### Video Transcript

In this video, we will learn how to
find the equations of tangents and normals to trigonometric, parametric, and
implicitly defined curves using derivatives. Letβs begin by recalling the basics
of tangents and normals. We know that at any given point on
a curve, the tangent to the curve has the same slope as the curve itself at that
point. The slope of a curve is given by
its first derivative, which may be dπ¦ by dπ₯ or π prime of π₯ or something else,
depending on how the equation of the curve is specified. We can obtain the equation of the
tangent to a curve at a given point by substituting the coordinates of this point,
which weβll call π₯ one, π¦ one, and the slope of the tangent, which weβll call π,
into the pointβslope form of the equation of the straight line, π¦ minus π¦ one
equals π π₯ minus π₯ one.

The way in which we find an
expression for the slope function, and hence evaluate the slope at the given point,
will depend on how the curve is defined. In the course of this video, weβll
see examples of how we can do this for trigonometric, parametric, and implicitly
defined curves. If instead we want to find the
equation of the normal to a curve at a given point, we recall that the normal is
perpendicular or orthogonal to the tangent at that point. We also recall that if two lines
are perpendicular, then the product of their slopes is negative one. Or, put another way, their slopes
are the negative reciprocals of one another.

We can therefore find the slope of
the normal to a curve at a given point by finding the negative reciprocal of the
slope of the tangent at that point. We then proceed in the same way by
substituting this slope and the coordinates of the point at which weβre finding the
normal into the pointβslope form of the equation of a straight line. Letβs begin with an example in
which weβll find the equation of the normal to a curve which is defined
implicitly.

Find the equation of the normal to
the curve three π¦ squared minus nine π¦π₯ plus seven π₯ squared equals one at the
point negative one, negative one.

The curve weβve been given has been
defined implicitly. It is a function of both π₯ and π¦
and cannot be easily rearranged to a form in which we have π¦ as a function of
π₯. To find the equation of the normal
to this curve, weβre going to use the pointβslope form of the equation of a straight
line. Thatβs π¦ minus π¦ one equals π π₯
minus π₯ one. We know the coordinates of a point
on the normal, the point negative one, negative one. But we need to calculate its
slope. We recall that a normal to a curve
is perpendicular to the tangent to the curve at that point. So first, we must calculate the
slope of the tangent. To do this, weβre going to need to
find the slope function of the curve. So we need to recall how we
differentiate a function which is defined implicitly.

Implicit differentiation is an
application of the chain rule. Recall that if π¦ is a function of
π’ and π’ is a function of π₯, then the chain rule states that dπ¦ by dπ₯ is equal
to dπ¦ by dπ’ multiplied by dπ’ by dπ₯. So the derivative of π¦ with
respect to π₯ is found by taking the derivative of π¦ with respect to π’ and then
multiplying this by the derivative of π’ with respect to π₯. Letβs consider then how we can
apply this to the equation of our curve. Weβll do this term by term,
starting with the first term three π¦ squared. As three π¦ squared is a function
of π¦ and we can consider π¦ to be a function of π₯, by the chain rule, we have that
the derivative with respect to π₯ of three π¦ squared is equal to the derivative
with respect to π¦ of three π¦ squared multiplied by dπ¦ by dπ₯.

By the power rule of
differentiation, the derivative with respect to π¦ of three π¦ squared is simply six
π¦. So we find that the derivative with
respect to π₯ of three π¦ squared is six π¦ dπ¦ by dπ₯. So, weβve differentiated the first
term in our equation implicitly. Differentiating the third term is
straightforward because it is a function of π₯ only. By the power rule of
differentiation, the derivative with respect to π₯ of seven π₯ squared is 14π₯. And differentiating the term on the
right-hand side of the equation is straightforward because it is a constant. And we know that the derivative of
any constant with respect to π₯ is simply zero.

We have one term left to
differentiate, negative nine π¦π₯. And this is a little more
complicated as it is a product involving both π¦ and π₯. Weβll need to use implicit
differentiation. But we also need to recall the
product rule. This states that for two
differentiable functions π’ and π£, the derivative with respect to π₯ of their
product π’π£ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. Weβll therefore let π’ equal
negative nine π¦ and π£ equal π₯. We then need to find their
individual derivatives with respect to π₯. dπ£ by dπ₯ is straightforward because π£
is a function of π₯ only. If π£ is equal to π₯, then dπ£ by
dπ₯ is equal to one.

