Question Video: Finding the Equation of a Tangent Line to a Vector-Valued Function

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 equal to 𝑒 to the power of 𝑠 plus one, 𝑒 to the power of two 𝑠 plus one, 𝑒 to the power of 𝑠 squared plus one.

We’re given a vector-valued function 𝐟 of 𝑥, and we’re asked to find a few things. First, we’re asked to find 𝐟 prime of 𝑠. That’s the derivative of our vector-valued function 𝐟 with respect to 𝑠. We also need to find the vector form of the equation of the tangent line to our vector-valued function 𝐟 of 𝑥 when 𝑠 is equal to zero. Let’s start with the first part of this question. We’re asked to find an expression for 𝐟 prime of 𝑠. And since 𝐟 is a vector-valued function, this means we need to differentiate each of our components separately. This means our derivative will be a vector-valued function.

Let’s start with the first component of 𝐟 of 𝑠. That’s 𝑒 to the power of 𝑠. We want to differentiate this with respect to 𝑥. We can do this term by term. We know the derivative of the exponential function is just equal to itself, and the derivative of a constant will just be equal to zero. So, the derivative of our first component function with respect to 𝑠 is 𝑒 to the power of 𝑠. We need to do exactly the same thing for our second component function. We need to differentiate it with respect to 𝑠. Once again, we can do this term by term.

To do this, we’ll recall one of our standard derivative results for exponential functions. For any real constant 𝑎, the derivative of 𝑒 to the power of 𝑎𝑠 with respect to 𝑠 is equal to 𝑎 times 𝑒 to the power of 𝑎𝑠. To differentiate the first term of this expression, we need to set our value of 𝑎 equal to two. This gives us two times 𝑒 to the power of two 𝑠. Once again, the derivative of the constant one is just equal to zero. Finally, we need to differentiate our third component function. We need to find the derivative of 𝑒 to the power of 𝑠 squared plus one with respect to 𝑠. There’s a few different options for doing this.

For example, we could do this directly by using the chain rule or we can recall the following result of using the chain rule on an exponential function. If 𝑔 of 𝑠 is a differentiable function, then the derivative of 𝑒 to the power of 𝑔 of 𝑠 with respect to 𝑠 is equal to 𝑔 prime of 𝑠 times 𝑒 to the power of 𝑔 of 𝑠. In this case, our function 𝑔 of 𝑠 will be 𝑠 squared. Of course, we know the derivative of 𝑠 squared with respect to 𝑠 is going to be two 𝑠. So, we can use this to differentiate our function. We get two 𝑠 times 𝑒 to the power of 𝑠 squared.

Now that we’ve differentiated each of the component functions of 𝐟 of 𝑠, we can find an expression for 𝐟 prime of 𝑠. 𝐟 prime of 𝑠 is the vector-valued function 𝑒 to the power of 𝑠, two 𝑒 to the power of two 𝑠, two 𝑠 times 𝑒 to the power of 𝑠 squared. And we’ll rewrite our third component function as two 𝑠 times 𝑒 to the power of 𝑠 squared. And this gives us our answer to the first part of our question. We found an expression for the vector-valued function 𝐟 prime of 𝑠.

We can now move on to the second part of our question. What do we mean by the vector form of the equation of the tangent line to a vector-valued function? First, we need to recall the vector form of the equation of a line. We recall the vector-valued function 𝐥 of 𝑡 is equal to 𝐥 of zero plus 𝑡 times 𝐯 traces out a line. And the important things to note about this is if we substitute 𝑡 is equal to zero into this equation, we see that our line must pass through the point 𝐥 zero. Second, the direction that our line is moving is entirely defined by our vector 𝐯.

Okay, so now we know the vector form of the equation of a line. How is this going to help us find the equation of our tangent line to our vector-valued function? We need to remember what a tangent line is. First, remember, we’re finding the tangent line to our vector-valued function at the point 𝐟 of zero. So because we want the tangent line to our curve at the point 𝐟 of zero, well, our tangent line should pass through the point 𝐟 of zero. Remember, though, we just want our tangent line to touch the curve at this point. In other words, we want the slopes to match, so we should choose our directional vector to be 𝐟 prime of zero. And since we have expressions for 𝐟 of 𝑠 and 𝐟 prime of 𝑠, we can find both of these vectors.

Let’s start with 𝐟 of zero. We just need to substitute 𝑠 is equal to zero into our expression for 𝐟 of 𝑠. This gives us 𝐟 of zero is 𝑒 to the zeroth power plus one, 𝑒 to the power of two time zero plus one, 𝑒 to the power of zero squared plus one. And we can evaluate this. 𝑒 to the zeroth power, 𝑒 to the power of two times zero, and 𝑒 to the power of zero squared all evaluate to give us one. So, 𝐟 zero is just equal to two, two, two. We now want to do the same to find 𝐟 prime of zero. We need to substitute 𝑠 is equal to zero into our expression for 𝐟 prime of 𝑠.

Doing this, we get 𝐟 prime of zero is equal to 𝑒 to the zeroth power, two times 𝑒 to the power of two times zero, two times zero times 𝑒 to the power of zero squared. And once again, we can just evaluate this component-wise. 𝑒 to the zeroth power, 𝑒 to the power of two times zero, and 𝑒 to the power of zero squared simplify to give us one. This means 𝐟 prime of zero is just equal to one, two, zero. Now, all we do is write these two vectors into the vector form of the equation of a line. This gives us the equation of our tangent line 𝐿 is two, two, two plus 𝑡 times one, two, zero.

Therefore, given the vector-valued function 𝐟 of 𝑠 is equal to 𝑒 to the power of 𝑠 plus one, 𝑒 the power of two 𝑠 plus one, 𝑒 to the power of 𝑠 squared plus one, we were able to show that 𝐟 prime of 𝑠 is 𝑒 to the power of 𝑠, two 𝑒 to the power two 𝑠, two 𝑠𝑒 to the power of 𝑠 squared. And the equation of our tangent line to this function when 𝑠 is equal to zero is given by two, two, two plus 𝑡 times one, two, zero.