Question Video: Evaluating the Derivative of a Vector-Valued Function at a Given Value Mathematics

Video Transcript

For the curve defined by the vector-valued equation 𝐫 of 𝑡 is equal to three 𝑡 squared minus five 𝑡 and square root of five 𝑡, find the value of 𝐫 prime of two.

We’re given a curve which is defined by a given vector-valued function. And we need to determine the value of 𝐫 prime evaluated at two. First, 𝐫 is a function of 𝑡. So this will be when 𝑡 is equal to two. And we know that 𝐫 prime will be the derivative of our vector-valued function with respect to 𝑡. So we first need to recall how we differentiate vector-valued functions. In particular, this is given as a column vector. In fact, we differentiate these in all the same way. We just differentiate it component-wise.

So let’s start by differentiating our first component function with respect to 𝑡. That’s three 𝑡 squared minus five 𝑡. And this is just a polynomial, so we can differentiate this term by term by using the power rule for differentiation. We get six 𝑡 minus five.

We want to do the same for our second component function. We need to differentiate the square root of five 𝑡 with respect to 𝑡. And we can do this by using our laws of exponents to rewrite root five 𝑡 as root five times 𝑡 to the power of one-half. We can then differentiate this by using the power rule for differentiation. And we do this once again by multiplying by our exponent of 𝑡 and then reducing this exponent by one. This gives us root five over two times 𝑡 to the power of negative one-half. And we can then simplify this by using our laws of exponents. We get root five divided by two root 𝑡.

Now that we’ve differentiated both of our component functions, we can find an expression for 𝐫 prime of 𝑡. Remember, to find 𝐫 prime of 𝑡, we just need to differentiate 𝐫 of 𝑡 component-wise with respect to 𝑡. This gives us that 𝐫 prime of 𝑡 is equal to six 𝑡 minus five and then root five divided by two root 𝑡. But remember, the question is asking us to evaluate 𝐫 prime of 𝑡 at two. So we need to substitute in 𝑡 is equal to two. Substituting in 𝑡 is equal to two, we get 𝐫 prime of two is equal to six times two minus five and then root five divided by two root two.

We just need to evaluate and simplify this expression. In our first component, six times two minus five is equal to seven. And in our second component, we can simplify by rationalizing our denominator. We’ll multiply both the numerator and the denominator through by root two. Doing this and then simplifying gives us root 10 divided by four.

Therefore, we were able to show if 𝐫 of 𝑡 is the vector-valued function three 𝑡 squared minus five 𝑡 and then root five 𝑡, we were able to show that 𝐫 prime of two is equal to seven root 10 over four.