### Video Transcript

Find an equation of the tangent to
the curve π₯ is equal to sin of π times π‘ and π¦ is equal to π‘ squared plus π‘ at
the point zero, two.

The question gives us a pair of
parametric equations. And it wants us to find an equation
of the tangent to the curve represented by this pair of parametric equations at the
point zero, two. To find an equation of a line, we
need to find the slope of that line. Since we want our line to be a
tangent to the curve, we can find the slope by using dπ¦ by dπ₯. However, weβre given π₯ and π¦ as a
pair of parametric equations.

We recall if π₯ is a function of π‘
and π¦ is a function of π‘, then we can find dπ¦ by dπ₯ by dividing the derivative
of π¦ with respect to π‘ by the derivative of π₯ with respect to π‘. So we set π of π‘ to be sin of
ππ‘ and π of π‘ to be π‘ squared plus π‘. Then by differentiating these with
respect to π‘, we can find an expression for dπ¦ by dπ₯ . We start by calculating
dπ¦ by dπ‘. Thatβs the derivative of π‘ squared
plus π‘ with respect to π‘.

We can differentiate this using the
power rule for differentiation. We multiply by the exponent and
then reduce the exponent by one. This gives us two π‘ plus one. We can do the same to find dπ₯ by
dπ‘. Itβs the derivative of sin of ππ‘
with respect to π‘. To differentiate this, we recall
that, for any constant π, the derivative of the sin of ππ‘ with respect to π‘ is
equal to π times the cos of ππ‘. Therefore, since π is just a
constant, the derivative of the sin of ππ‘ with respect to π‘ is just equal to π
times the cos of ππ‘.

Now that we found dπ¦ by dπ‘ and
dπ₯ by dπ‘, we can find an expression for dπ¦ by dπ₯. Itβs equal to two π‘ plus one
divided by π times the cos of ππ‘. We see that our expression for dπ¦
by dπ₯ is in terms of π‘. This means it tells us the slope of
the tangent to the curve at the point π‘. However, we want to find the slope
of the tangent at the point zero, two. So we need to find the value of π‘
which gives us the point zero, two. The curve being at the point zero,
two is the same as saying its π₯-coordinate is equal to zero and its π¦-coordinate
is equal to two.

To find the values of π‘ where our
curve is at the point zero, two, weβll set our π¦-value equal to two, giving us the
equation two is equal to π‘ squared plus π‘. We then subtract two from both
sides of this equation, giving us that zero is equal to π‘ squared plus π‘ minus
two. And we see in this case we can
fully factor the quadratic. Itβs equal to π‘ plus two
multiplied by π‘ minus one.

Since we want this to be equal to
zero, one of the factors must be equal to zero. Solving these gives us π‘ is equal
to negative two or π‘ is equal to one. We need to check that these values
of π‘ give us an π₯-coordinate of zero. We see when π‘ is equal to one, π₯
is equal to sin of π times one, which is just equal to zero. And we also see when π‘ is equal to
negative two, π₯ is equal to the sin of negative two π, which is also equal to
zero. So we have two different values of
π‘ at which our parametric curve is at the point zero, two.

At first, this might be
confusing. We have several different values of
π‘. We donβt know which one we should
pick. Sometimes our values of π‘ are
restricted. However, thatβs not given in this
case. In this case, weβre asked to find
an equation of the tangent. So it doesnβt actually matter which
value of π‘ we pick. We will always get an equation for
the tangent. However, itβs worth noting at this
point that the value of π‘ that you choose could possibly give completely different
equations for the tangent.

For example, consider a parametric
curve which gives a figure-eight pattern as shown in the sketch. Then, if we were asked to find a
tangent at the origin to this curve, we can see that there are two
possibilities. Both lines share a slope with
different parts of the curve at the origin. In fact, both of these are tangent
lines to the curve at the origin. And which tangent line you get
would depend on which value of the parameterization variable you choose.

In our case, since weβre just asked
to find an equation of the tangent, it doesnβt matter which value of π‘ we
choose. Letβs start by checking π‘ is equal
to one. We have the slope of our curve when
π‘ is equal to one is equal to two times one plus one divided by π times the cos of
π times one, which we can calculate to give us negative three divided by π.

To be thorough, letβs check what
the value of our slope wouldβve been if we had chosen π‘ is equal to negative two
instead. We wouldβve gotten a slope of two
times negative two plus one all divided by π times the cos of π times negative
two. And we wouldβve gotten, again, the
slope to be equal to negative three divided by π. So in this case, we can see the
slope is the same regardless of which value of π‘ we wouldβve gotten. This means our equation of the
tangent will be the same in both cases.

To help us find the equation of our
tangent line to the curve at the point zero, two, we recall the equation of a line
which passes through the point π₯ one, π¦ one with a slope of π is given by π¦
minus π¦ one is equal to π times π₯ minus π₯ one.

We already found slope of our
tangent line. Itβs equal to negative three
divided by π. And we already know our tangent
line passes through the point zero, two. So weβll set π₯ one equal to zero
and π¦ one equal to two. This gives us π¦ minus two is equal
to negative three over π times π₯ minus zero. Simplifying this and rearranging to
make π¦ the subject gives us π¦ is equal to negative three over π times π₯ plus
two.

Therefore, weβve shown that an
equation of the tangent to the curve π₯ is equal to the sin of ππ‘ and π¦ is equal
to π‘ squared plus two at the point zero, two is π¦ is equal to negative three over
π π₯ plus two.