Calculate 𝑓 prime of 𝑠 and find
the vector form of the equation of the tangent line at 𝑓 of zero for 𝑓 of 𝑠
equals 𝑠 plus one, 𝑠 squared plus one, 𝑠 cubed plus one.
In this question, we’ve been given
a rule for 𝑓 in terms of its 𝑥-, 𝑦-, and 𝑧-coordinates. We can alternatively think of this
as a vector-valued function such that the vector 𝐟 of 𝑠 is equal to 𝑠 plus one,
𝑠 squared plus one, 𝑠 cubed plus one, where 𝑠 plus one is the 𝐢-component, 𝑠
squared plus one is the 𝐣-component, and 𝑠 cubed plus one is the 𝐤-component.
Remember then, to find the
derivative of a vector-valued function, we simply take the derivative of each
component. This means we’re going to
individually differentiate with respect to 𝑠, 𝑠 plus one, 𝑠 squared plus one, and
𝑠 cubed plus one. That gives us one, two 𝑠, and
three 𝑠 squared, respectively. Taking this back into the rule for
the 𝑥-, 𝑦-, and 𝑧-coordinates, and we find 𝑓 prime of 𝑠 to be equal to one, two
s, three 𝑠 squared.
We do still need to find the vector
form of the equation of the tangent line at 𝑓 of zero. So, we recall that a line that
passes through a point 𝑥 nought, 𝑦 nought, 𝑧 nought with the direction vector 𝐯
is given by the equation 𝐫 equals 𝐫 nought plus 𝑡 times 𝐯. So, we’ll 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, 𝑠, is equal to zero. So, let’s substitute 𝑠 equals zero
into our derivative. That’s one, zero, zero or one 𝐢
plus zero 𝐣 plus zero 𝐤.
We can also find that the point our
tangent line passes through by substituting 𝑠 equals zero into the original
function. That’s 𝑓 of zero. When we do, we obtain 𝑓 of zero to
be zero plus one, zero plus one, zero plus one. That’s one, one, one. And so, the vector form of the
equation of the tangent line at 𝐟 zero, let’s call that 𝐿, is one, one, one plus
𝑡 times one, zero, zero.