Video Transcript
What is the equation of the normal
to the curve 𝑦 equals four 𝑥 squared over three plus root two 𝑥 at a half a
quarter?
Let’s begin by thinking about what
we actually understand by this word “the normal.” Let’s say we have this curve given,
𝑓 of 𝑥. The tangent to the curve might look
something like this. The normal is the line that’s
perpendicular to the tangent at this point. So in order to be able to find the
equation of the normal to the curve, we need to find first the gradient of the
tangent to the curve. And we can find the gradient of the
tangent by differentiating our equation for the curve.
But how do we differentiate four 𝑥
squared over three plus root two 𝑥? Well, we use the quotient
function. This says that for two
differentiable functions 𝑢 and 𝑣, the derivative of that quotient is 𝑣 times d𝑢
by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣 squared. So we let 𝑢 be equal to four 𝑥
and 𝑣 be equal to three plus root two 𝑥. We’re going to need to
differentiate each of these functions with respect to 𝑥.
Well, the derivative of four 𝑥
squared with respect to 𝑥 is two times four 𝑥 to the power of one, which is just
eight 𝑥. But before we can differentiate 𝑣
with respect to 𝑥, we’re going to need to rewrite it a little bit. We’re going to rewrite it as three
plus two 𝑥 all to the power of a half. And then, we’ll rewrite it a little
further. And we say that two 𝑥 to the power
of a half is the same as two to the power of a half times 𝑥 to the power of a
half. And since two to the power of a
half is simply a constant, we can now differentiate this as normal.
The derivative of three with
respect to 𝑥 is zero. And then, the derivative of two to
the power of a half times 𝑥 to the power of a half is a half times two to the power
of a half times 𝑥 to the power of negative a half. We know that one over two is the
same as one over two to the power of one. And then, two to the power of a
half divided by two to the power of one is two to the power of negative a half. Remember when the bases are the
same and we’re dividing, we can subtract the exponents. So d𝑣 by d𝑥 is equal to two to
the power of negative a half times 𝑥 to the power of negative a half, which can
further be rewritten as two 𝑥 to the power of negative one-half.
Let’s substitute everything we now
have into our formula for the quotient rule. It’s 𝑣 times d𝑢 by d𝑥 minus 𝑢
times d𝑣 by d𝑥 all over 𝑣 squared. Now, we could try and simplify this
a little. But remember our aim is to find the
gradient of the tangent at the point one-half, one-quarter. So in fact, we can go ahead and
substitute 𝑥 is equal to one-half to find the gradient of the tangent at this
point. And we get three plus two times a
half to the power of a half times eight multiplied by one-half minus four times a
half squared times two times a half to the power of negative a half. And that’s all over three plus two
times a half to the power of one-half squared.
Of course, two times one-half is
just one. We know eight times one-half is
four. And then, since one-half squared is
a quarter, we have four times one-quarter which is one. A half to the power of anything is
one. So we can simplify this a
little. And we get four times four minus
one times one all over four squared. And we see that the gradient of the
tangent to the curve at one-half, one-quarter is fifteen sixteenths.
So how does this help us find the
equation of the normal to the curve? Well, we know that the normal is
perpendicular to the tangent. And we know that the product of the
gradients of two lines which are perpendicular is negative one. If 𝑚 one and 𝑚 two are the
gradients of our lines which are perpendicular to one another, we can rearrange our
formula. We can say that 𝑚 two is equal to
negative one divided by 𝑚 one. So let’s now use this information
to find the gradient of the normal to the curve. Let’s clear some space.
Since the gradient of our tangent
is fifteen sixteenths, the gradient of the normal at this point is negative one over
fifteen sixteenths, which is negative sixteen fifteenths. You might have seen this called the
negative reciprocal. And now that we know the gradient
of the normal to the curve and we know it passes through the point whose Cartesian
coordinates are one-half, one-quarter, we can substitute everything we have into the
formula for the equation of a straight line.
At our point, 𝑦 is one-quarter and
𝑥 is one-half. So our equation becomes 𝑦 minus a
quarter equals negative 16 over 15 multiplied by 𝑥 minus one-half. To make this more aesthetically
pleasing, we can rearrange it and write it in the form 𝑎𝑦 plus 𝑏𝑥 equals 𝑐,
where 𝑎, 𝑏, and 𝑐 are constants. There are a number of ways we can
do this. But I’m going to distribute our
parentheses on the right-hand side. And when I do, our equation becomes
𝑦 minus a quarter equals negative sixteen fifteenths 𝑥 plus eight fifteenths.
We’re now going to multiply through
by the lowest common multiple of four and 15. That’s 60. And that gives me 60𝑦 minus 15
equals negative 64𝑥 plus 32. To get it in the form 𝑎𝑦 plus 𝑏
𝑥 equals 𝑐, I’m going to add 15 and 64𝑥 to both sides of my equation. And we see that the equation of the
normal to the curve at a half, one-quarter is 60𝑦 plus 64𝑥 equals 47.