For dπ’ by dπ₯ though, as π’ is a
function of π¦, weβre going to need to use implicit differentiation again. By the chain rule, we have that the
derivative with respect to π₯ of negative nine π¦ is equal to the derivative with
respect to π¦ of negative nine π¦ multiplied by dπ¦ by dπ₯. The derivative with respect to π¦
of negative nine π¦ is simply negative nine. So we have that dπ’ by dπ₯ is equal
to negative nine dπ¦ by dπ₯. We then need to substitute each of
these expressions into the product rule. We have that the derivative with
respect to π₯ of negative nine π¦π₯ is equal to π’ times dπ£ by dπ₯. Thatβs negative nine π¦ multiplied
by one. Then we add π£ times dπ’ by
dπ₯. Thatβs π₯ multiplied by negative
nine dπ¦ by dπ₯. And this will simplify to negative
nine π¦ minus nine π₯ dπ¦ by dπ₯.

So weβve now differentiated the
entire equation, giving six π¦ dπ¦ by dπ₯ minus nine π¦ minus nine π₯ dπ¦ by dπ₯
plus 14π₯ is equal to zero. We want to evaluate the slope
function at a particular point. So first, we need to rearrange this
equation to give an explicit expression for dπ¦ by dπ₯. In other words, we need to make dπ¦
by dπ₯ the subject of this equation. We begin by collecting the terms
that do involve dπ¦ by dπ₯ on one side of the equation and the terms that donβt on
the other. So we have six π¦ dπ¦ by dπ₯ minus
nine π₯ dπ¦ by dπ₯ is equal to nine π¦ minus 14π₯. We can then factor the left-hand
side of the equation to give six π¦ minus nine π₯ multiplied by dπ¦ by dπ₯ is equal
to nine π¦ minus 14π₯.

And to find an expression for dπ¦
by dπ₯ which is in terms of both π₯ and π¦, we divide both sides of the equation by
six π¦ minus nine π₯. So we have dπ¦ by dπ₯ is equal to
nine π¦ minus 14π₯ over six π¦ minus nine π₯. We have an expression for the
general slope function of the curve. But we need to evaluate the slope
at a particular point, the point negative one, negative one. We therefore need to substitute the
value negative one for both π₯ and π¦. We have negative nine plus 14 in
the numerator and negative six plus nine in the denominator, which simplifies to the
fraction five over three.

Remember though that this is the
slope of the tangent to the curve at the point negative one, negative one, whereas
weβre looking to find the equation of the normal. But we know that the slope of the
normal is the negative reciprocal of the slope of the tangent. So we have that the slope of the
normal is equal to negative three-fifths. We can invert the fraction and
change the sign.

As we now know the coordinates of a
point on the normal and its slope, we can find its equation using the pointβslope
form of the equation of a straight line. We have π¦ minus negative one is
equal to negative three-fifths π₯ minus negative one. Of course, thatβs just π¦ plus one
equals negative three-fifths π₯ plus one. We can multiply through by five and
distribute the parentheses on the right-hand side to give five π¦ plus five equals
negative three π₯ minus three. And finally, weβll group all of the
terms on the left-hand side of the equation. Using implicit differentiation
then, we found that the equation of the normal to the given curve at the point
negative one, negative one is five π¦ plus three π₯ plus eight equals zero.

Letβs now consider an example in
which we find the equation of the tangent to a curve which is defined
parametrically.

Find the equation of the tangent to
the curve π₯ equals π‘ cubed plus one, π¦ equals π‘ to the fourth power plus π‘ at
the point corresponding to the value π‘ equals negative one.

The equation of this curve has been
given in parametric form. π₯ and π¦ are both given as
functions of a third variable π‘. We could write π₯ is equal to some
function π of π‘ and π¦ is equal to another function π of π‘, if we wish. To find the equation of the
tangent, we can use the pointβslope form of the equation of a straight line. So we need to know two things: the
coordinates of a point that lies on the tangent and its slope. Letβs begin by finding the
coordinates for the point at which weβre finding the tangent. Weβre told that this point
corresponds to the value π‘ equals negative one. So we can find the coordinates of
this point by substituting π‘ equals negative one into the parametric equations for
π₯ and π¦.

When π‘ equals negative one, π₯ is
equal to negative one cubed plus one, which is negative one plus one, which is
zero. And π¦ is equal to negative one to
the fourth power plus negative one. Thatβs one minus one, which is also
equal to zero. So we now know that the tangent
weβre looking for is the tangent at the origin, the point zero, zero.

Next, we need to find the slope of
the tangent, which we can do by using parametric differentiation to find the slope
of the curve at this point. We recall that if π₯ and π¦ are
each functions of a third parameter π‘, then dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ over
dπ₯ by dπ‘. We need to find the individual
derivatives of π₯ and π¦ with respect to π‘, which we can do using the power rule of
differentiation. dπ₯ by dπ‘ will be equal to three π‘ squared, and dπ¦ by dπ‘ is
equal to four π‘ cubed plus one. Substituting each of these
expressions into our formula for dπ¦ by dπ₯, we have that dπ¦ by dπ₯ is equal to
four π‘ cubed plus one over three π‘ squared. This is, of course, in terms of the
parameter π‘.

