Video Transcript
For the curve defined by the
vector-valued equation 𝐫 of 𝑡 is equal to the sin of 𝜋 by three 𝑡 𝐢 plus two 𝑡
cubed plus three 𝑡 squared plus 𝑡 𝐣, find the value of 𝐫 prime of one.
The question gives us the
vector-valued function 𝐫 of 𝑡. And it wants us to find the
derivative of 𝐫 of 𝑡 with respect to 𝑡 when 𝑡 is equal to one. Remember, since 𝐫 of 𝑡 is a
vector-valued equation, we’re going to need to differentiate this
component-wise. In other words, we need to
differentiate each component separately.
Let’s start by finding the
derivative of our 𝑥-component with respect to 𝑡. That’s the derivative of the sine
of 𝜋 by three 𝑡 with respect to 𝑡. To differentiate this, recall for
any constant 𝑎, the derivative of the sin of 𝑎𝑡 with respect to 𝑡 is equal to 𝑎
times the cos of 𝑎𝑡. In our case, the value of 𝑎 is 𝜋
by three. So, we get that our derivative is
equal to 𝜋 by three times the cos of 𝜋 by three 𝑡.
We now want to do the same with our
second component. In other words, we want to find the
derivative of two 𝑡 cubed plus three 𝑡 squared plus 𝑡 with respect to 𝑡. We can, of course, do this term by
term by using the power rule for differentiation. We multiply by our exponent of 𝑡
and then reduce this exponent by one. This gives us six 𝑡 squared plus
six 𝑡 plus one.
We can now use these two
expressions to find an expression for the derivative of 𝐫 of 𝑡 with respect to
𝑡. We get that 𝐫 prime of 𝑡 is the
derivative of our first component. That’s 𝜋 by three times the cos of
𝜋 by three 𝑡 times 𝐢 plus the derivative of our second component with respect to
𝑡. That’s six 𝑡 squared plus six 𝑡
plus one times 𝐣.
But remember, the question is
asking us to find 𝐫 prime of one. So, it wants us to evaluate this
when 𝑡 is equal to one. So, we just need to substitute 𝑡
is equal to one into this expression for 𝐫 prime of 𝑡. Substituting 𝑡 is equal to one
into this expression, we get 𝐫 prime of one is equal to 𝜋 by three times the cos
of 𝜋 times one over three times 𝐢 plus six times one squared plus six times one
plus one 𝐣.
And we can just simplify this
expression. First, we know the cos of 𝜋 by
three is equal to one-half. And then, 𝜋 by three multiplied by
one-half is 𝜋 by six. So, we can simplify our first
component to see that we get 𝜋 by six 𝐢. Next, we can calculate our second
component to get 13. So, this simplifies to give us 𝜋
by six 𝐢 plus 13𝐣.
Therefore, we were able to show the
derivative of the vector-valued function 𝐫 of 𝑡 is equal to the sin of 𝜋 by three
𝑡 𝐢 plus two 𝑡 cubed plus three 𝑡 squared plus 𝑡 𝐣 with respect to 𝑡 when 𝑡
is equal to one is 𝜋 by six 𝐢 plus 13𝐣.
