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Consider the parametric curve π₯ is equal to the cos cubed of π and π¦ is equal to the sin cubed of π.
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Determine whether this curve is concave up, down, or neither at π is equal to π by six.
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The question gives us a curve which is defined by a pair of parametric equations.
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Weβre given π₯ in terms of π and π¦ in terms of π.
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We need to determine the concavity of this curve at the point where π is equal to π by six.
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To start, we need to recall what is meant by the concavity of a curve.
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The concavity tells us whether the tangent lines to the curve lie above or below our curve.
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And in particular, we can measure this by using our second derivative.
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We know if d two π¦ by dπ₯ squared is greater than zero at a point, then our curve is concave upward at this point.
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And we also know if d two π¦ by dπ₯ squared is less than zero at a point, then our curve is concave downwards at this point.
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So, we can check the concavity of our curve by looking at d two π¦ by dπ₯ squared.
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However, we canβt differentiate π¦ with respect to π₯ directly in this case because weβre given our curve as a parametric curve.
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Weβre not given π¦ in terms of π₯.
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Weβre given π¦ in terms of π and π₯ in terms of π.
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So, weβre going to need to use our rules for differentiating parametric curves.
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The first thing weβll recall is an application of the chain rule.
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If π¦ is a function of π and π₯ is a function of π, then we can find by using the chain rule dπ¦ by dπ₯ is equal to dπ¦ by dπ divided by dπ₯ by dπ.
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And this is, of course, only valid when our denominator dπ₯ by dπ is not equal to zero.
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Of course, we need to find an expression for the second derivative of π¦ with respect to π₯.
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Thatβs the derivative of dπ¦ by dπ₯ with respect to π₯.
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And once again, we can do this by using the chain rule.
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We get d two π¦ by dπ₯ squared is equal to the derivative of dπ¦ by dπ₯ with respect to π divided by dπ₯ by dπ.
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And just as before, this wonβt be valid when our denominator dπ₯ by dπ is equal to zero.
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So, to find our expression for d two π¦ by dπ₯ squared, we need to find an expression for dπ¦ by dπ₯.
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But to find an expression for dπ¦ by dπ₯, we need to differentiate π¦ with respect to π and π₯ with respect to π.
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And weβre given π¦ and π₯ as functions of π.
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So, we can do this.
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Letβs start by finding an expression for dπ₯ by dπ.
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We know π₯ is the cos cubed of π, so we need to differentiate the cos cubed of π with respect to π.
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Now, thereβs a couple of different ways we can evaluate this derivative.
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For example, we could use the chain rule.
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However, in this case, we can use the general power rule.
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Recall, the general power rule tells us for any real number π and differentiable function π of π, the derivative of π of π to the πth power with respect to π is equal to π times π prime of π multiplied by π of π to the power of π minus one.
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In our case, we want to differentiate the cos cubed of π.
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So, weβll set π of π to be the cos of π and π to be equal to three.
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Then, to use the general power rule, we need to find an expression for π prime of π.
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Of course, this is a standard trigonometric derivative result.
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The derivative of the cos of π with respect to π is negative the sin of π.
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Now, we can apply the general power rule.
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We get dπ₯ by dπ is equal to three times negative the sin of π multiplied by the cos of π raised to the power of three minus one.
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And, of course, we can simplify this to give us negative three sin of π times the cos squared of π.
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Now, weβll clear our working and do something very similar to find an expression for dπ¦ by dπ.
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This time, we need to differentiate the sin cubed of π with respect to π.
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Again, we could do this by using the chain rule, but weβll use the general power rule.
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This time, weβre differentiating the sin cubed of π, so weβll set π of π to be the sin of π and our value of π equal to three.
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This time, we need to use the fact that the derivative of the sin of π with respect to π is equal to the cos of π.
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So, we find π prime of π is the cos of π.
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Now, by applying the general power rule, we get dπ¦ by dπ is equal to three cos of π multiplied by the sin of π raised to the power of three minus one.
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And, of course, we can simplify this to get three cos of π sin squared of π.
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Now that we found expressions for dπ¦ by dπ and dπ₯ by dπ, letβs clear our working and use these to find an expression for dπ¦ by dπ₯.
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So, by using our formula, we get dπ¦ by dπ₯ is equal to dπ¦ by dπ divided by dπ₯ by dπ.
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Then, we substitute in our expressions for dπ¦ by dπ and dπ₯ by dπ.
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We get dπ¦ by dπ₯ is equal to three cos of π sin squared of π all divided by negative three sin of π cos squared of π.
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And we can simplify this expression.
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First, weβll cancel the shared factor of three in our numerator and our denominator.
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Next, weβll cancel the shared factor of the sin of π in our numerator and our denominator.
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Finally, weβll cancel the shared factor of the cos of π in our numerator and our denominator.
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This leaves us with the sin of π divided by negative the cos of π.
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And we can rewrite this as negative the tan of π.
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Weβre now ready to start finding our expression for our second derivative of π¦ with respect to π₯.
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Letβs start by finding the numerator of our formula.
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The numerator of this expression is the derivative of dπ¦ by dπ₯ with respect to π.
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But we already showed dπ¦ by dπ₯ is equal to negative the tan of π.
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So, we need to differentiate negative the tan of π with respect to π.
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And this is one of our standard trigonometric derivative results.
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The derivative of the tan to π with respect to π is equal to the sec squared of π.
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So, by using this, weβve shown the derivative of dπ¦ by dπ₯ with respect to π is equal to negative the sec squared of π.
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Now, all we need to do to find an expression for our second derivative of π¦ with respect to π₯ is divide this expression by dπ₯ by dπ.
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But we already found dπ₯ by dπ.
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Itβs negative three sin of π times the cos squared of π.
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So, by using our formula and substituting in the expressions we found for the numerator and the denominator, we get d two π¦ by dπ₯ squared is equal to negative the sec squared of π divided by negative three sin of π multiplied by the cos squared of π.
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And we can simplify this slightly since negative one divided by negative one is just equal to one.
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Now, we could start simplifying this expression even more.
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But remember, weβre only interested in the concavity of this curve at a point.
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And to find this, weβre only interested in if our output when π is equal to π by six is positive or negative.
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Well, we know the sec squared of π is a square; it will always be greater than or equal to zero.
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Similarly, the cos squared of π is also a square.
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It will be greater than or equal to zero for any input value of π.
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So, in actual fact, the only part of this expression which can be negative is the sin of π.
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So, we can just evaluate this.
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The sin of π by six, we know, is equal to one-half.
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And remember, this formula is only valid if dπ₯ by dπ is not equal to zero.
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So, we should check that the cos of π by six is also not equal to zero.
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Of course, we know the cos of π by six is the square root of three divided by two.
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So, this is not equal to zero.
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So, what weβve shown is when π is equal to π by six, the numerator of d two π¦ by dπ₯ squared is positive.
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Itβs a nonzero number squared.
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Similarly, the denominator is also positive.
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Itβs the product of three nonzero positive numbers.
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So, when π is equal to π by six, d two π¦ by dπ₯ squared is the quotient of two positive numbers.
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Therefore, itβs positive.
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And remember, if d two π¦ by dπ₯ squared is positive at a point, then we know that our curve is concave upward at this point.
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Therefore, given the parametric curve π₯ is equal to the cos cubed of π and π¦ is equal to the sin cubed of π.
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We were able to show that the curve is concave upward at the point where π is equal to π by six.