than or equal to
zero. When is the velocity of the
particle increasing?

In this question, weβve been given
a function for velocity in terms of time. And, weβre looking to work out when
the velocity of that particle is increasing. Well, the velocity of something
will be increasing when its acceleration is greater than zero. And so, next, we recall that
acceleration is the rate of change of velocity of the object with respect to
time. In derivative form, π is the
derivative of π£ with respect to π‘.

So, we need to work out when the
first derivative of our function π£ with respect to π‘ is greater than zero. And actually, this makes a lot of
sense. We know that we have some function
π. When its derivative is greater than
zero, itβs increasing. And, when its first derivative is
less than zero, it must be a decreasing function. So, it makes sense to begin by
differentiating our function π£ with respect to π‘. Thatβs negative π‘ squared minus
68π‘ plus 63.

Now, this is still a little bit
nasty, so weβre going to separate this somewhat. We know that we can take out any
constant factors and focus on differentiating the function itself. So, letβs take out this common
factor of negative one. We also know that the derivative of
the sum of a number of functions is equal to the sum of the derivatives of each of
those functions. So, weβre going to differentiate π‘
squared, negative 68π‘, and 63, individually.

And then, we recall that for real
constants π and π, the derivative of π times π₯ to the power of π is πππ₯ to
the power of π minus one. In other words, we multiply the
entire expression by the exponent and then reduce that exponent by one. This means the derivative of π‘
squared is two π‘ to the power of one or two π‘. The derivative of negative 68π‘ is
one times negative 68π‘ to the power of zero. Well, π‘ to the power of zero is
one. So, its derivative is just negative
68. And, the derivative of any constant
is zero. We finally distribute our
parentheses and we see that our first derivative of the function π£ with respect to
π‘ is negative two π‘ plus 68.

All thatβs left is to find the
interval of increase. We set the expression for the
derivative as an inequality greater than zero and solve for π‘. Now, we could subtract 68 from both
sides, but then we end up with a negative coefficient of π‘. So instead, letβs add two π‘ to
both sides. When we do, we find that 68 is
greater than two π‘. And finally, we divide through by
two. And, we see that 34 is greater than
π‘. In other words, π‘ is less than
34. The velocity of the particle is
increasing when π‘ is less than 34.

Now, letβs just go back to our
inequality and double check that we would have got the same answer had we indeed
subtracted 68 to start. We would have obtained negative two
π‘ is greater than negative 68. Then, we divide both sides by
negative two. But of course, when we divide or
multiply through in an inequality by a negative, we must reverse the inequality
symbol. And so, once again, we do indeed
end up with π‘ being less than 34.