### Video Transcript

Find an equation of the tangent to the curve π₯ equals one plus root π‘, π¦ equals π
to the power of π‘ squared at the point two, π.

Now in this question, what weβre trying to do is find the equation of the tangent to
the curve. Now there is actually a relationship that we know between a tangent and our
curve. And that is that they actually have the same slope. So therefore, the first thing we want to do is actually find the slope function of
our curve so we can actually determine the slope of our curve.

So to enable us to actually find slope function, what we need to think about is the
fact that we actually have parametric equations. And if we actually have parametric equations, we actually have a general form to help
us find what the slope function is going to be. And the relationship we know is that ππ¦ ππ₯, so our slope function, is equal to
ππ¦ ππ‘ divided by ππ₯ ππ‘. And a good thing about this is it allows us deal with each other parametric equation
separately, so differentiate them, and then bring it together to find our slope
function.

So the first thing we gonna start with is actually π₯ equals one plus root π‘. And what we want to do is actually differentiate this to allow us to find ππ₯
ππ‘. But before differentiate it, what Iβm gonna do is actually rewrite it with an
exponent instead of our root. And when we do that, we get π₯ is equal to one plus π‘ to the power of a half. And when we differentiate this expression, what weβre gonna get is ππ₯ ππ‘ is equal
to a half π‘ to the power of negative a half.

And just to remind us how we did that, we had one multiplied by a half, so our
coefficient multiplied by a half which was our exponent. This is for the second term, which gives us a half. And then π‘ to the power of, and then youβve got a half minus one because we subtract
one from the exponent which gives us a negative half. And the first term was just one. And if you differentiate one, it goes to zero. So weβre left with ππ₯ ππ‘ is equal to a half π‘ to the power of negative a
half.

Okay, great! Now letβs move on to π¦. So what we have is that π¦ is equal to π to the power of π‘ squared. Now to enable us to actually differentiate this, what we actually have is a rule: and
thatβs if π¦ is equal to π to the power of ππ₯, a function, then the derivative is
gonna be equal to the derivative of ππ₯ multiplied by π to the power of ππ₯. And we actually get this from the chain rule.

Okay, so letβs apply this to ours and find ππ¦ ππ‘. So therefore, if we actually differentiate, weβre gonna get get ππ¦ ππ‘ is equal to
two π‘. And we got two π‘ because if you differentiate π‘ squared, so our ππ₯, you get two
π‘. And then this is multiplied by π to the power of π‘ squared. Okay, great! So now Iβve got ππ₯ ππ‘ and ππ¦ ππ‘, we can actually find a slope function using
the relationship we spoke about earlier.

So therefore, what weβre gonna get is our slope function ππ¦ ππ₯ is actually gonna
be equal to two π‘π to the power of π‘ squared divided by one over two π‘ to the
power of a half. We got that second bit because actually what we did is we had a half π‘ to the power
of negative a half. Well if itβs negative a half, itβs one over two to power of a half. Okay, great! And then actually using the rule of fractions when we divide, what we can say is itβs
gonna be equal to two π‘π to the power of π‘ squared multiplied by two π‘ to the
power of a half.

And thatβs because when we actually divide by a fraction, we multiply by the
reciprocal of that fraction. Okay, so we now have the slope function. What we need to do is actually find out what the slope is. But to enable to do that, we need to know what the π‘ value is gonna be at this
point. Now in order to find the π‘ value, what we need to do is actually substitute our
value of our point. So weβve got π₯ equals two and π¦ equals π into our parametric equations. So if we start with our π₯-value, which is two, we substitute that into π₯ equals one
plus root π‘.

And what we get is one plus root π‘ is equal to two. Iβve just flipped around the other way cause itβs easier to solve. And then subtract one from each side. When I do that, I get root π‘ is equal to one. And then what I need to do to find π‘ is actually square both sides. When I square both sides, I get π‘ is equal to one. Okay, great! So Iβve got π‘ is equal to one. So this looks like this is gonna be our π‘ value but what Iβm gonna do now is
actually check it by using our value for π¦, which is π.

So what we have is π to the power of π‘ squared is equal to π. Well writing π is like writing π to the power of one. And therefore, as we actually have the bases the same on each side of our equation,
what we can actually do is equate our exponents. So therefore, what weβre gonna have is π‘ squared is equal to one. So therefore, what we can see is that a solution is also gonna be π‘ equals one. So therefore, yep we found our π‘ value. Itβs going to be one.

So what we can do now is actually substitute π‘ equals one back into our slope
function. And this will find us a value for the slope. So then when I substitute in one for all values of π‘, I get π‘ multiplied by one π
to the power of one squared multiplied by two and then multiplied by root one. We got root one because π‘ to the power of a half is the same as root of π‘, which
gonna be equal to π multiplied by two.

So therefore, we found the slope. And the slope is gonna be equal to four π. Okay, great! So now what do we do? So now what we can do is actually solve the problem and actually find the equation of
the tangent to the curve. So because itβs equation of a tangent to the curve, we know that the tangentβs going
to be a straight line. So therefore, we can actually use the general form for a straight line, which is π¦
equals ππ₯ plus π where π is the slope and π is the π¦- intercept.

So therefore we can actually substitute in the four π that we found the slope, so
our π value, and we can say that π¦ is equal to four ππ₯ plus π. But we need to find out what π is. And to do this, we can actually substitute in the value we have for the point because
the point is actually a point on both the tangent and the curve. And thatβs two, π. So π₯ equals two and π¦ equals π. So when we do that, we get π is equal to four π multiplied by two plus π.

So therefore π is equal to eight π plus π. So therefore what we can do is actually to subtract eight π from both sides of the
equation. And when we do this, weβll get negative seven π is equal to π. So weβve got our value now for our π. So therefore, what we can say is that if we actually substitute that back into our
equation for the line, we can say that the equation of the tangent to the curve π₯
equals one plus root π‘, π¦ equals π to the power of π‘ squared at the point two,
π is π¦ equals four ππ₯ minus seven π. And we got that because four π was our slope and negative seven π was our value for
π, the π¦-intercept.