This slide presents the parametric differentiation rule $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ together with the key restriction that $\frac{dx}{dt}\neq 0$. It emphasizes that the slope with respect to $x$ is obtained by comparing how both coordinates change with the same parameter. Source

When a curve is given parametrically, the key derivative is found by comparing how $x$ and $y$ change with the same parameter. This converts a two-function description into a familiar rate of change.

Understanding the setup

A parametric curve is described by giving both coordinates as functions of a third variable, usually $t$.

This figure illustrates a circle parameterized by $x=a\cos t$ and $y=a\sin t$, with specific points labeled by their parameter values (e.g., $t=0$ and $t=\pi/2$). The arrow indicates the direction the curve is traced as $t$ increases, highlighting that parametric equations encode both the geometric path and its orientation. Source

Instead of writing $y$ directly as a function of $x$, you write separate expressions for $x$ and $y$.

Because both coordinates depend on $t$, you usually cannot find $\dfrac{dy}{dx}$ by taking an ordinary derivative of $y$ with respect to $x$ right away. The derivative must be built from the way each coordinate changes with the parameter.

Why ordinary differentiation is not enough

For a parametrically defined curve, the quantities $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ tell you how $x$ and $y$ change as $t$ changes. If $x$ is changing and $y$ is also changing, then the rate of change of $y$ with respect to $x$ is found by comparing those two rates.

This is the essential AP Calculus BC idea: differentiate both coordinate functions with respect to the parameter, then divide. The formula is valid only when the denominator is not zero.

This formula is a chain-rule result, not just a shortcut. It tells you how fast $y$ changes compared with how fast $x$ changes at the same parameter value.

How to find $\dfrac{dy}{dx}$

The standard process is short but must be done carefully:

• Write the parametric equations clearly, such as $x=f(t)$ and $y=g(t)$.

• Differentiate $x$ with respect to $t$ to get $\dfrac{dx}{dt}$.

• Differentiate $y$ with respect to $t$ to get $\dfrac{dy}{dt}$.

• Form the quotient $\dfrac{dy/dt}{dx/dt}$.

• Simplify if possible.

• If the problem asks for the derivative at a specific value of $t$, substitute that value only after finding the derivative formula unless there is a clear reason to do otherwise.

In many problems, the final derivative is left in terms of $t$. That is normal. You do not need to eliminate the parameter unless the question specifically requires the answer in terms of $x$ or $y$.

Interpreting the quotient

The sign and size of $\dfrac{dy}{dx}$ come from the relationship between the two parameter derivatives.

• If $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ have the same sign, then $\dfrac{dy}{dx}$ is positive.

• If they have opposite signs, then $\dfrac{dy}{dx}$ is negative.

• If $\left|\dfrac{dx}{dt}\right|$ is very small while $\dfrac{dy}{dt}$ is not, the value of $\dfrac{dy}{dx}$ can become very large in magnitude.

So the derivative is not telling you how $y$ changes as time passes or as the parameter changes. It tells you how $y$ changes per unit change in $x$.

Conditions for using the formula

The most important restriction is that $\dfrac{dx}{dt}$ cannot equal $0$ when you divide. If it does, the quotient formula does not produce a valid value for $\dfrac{dy}{dx}$ at that parameter value.

This restriction matters because parametric equations can behave differently from ordinary functions. A single curve may:

• pass through the same point more than once,

• reverse direction as $t$ changes,

• have different behavior at different parameter values that produce the same location.

That means the parameter value matters. If a problem asks for $\dfrac{dy}{dx}$ at a given point, you may need to identify which value or values of $t$ correspond to that point before evaluating the derivative.

Reading AP-style questions carefully

On the AP exam, prompts often vary slightly. You may be asked to:

• find $\dfrac{dy}{dx}$ in terms of $t$,

• find $\dfrac{dy}{dx}$ at a specific parameter value,

• determine where $\dfrac{dy}{dx}=0$,

• determine where $\dfrac{dy}{dx}$ is undefined.

Each version uses the same core idea. The difference is what you do after forming the quotient.

If $\dfrac{dy}{dx}=0$, then the numerator must be $0$ while the denominator is nonzero. If $\dfrac{dy}{dx}$ is undefined by this formula, then the denominator is $0$. These distinctions are easy to lose if you rush.

Common errors to avoid

Students often know the formula but lose points through small mistakes. Watch for these issues:

• Switching the order of the derivatives: $\dfrac{dy}{dx}$ is $\dfrac{dy/dt}{dx/dt}$, not the other way around.

• Using different parameter values: both derivatives must be evaluated at the same value of $t$.

• Forgetting the denominator restriction: a quotient is only valid when $\dfrac{dx}{dt}\ne 0$.

• Confusing $\dfrac{dy}{dt}$ with $\dfrac{dy}{dx}$: they measure different kinds of change.

• Substituting too early: differentiating first usually reduces algebra errors.

• Ignoring multiple parameter values for one point: the same rectangular point can sometimes come from more than one value of $t$.

What your final answer should look like

A strong AP response is usually concise and algebraically clean. It should:

• show both $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$,

• form the correct quotient,

• simplify correctly,

• include any needed evaluation at a specified value of $t$,

• respect the condition that $\dfrac{dx}{dt}\ne 0$.