### Video Transcript

Calculate π prime of π and find
the vector form of the equation of the tangent line at π of zero for π of π
equals π plus one, π squared plus one, π cubed plus one.

In this question, weβve been given
a rule for π in terms of its π₯-, π¦-, and π§-coordinates. We can alternatively think of this
as a vector-valued function such that the vector π of π is equal to π plus one,
π squared plus one, π cubed plus one, where π plus one is the π’-component, π
squared plus one is the π£-component, and π cubed plus one is the π€-component.

Remember then, to find the
derivative of a vector-valued function, we simply take the derivative of each
component. This means weβre going to
individually differentiate with respect to π , π plus one, π squared plus one, and
π cubed plus one. That gives us one, two π , and
three π squared, respectively. Taking this back into the rule for
the π₯-, π¦-, and π§-coordinates, and we find π prime of π to be equal to one, two
s, three π squared.

We do still need to find the vector
form of the equation of the tangent line at π of zero. So, we recall that a line that
passes through a point π₯ nought, π¦ nought, π§ nought with the direction vector π―
is given by the equation π« equals π« nought plus π‘ times π―. So, weβll need two things, a point
that our line passes through and its direction of travel. We can find the latter by
evaluating the derivative when the parameter value, π , is equal to zero. So, letβs substitute π equals zero
into our derivative. Thatβs one, zero, zero or one π’
plus zero π£ plus zero π€.

We can also find that the point our
tangent line passes through by substituting π equals zero into the original
function. Thatβs π of zero. When we do, we obtain π of zero to
be zero plus one, zero plus one, zero plus one. Thatβs one, one, one. And so, the vector form of the
equation of the tangent line at π zero, letβs call that πΏ, is one, one, one plus
π‘ times one, zero, zero.