Hey guys, in this video I want to go over this calculus 3 problem. It says, find the unit normal
vector at the given point and then we're given a vector function r(t) in three dimensional space
and a point in three dimensional space, which is (1, 0, 0). Let's go ahead and get started with
solving this problem. The unit normal vector N(t) is equal to T prime over the magnitude of T prime,
where T is equal to the tangent vector and the tangent vector T has its own formula,
which is r prime over the magnitude of r prime and r refers to the vector function,
which we're given. First we have to solve for T and then we're going to plug in T prime and the
magnitude of T prime to get our final solution. So r prime of t is equal to the derivatives of the
three components of r(t). The derivative of cosine t is equal to negative sine t. The derivative of
sine t is cosine t. And the derivative of the natural log of cosine t is equal to negative sine
t over cosine t. This is also equal to negative tangent t. Then we have to find the magnitude of r
prime. The magnitude of r prime t is equal to the square root of the sum of the squares of each of
the three components. That's going to be equal to sine squared t plus cosine squared t plus
tangent squared t. Sine squared t plus cosine squared t is equal to one, so we can simplify
this as being equal to one plus tangent squared t. And that can be simplified as well as being equal
to secant squared t and the square root of secant squared t is equal to secant of t, or that's also
the same as saying one over cosine t. Now that we have r prime and the magnitude of r prime, we can
plug in both those equations to solve for T. So T is going to be equal to--for the first component,
it's going to be negative sine t over secant t, which is equal to one over cosine t. That's equal
to times cosine t. For the second component, it's going to be cosine squared t and for the
third component the cosines are going to cancel out, so you're going to have negative sine t.
Now that I know the three components for T, I'm going to solve for N(t). I have to find T prime
and the magnitude of T prime. Again for finding T prime, I'm just going to take the derivative of
each of the three components of T. Negative sine t times cosine t, I'm going to use the product rule.
That's going to be equal to negative sine t times negative sine t plus negative cosine
t times cosine t, for the first component. For the second component, I'm going to get
negative 2 cosine t times sine t and for the third component, I'm going to get negative cosine t.
Instead of solving for the magnitude of T prime since that's going to be a long calculation,
I'm now going to use the other information that I'm given in the problem, which is that we're
solving for the unit vector at the point (1, 0, 0). I'm going to now use this information
for finding the magnitude of T prime. We're given in the problem the point (1, 0, 0) and then if we
think of r(t) as the position function in three dimensions, then we want to find the time t where
r of t is going to give me the position (1, 0, 0). We see for, for example, the first component,
I want position 1, so the time t that would give me position one in r(t) is going to be t
is equal to zero. And if we check for the three components, we're going to see that time of t
equal to zero gives us this point in space. Cosine of zero is equal to 1, sine of 0 is equal to 0,
and the natural log of cosine of 0 is also equal to 0. This point is going to be in this r(t) when
t is equal to 0. Instead of finding the magnitude of T prime for any value t, I'm first going to
find T prime at time t is equal to zero, so T prime of zero. That's going to be equal to--if
we plug in zero for the t's here, that's going to be zero minus one for the first component. For
the second component, we're going to get zero, and for the third component, we're going to get
negative one. At time t is equal to zero, T prime is equal to <-1, 0, -1>. And now it's a lot easier
for us to find the magnitude of T prime because we know we're finding it for time t is equal to 0.
That's going to be equal to the square root of the sums of the squares of the components. That's
going to be the square root of negative 1 squared plus 0 squared plus the square of negative 1.
That's just going to be equal to the square root of 2. Now we can plug in to our final answer, so
N(t) at the time t is equal to 0 is going to be equal to <-1, 0, -1> divided by the square root
of two. Or the final answer is the negative one divided by the square root of two, zero, negative
one divided by the square root of two. And this is our final answer and we're done with this problem.
*chiptune music*