### Video Transcript

Consider the parametric curve π₯ is equal to π minus the sin of π and π¦ is equal to one minus the cos of π. Determine whether this curve is concave upward, concave downward, or neither at π is equal to π by six.

Weβre given a curve which is defined parametrically. Weβre given π₯ in terms of π and π¦ in terms of π. And we need to determine the concavity of this curve when π is equal to π by six. Recall the concavity of a curve means whether its tangent lines lie above or below the curve. We can determine this by looking at the second derivative of our function. In particular, we know if the second derivative of π¦ with respect to π₯ at a certain point is greater than zero, then our curve is concave upward at this point. And alternatively, if the second derivative of π¦ with respect to π₯ is negative at this point, then our curve will be concave downward at this point.

So to determine the concavity of our curve, we need to find an expression for d two π¦ by dπ₯ squared. But we canβt do this directly. Our curve is defined parametrically. π¦ is not given as a function of π₯. So instead, weβre going to need to use our rules for differentiating parametric curves. First, we need to recall if π¦ is given as a function of π and π₯ is given as a function of π. Then by using the chain rule, weβll have dπ¦ by dπ₯ is equal to dπ¦ by dπ divided by dπ₯ by dπ. This assumes that the denominator dπ₯ by dπ is not equal to zero.

But in this case, weβre looking for the second derivative of π¦ with respect to π₯. And to do this, recall that d two π¦ by dπ₯ squared is equal to the derivative of dπ¦ by dπ₯ with respect to π₯. Weβll then use the chain rule again. And this gives us the following expression for d two π¦ by dπ₯ squared. Itβs equal to the derivative of dπ¦ by dπ₯ with respect to π divided by dπ₯ by dπ. And once again, this will only be valid if our denominator is not equal to zero. So now we can see what we need to find an expression for d two π¦ by dπ₯ squared. We need to find an expression for dπ¦ by dπ₯, which means we need to differentiate π¦ with respect to π and differentiate π₯ with respect to π.

Letβs start by finding an expression for dπ₯ by dπ. Thatβs the derivative of π minus the sin of π with respect to π. Of course, we can do this term by term. π is a linear function. So the derivative of π with respect to π is the coefficient of π, which in this case is one. Next, we should know the derivative of negative sin of π with respect to π is negative the cos of π. This is a standard trigonometric derivative result which we should commit to memory. We can now do the same to find an expression for dπ¦ by dπ. Thatβs the derivative of one minus the cos of π with respect to π.

Again, we do this term by term. The derivative of the constant one with respect to π is equal to zero. Next, the derivative of negative the cos of π is equal to the sin of π. So weβve now found expressions for dπ¦ by dπ and dπ₯ by dπ. So we can use our formula to find an expression for dπ¦ by dπ₯. So we have dπ¦ by dπ₯ is equal to dπ¦ by dπ divided by dπ₯ by dπ. And substituting in our expressions for these derivatives, we get the sin of π divided by one minus the cos of π. But remember, to find the concavity of our curve, we need an expression for d two π¦ by dπ₯ squared. So we need to differentiate this expression with respect to π₯. And weβve already found a formula for this by using the chain rule.

We need to differentiate this expression with respect to π and then divide this by dπ₯ by dπ. And in this case, we can see dπ¦ by dπ₯ is the quotient of two functions. So to differentiate this, weβll use the quotient rule. We recall the quotient rule tells us the derivative of the quotient of two functions π’ over π£ is equal to π’ prime π£ minus π£ prime π’ all divided by π£ squared. So to find the derivative of dπ¦ by dπ₯ with respect to π, weβll set π’ to be the function in our numerator. Thatβs the sin of π. And weβll set π£ to be the function in our denominator. Thatβs one minus the cos of π.

Then, to use the quotient rule, we need to find expressions for π’ prime and π£ prime. Letβs start with π’ prime. Thatβs the derivative of the sin of π with respect to π. We know this is equal to the cos of π. Similarly, we need to find an expression for π£ prime. Thatβs the derivative of one minus the cos of π with respect to π. And we can do this term by term. First, we know the derivative of the constant one is just equal to zero. Next, the derivative of negative the cos of π is just equal to the sin of π. Weβre now ready to evaluate this derivative by using the quotient rule. We get that this is equal to π’ prime π£ minus π£ prime π’ divided by π£ squared.

Substituting in our expressions for π’, π£, π’ prime, and π£ prime, we get the derivative of dπ¦ by dπ₯ with respect to π is equal to the cos of π times one minus the cos of π minus the sin of π multiplied by the sin of π all divided by one minus the cos of π all squared. And at this point, we might be tempted to start simplifying. But remember, we only need to find the concavity of our curve when π is equal to π by six. So weβre just going to be substituting π is π by six into an expression. So we donβt actually need to simplify.

Instead, letβs use our formula and divide this by dπ₯ by dπ. Weβve already found an expression for dπ₯ by dπ. Itβs equal to one minus the cos of π. So by using our formula, we have that d two π¦ by dπ₯ squared is equal to the derivative of dπ¦ by dπ₯ with respect to π divided by dπ₯ by dπ. We just found an expression for the numerator of this expression by using the quotient rule. And in fact, weβve already found an expression for the denominator of this expression. So we need to divide this entire expression by one minus the cos of π.

And instead of doing this, we can just multiply by the reciprocal of one minus the cos of π. Then because this is where d appears in our denominator, we can just increase this exponent by one. This gives us d two π¦ by dπ₯ squared is equal to the cos of π times one minus the cos of π minus the sin of π multiplied by the sin of π all divided by one minus the cos of π all cubed. And now weβre almost done. All we need to do is substitute our value of π equal to π by six into this expression. First, letβs clear some space.

Now all thatβs left to do is substitute the value of π is equal to π by six into this expression. Substituting this value in, we get the cos of π by six multiplied by one minus the cos of π by six minus the sin of π by six times the sin of π by six all divided by one minus the cos of π by six all cubed. And remember, we donβt need to evaluate this answer exactly. We only need to know if this gives a positive or negative output. And if we evaluate this expression, we can see to one decimal place, we get negative 55.7. This is a negative output. So we can conclude that our curve must be concave downward when π is equal to π by six.

Therefore, we were able to show given the parametric curve π₯ is equal to π minus the sin of π and π¦ is equal to one minus the cos of π. We were able to show that the curve must be concave downward when the value of π is equal to π by six.