AP®MathsCollege BoardCalculus BCFlashcardsUnit 5: Analytical Applications of DifferentiationBehaviors of Implicit Relations

Behaviors of Implicit Relations (College Board AP® Calculus BC): Flashcards

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FrontSecond Derivatives of Implicit Functions
How do you find $\frac{d^{2} y}{d x^{2}}$ for a curve given implicitly?

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How do you find $\frac{d^{2} y}{d x^{2}}$ for a curve given implicitly?
Find $\frac{d y}{d x}$ first, then differentiate that whole expression with respect to $x$ again.
The right-hand side usually needs the quotient rule, and any $y$ inside it still picks up a factor of $\frac{d y}{d x}$ .
Fill in the missing derivative in the last step of an implicit second derivative:
if the expression for $\frac{d^{2} y}{d x^{2}}$ still contains $______$ then substitute the expression you already found for it
The completed instruction is: if the expression for $\frac{d^{2} y}{d x^{2}}$ still contains $\frac{d y}{d x}$ then substitute the expression you already found for it.
That turns the answer into one written in $x$ and $y$ alone, which is usually the form wanted.
A curve has $\frac{d y}{d x} = \frac{2 - x}{y}$ . Find $\frac{d^{2} y}{d x^{2}}$ in terms of $x$ and $y$ .
It is $\frac{d^{2} y}{d x^{2}} = - \frac{y^{2} + {(2 - x)}^{2}}{y^{3}}$ .
The quotient rule gives $\frac{- y - (2 - x) \frac{d y}{d x}}{y^{2}}$ , and substituting $\frac{d y}{d x} = \frac{2 - x}{y}$ then simplifying produces that answer.
True or False?
Differentiating $\frac{d y}{d x} = \frac{2 - x}{y}$ again needs no chain rule, because the numerator contains no $y$ .
False.
The denominator is $y$ , and differentiating $y$ with respect to $x$ still brings in a factor of $\frac{d y}{d x}$ .
Every $y$ in the expression carries that factor, wherever in the expression it sits.
A curve has $\frac{d y}{d x} = \frac{\cos x}{\sin y}$ . What is $\frac{d^{2} y}{d x^{2}}$ in terms of $x$ and $y$ ?
It is $\frac{d^{2} y}{d x^{2}} = - \frac{\sin x \sin^{2} y + \cos^{2} x \cos y}{\sin^{3} y}$ .
The quotient rule gives $\frac{- \sin x \sin y - \cos x \cos y \frac{d y}{d x}}{\sin^{2} y}$ , and substituting for $\frac{d y}{d x}$ then writing both terms over $\sin^{3} y$ gives that result.
Why can two correct answers for $\frac{d^{2} y}{d x^{2}}$ look different?
Because the same expression can be written in several ways, for instance with a negative sign factored out of the numerator or left inside it.
Check by rearranging one into the other, rather than assuming yours is wrong.
Fill in the two missing words in the definition of a critical point on an implicit relation:
a critical point occurs where the $______$ is equal to zero, or where it does not $______$ at all
The completed definition is: a critical point occurs where the derivative is equal to zero, or where it does not exist at all.
It is exactly the same definition as for any other function, so nothing new has to be learned for an implicit relation.
Where is the tangent to an implicitly defined curve horizontal?
At any point on the curve where $\frac{d y}{d x} = 0$ .
When the derivative is a quotient, that means solving for a zero numerator, and then checking that the denominator is not also zero there.
Where is the tangent to an implicitly defined curve vertical?
At any point where $\frac{d x}{d y} = 0$ , which is the same as $\frac{d y}{d x}$ having a zero denominator and a non-zero numerator.
The slope is unbounded there rather than zero.
True or False?
Locating a horizontal tangent on an implicit relation needs the $y$ -coordinate as well as the $x$ -coordinate.
True.
The derivative of an implicit relation is usually an expression in both $x$ and $y$ , so setting the numerator to zero gives $x$ , and the original equation is then needed to find $y$ .
The answer is a point, not just an $x$ value.
Find where the tangent to $2 y^{2} - 6 = y \sin 2 x$ is horizontal, for $0 \leq x \leq \frac{π}{2}$ and $y > 0$ .
At $(\frac{π}{4}, 2)$ .
Implicit differentiation gives $\frac{d y}{d x} = \frac{2 y \cos 2 x}{4 y - \sin 2 x}$ , so the numerator is zero when $\cos 2 x = 0$ , giving $x = \frac{π}{4}$ .
Substituting back gives $2 y^{2} - y - 6 = 0$ and so $y = 2$ , and the denominator is 7 there rather than zero.
When you evaluate $\frac{d^{2} y}{d x^{2}}$ at a critical point of an implicit relation, what three things do you substitute?
The $x$ -coordinate, the $y$ -coordinate, and the value of $\frac{d y}{d x}$ , which is zero at a critical point.
An implicit second derivative usually still contains $\frac{d y}{d x}$ , so all three are needed before its sign can be read off.

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