### Video Transcript

Find the length of the curve of the parametric equations π₯ is equal to two multiplied by the inverse of sin of π‘ and π¦ is equal to the natural logarithm of one minus π‘ squared, where π‘ is between zero and a half.

We recall our formula for calculating the length of a parametric curve from π‘ equals πΌ to π‘ equals π½ is equal to the integral from πΌ to π½ with respect to π‘ of the square root of the derivative of π₯ with respect to π‘ squared plus the derivative of π¦ with respect to π‘ squared. We see from the question that our parametric equation is defined from π‘ equals zero to π‘ equals a half. So, in our formula, we will have that πΌ is equal to zero and that π½ is equal to a half. We can also see from our formula that we will need to calculate the derivative of π₯ with respect to π‘ and the derivative of π¦ with respect to π‘.

We start by finding the derivative of π₯ with respect to π‘ which is equal to the derivative of two multiplied by the inverse sin of π‘ with respect to π‘. We recall that the derivative of the inverse sin of π₯ with respect to π₯ is equal to one divided by the square root of one minus π₯ squared. Meaning, we can calculate the derivative of π₯ with respect to π‘ as two divided by the square root of one minus π‘ squared. Now, to find the derivative of π¦ with respect to π‘, we need to differentiate the parametric equation for π¦ with respect to π‘. Which gives us the derivative of the natural logarithm of one minus π‘ squared with respect to π‘.

We now recall the chain rule which says that the derivative of the composition of two functions π and π is equal to the derivative of π evaluated at π multiplied by the derivative of π. In our derivative of π¦ with respect to π‘, we have an inner function of π is equal to one minus π‘ squared and an outer function π is equal to the natural logarithm of π‘. Evaluating our derivative of π¦ with respect to π‘ by using the chain rule gives us the derivative of π with respect to π‘ evaluated at π multiplied by the derivative of π with respect to π‘. We see the derivative of π with respect to π‘ is just a derivative of our natural log function with respect to π‘, which we know is just equal to one divided by π‘.

Therefore, we can conclude that the derivative of π with respect to π‘, evaluated at π of π‘, is just equal to one divided by one minus π‘ squared. Next, we have that the derivative of π with respect to π‘ is equal to the derivative of one minus π‘ squared with respect to π‘. Which we know, by the power rule for differentiation, is equal to negative two π‘. Combining this gives us that our derivative of π¦ with respect to π‘ is equal to one divided by one minus π‘ squared multiplied by negative two π‘. So letβs clear some space.

Weβre now ready to start substituting this information into our formula, giving us the length of our parametric equations. Where π‘ is between zero and a half is equal to the integral from zero to a half with respect to π‘ of the square root of two divided by the square root of one minus π‘ squared squared plus negative two π‘ divided by one minus π‘ squared squared. We can square the first set of parentheses to get four divided by one minus π‘ squared. And we can evaluate the square in our second set of parentheses to get four π‘ squared divided by one minus π‘ squared squared. Giving us a new integrand of the square root of four divided by one minus π‘ squared plus four π‘ squared divided by one minus π‘ squared squared.

We can then multiply both the numerator and the denominator of our first fraction by one minus π‘ squared. Which we can simplify to four minus four π‘ squared divided by one minus π‘ squared squared. Weβre now ready to add the two fractions in our integrand together. And we see that the negative four π‘ squared and the four π‘ squared cancel. Giving us a new integral from zero to a half of the square root of four divided by one minus π‘ squared squared with respect to π‘. At this point, we can notice that four divided by one minus π‘ squared squared is equal to two divided by one minus π‘ squared, all squared.

This means that we can now evaluate our square root, giving us the new integral from zero to a half of two divided by one minus π‘ squared with respect to π‘. To evaluate an integral in this form, we could attempt to use partial fractions. We see that our denominator, one minus π‘ squared, is a difference between squares, so is equal to one minus π‘ multiplied by one plus π‘. This means using partial fractions, we can write two divided by one minus π‘ multiplied by one plus π‘ as π΄ divided by one minus π‘ plus π΅ divided by one plus π‘. We then multiply both sides of this equation by our denominator, one minus π‘ multiplied by one plus π‘. Giving us that two is equal to π΄ multiplied by one plus π‘ plus π΅ multiplied by one minus π‘. Which we will write as equivalent because we know this equation must be true for all values of π‘.

We now see that when π‘ is equal to one, we will have that two must be equal to π΄ multiplied by one plus one plus π΅ multiplied by one minus one. Which if we calculate and simplify, we will get that the π΄ must be equal to one. Similarly, when π‘ is equal to negative one, we will have that two is equal to π΄ multiplied by one minus one plus π΅ multiplied by one minus negative one. Which if we calculate, we will see that π΅ must be equal to one. Therefore, by using partial fractions, we have a new integral for our length of our curve. Which is the integral from zero to a half of one divided by one plus π‘ plus one divided by one minus π‘ with respect to π‘.

Weβre now ready to start evaluating this integral. We recall that the integral of a derivative π prime of π₯ divided by π of π₯ is equal to the natural logarithm of the modulus of π of π₯ plus a constant of integration. We see that our function, one divided by one plus π‘, is already in the form of π prime divided by π. And we can easily change one divided by one minus π‘ to be in the form π prime divided by π by changing the numerator to negative one and multiplying the entire fraction by negative one. So we can use our integral formula to evaluate these integrals as the natural logarithm of the modulus of one plus π‘ minus the natural logarithm of the modulus of one minus π‘. Where we ignore our constant of integration because it will cancel out as our integral is definite.

We now evaluate this at the limits of our integral. Giving us the natural logarithm of the modulus of one plus a half minus the natural logarithm of the modulus of one minus a half minus the natural logarithm of the modulus of one plus zero minus the natural logarithm of the modulus of one minus zero. We know that the natural logarithm of one is equal to zero. So both of these terms are equal to zero. We know that one plus a half is equal to three over two and one minus a half is equal to a half. Next, we know that both three over two and a half are positive. So the modulus of three over two is equal to three over two. And the modulus of a half is equal to a half.

Finally, we can use our rule for logarithms, which says that the log of π₯ minus the log of π¦ is equal to the log of π₯ divided by π¦. Giving us the natural logarithm of three over two divided by a half. Which we can evaluate to be equal to the natural logarithm of three. Meaning that what we have shown is that the length of the curve with parametric equations π₯ is equal to two multiplied by the inverse sin of π‘. And π¦ is equal to the natural logarithm of one minus π‘ squared, where π‘ is between zero and a half. Is equal to the natural logarithm of three.