Video: Evaluating a Limit Using the Definition of Differentiation

Video Transcript

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.