Video: Determining the Type of Concavity of a Parametric Curve

Consider the parametric curve π‘₯ = πœƒ βˆ’ sin πœƒ and 𝑦 = 1 βˆ’ cos πœƒ. Determine whether this curve is concave upward, concave downward, or neither at πœƒ = πœ‹/6.

07:12

Video Transcript

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

We’re given a curve which is defined parametrically. We’re given π‘₯ in terms of πœƒ and 𝑦 in terms of πœƒ. And we need to determine the concavity of this curve when πœƒ is equal to πœ‹ 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 𝑦 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 𝑦 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 𝑦 by dπ‘₯ squared. But we can’t do this directly. Our curve is defined parametrically. 𝑦 is not given as a function of π‘₯. So instead, we’re going to need to use our rules for differentiating parametric curves. First, we need to recall if 𝑦 is given as a function of πœƒ and π‘₯ is given as a function of πœƒ. Then by using the chain rule, we’ll have d𝑦 by dπ‘₯ is equal to d𝑦 by dπœƒ divided by dπ‘₯ by dπœƒ. This assumes that the denominator dπ‘₯ by dπœƒ is not equal to zero.

But in this case, we’re looking for the second derivative of 𝑦 with respect to π‘₯. And to do this, recall that d two 𝑦 by dπ‘₯ squared is equal to the derivative of d𝑦 by dπ‘₯ with respect to π‘₯. We’ll then use the chain rule again. And this gives us the following expression for d two 𝑦 by dπ‘₯ squared. It’s equal to the derivative of d𝑦 by dπ‘₯ with respect to πœƒ divided by dπ‘₯ by dπœƒ. 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 𝑦 by dπ‘₯ squared. We need to find an expression for d𝑦 by dπ‘₯, which means we need to differentiate 𝑦 with respect to πœƒ and differentiate π‘₯ with respect to πœƒ.

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

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

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

Then, to use the quotient rule, we need to find expressions for 𝑒 prime and 𝑣 prime. Let’s start with 𝑒 prime. That’s the derivative of the sin of πœƒ with respect to πœƒ. We know this is equal to the cos of πœƒ. Similarly, we need to find an expression for 𝑣 prime. That’s the derivative of one minus the cos of πœƒ with respect to πœƒ. 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œƒ is just equal to the sin of πœƒ. We’re now ready to evaluate this derivative by using the quotient rule. We get that this is equal to 𝑒 prime 𝑣 minus 𝑣 prime 𝑒 divided by 𝑣 squared.

Substituting in our expressions for 𝑒, 𝑣, 𝑒 prime, and 𝑣 prime, we get the derivative of d𝑦 by dπ‘₯ with respect to πœƒ is equal to the cos of πœƒ times one minus the cos of πœƒ minus the sin of πœƒ multiplied by the sin of πœƒ all divided by one minus the cos of πœƒ 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œƒ is equal to πœ‹ by six. So we’re just going to be substituting πœƒ is πœ‹ by six into an expression. So we don’t actually need to simplify.

Instead, let’s use our formula and divide this by dπ‘₯ by dπœƒ. We’ve already found an expression for dπ‘₯ by dπœƒ. It’s equal to one minus the cos of πœƒ. So by using our formula, we have that d two 𝑦 by dπ‘₯ squared is equal to the derivative of d𝑦 by dπ‘₯ with respect to πœƒ divided by dπ‘₯ by dπœƒ. We just found an expression for the numerator of this expression by using the quotient rule. And in fact, we’ve 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œƒ.

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

Now all that’s left to do is substitute the value of πœƒ is equal to πœ‹ by six into this expression. Substituting this value in, we get the cos of πœ‹ by six multiplied by one minus the cos of πœ‹ by six minus the sin of πœ‹ by six times the sin of πœ‹ by six all divided by one minus the cos of πœ‹ by six all cubed. And remember, we don’t 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œƒ is equal to πœ‹ by six.

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

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.