Video Transcript
For the given function 𝐫 of 𝑡 equals two 𝑒 to the power of 𝑡 minus one 𝐢 plus cos of 𝑡𝜋 𝐣 minus the natural log of 𝑡 𝐤, evaluate 𝐫 of one.
In this question, we’ve been given a vector-valued function. It’s a function whose range is a vector or set of vectors and whose domain is a subset of the real numbers. We’re being asked to evaluate 𝐫 of one. In other words, we need to let 𝑡 be equal to one. So we replace 𝑡 with one in each part of our function.
So the component for 𝐢 becomes two 𝑒 to the power of one minus one. The component of 𝐣 is cos of one 𝜋. And the component for 𝐤 is negative the natural log of one. Now of course, two 𝑒 to the power of one minus one is two 𝑒 to the power of zero. And cos of one 𝜋 is equal to cos of 𝜋. So we have two 𝑒 to the power of zero 𝐢 plus cos of 𝜋 𝐣 minus the natural log of one 𝐤.
But of course, anything to the power of zero is one. And so the component of 𝐢 here is two. We know cos of 𝜋 to be equal to negative one. So the component for 𝐣 is negative one. And we can write 𝐫 of one as two 𝐢 minus 𝐣. But we also know that the natural log of one is zero. So 𝐫 of one is two 𝐢 minus one 𝐣 plus zero 𝐤, which is simply two 𝐢 minus 𝐣.