To find the slope at the point
where π‘ is equal to negative one, we substitute π‘ equals negative one into our
expression for dπ¦ by dπ₯. This gives negative four plus one
over three multiplied by one. Thatβs negative three over three,
which is equal to negative one. And so we find that the slope of
this tangent is negative one. We now know that this tangent
passes through the origin and has a slope of negative one. We can go through the formal
process of substituting these values into the pointβslope form of the equation of a
straight line if we wish. Or we may be able to recognize the
equation of this tangent. Itβs π¦ equals negative π₯.

In our final example, weβll find
the equation of the normal to a curve whose definition involves trigonometric and
reciprocal trigonometric functions.

Find the equation of the normal to
the curve π¦ equals eight cos π₯ minus three sec π₯ at π₯ equals π by three.

To find the equation of the normal
to any curve, we need to know the coordinates of a point that lies on the normal and
its slope. Weβre asked for the normal at the
point where π₯ is equal to π by three. We can find the π¦-value at this
point by substituting π₯ equals π by three into the equation of the curve. cos of π by three is one-half, so
sec of π by three is two. We have eight multiplied by a half
minus three multiplied by two. Thatβs four minus six, which is
equal to negative two. So, the coordinates of the point at
which weβre finding the normal are π by three, negative two.

Next, we need to find the slope of
the normal. Recall that the normal at any point
on a curve is perpendicular or orthogonal to the tangent at that same point. And hence, their slopes are the
negative reciprocals of one another. The slope of the tangent is the
same as the slope of the curve itself. And we can find the slope function
of the curve using differentiation. We need to recall two general rules
for differentiating trigonometric functions. These rules only apply when the
angle π₯ is measured in radians. Firstly, the derivative with
respect to π₯ of cos π₯ is negative sin π₯. And secondly, the derivative with
respect to π₯ of sec π₯ is sec π₯ tan π₯. We should be familiar with the
derivatives of the three trigonometric functions sin, cos, and tan as well as their
reciprocals csc, sec, and cot.

Applying these results then, we
have that dπ¦ by dπ₯ is equal to eight multiplied by negative sin π₯ minus three
multiplied by sec π₯ tan π₯. So we have the general slope
function of the curve. We need to evaluate this at the
point where π₯ is equal to π by three. We have negative eight sin π by
three minus three sec π by three tan π by three. Evaluating on a calculator or
recalling these results from memory, we have negative eight multiplied by root three
over two minus three multiplied by two multiplied by root three. Thatβs negative four root three
minus six root three, which simplifies to negative 10 root three.

Remember though that this is the
slope of the tangent at the point where π₯ equals π by three. The slope of the normal is the
negative reciprocal of this. We can cancel the negatives and
multiply both the numerator and denominator by root three in order to rationalize
the denominator. So we find that the slope of the
normal is root three over 30.

Finally, weβre able to calculate
the equation of the normal by substituting the coordinates of the point and the
slope we found into the pointβslope form of the equation of a straight line. We have π¦ minus negative two
equals root three over 30 multiplied by π₯ minus π by three. Simplifying on the left-hand side
and then distributing the parentheses on the right-hand side, we have π¦ plus two is
equal to root three π₯ over 30 minus root three π over 90. Finally, weβll group all the terms
on the same side of the equation. And we have that the equation of
the normal to the given curve at the point where π₯ equals π by three is π¦ minus
root three π₯ over 30 plus root three π over 90 plus two equals zero.

Letβs now summarize the key points
from this video. Firstly, we recalled the basics of
tangents and normals. The tangent to a curve at any given
point has the same slope as the curve itself at that point. The normal is perpendicular to the
tangent, and the product of their slopes are negative one, which means that the
slopes are the negative reciprocals of one another. We can find the slope function of a
curve using differentiation. And the method we use will depend
on how the curve is defined. For a curve that has been defined
implicitly, we use implicit differentiation, which is an application of the chain
rule. If π¦ is a function of π’ and π’ is
a function of π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ multiplied by dπ’ by
dπ₯.

For a curve defined parametrically
in terms of a third parameter π‘, dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ over dπ₯ by
dπ‘. We also need to recall the
derivatives of trigonometric functions where the argument is given in radians. The derivative with respect to π₯
of sin π₯ is cos π₯. The derivative with respect to π₯
of cos π₯ is negative sin π₯. And the derivative with respect to
π₯ of tan π₯ is sec squared π₯. And finally, we should also recall
the standard results for the derivatives of the three reciprocal trigonometric
functions. These results and techniques enable
us to find the equations of tangents and normals to the graphs of more advanced
functions.