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Amy Ferguson Moncure

Calculus BC ยท Unit 9

Derivatives of Parametric Equations

Divide dy/dt by dx/dt. A horizontal tangent needs a zero numerator and a nonzero denominator.

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Key Ideas Easy Medium Hard

\(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\)

What to Remember

Slope is a quotient of rates

\(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\), provided \(dx/dt\neq 0\).

Horizontal and vertical tangents

Horizontal: \(dy/dt=0\) and \(dx/dt\neq 0\).

Vertical: \(dx/dt=0\) and \(dy/dt\neq 0\).

If both are zero, the point needs a separate check.

A tangent line needs a point and a slope

Use the same value of \(t\) for \(x\), \(y\), and \(dy/dx\).

t is not the slope

The parameter orders the points. The slope comes from the two derivatives.

Level 1

Easy

Both derivatives are simple, and dx/dt is not zero.

Example
\(x=t^2\) and \(y=t^3-3t\). Find \(dy/dx\) at \(t=2\).

\(\dfrac{dx}{dt}=2t\)

\(\dfrac{dy}{dt}=3t^2-3\)

\(\dfrac{dy}{dx}=\dfrac{3t^2-3}{2t}\)

At \(t=2\): \(\dfrac{3\cdot 4-3}{4}\).

\(\dfrac{9}{4}\)

Answer: \(\dfrac{9}{4}\)

1
\(x=3t\) and \(y=6t-1\). Find \(dy/dx\).
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\(\dfrac{dy/dt}{dx/dt}=\dfrac{6}{3}\)

\(2\)

Answer: \(2\)

2
\(x=\cos t\) and \(y=\sin t\). Find \(dy/dx\) at \(t=\pi/3\).
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\(\dfrac{dy}{dx}=\dfrac{\cos t}{-\sin t}=-\cot t\)

At \(\pi/3\): \(-\cot\dfrac{\pi}{3}\).

\(-\dfrac{1}{\sqrt{3}}\)

Answer: \(-\dfrac{1}{\sqrt{3}}\)

3
\(x=t^2+1\) and \(y=4t\). Find the slope at \(t=-1\).
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\(\dfrac{dy}{dx}=\dfrac{4}{2t}=\dfrac{2}{t}\)

At \(t=-1\): \(-2\).

Answer: \(-2\)

4
For \(x=t^2\) and \(y=t\), where is the tangent horizontal?
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\(dy/dt=1\), which is never 0.

There is no horizontal tangent.

Answer: nowhere

5
For the same curve, where is the tangent vertical?
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\(dx/dt=2t=0\) at \(t=0\).

\(dy/dt=1\neq 0\).

The point is \((0,0)\).

Answer: vertical tangent at \((0,0)\)

Level 2

Medium

Find the point, then the slope, at the same t.

Example
\(x=t^2\) and \(y=t^3-3t\). Find the points with a horizontal tangent.

\(dy/dt=3(t^2-1)=0\) at \(t=\pm 1\).

At \(t=1\), \(dx/dt=2\neq 0\). Point: \((1,-2)\).

At \(t=-1\), \(dx/dt=-2\neq 0\). Point: \((1,2)\).

Answer: \((1,-2)\) and \((1,2)\)

1
For that same curve, find the point with a vertical tangent.
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\(dx/dt=2t=0\) at \(t=0\).

\(dy/dt=-3\neq 0\).

Point: \((0,0)\).

Answer: \((0,0)\)

2
Write the tangent line to \(x=t^2\), \(y=t^3-3t\) at \(t=2\).
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Point: \(x=4\), \(y=8-6=2\).

Slope: \(\dfrac{9}{4}\).

\(y-2=\dfrac{9}{4}(x-4)\)

Answer: \(y-2=\dfrac{9}{4}(x-4)\)

3
\(x=e^t\) and \(y=e^{2t}\). Find \(dy/dx\) in terms of \(t\), then at \(t=0\).
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\(\dfrac{dy}{dx}=\dfrac{2e^{2t}}{e^t}=2e^{t}\)

At \(t=0\): \(2\).

Answer: \(2e^{t}\), and 2 at \(t=0\)

4
At a parameter value, \(dx/dt=0\) and \(dy/dt=0\). What can you say about the tangent?
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The slope formula divides by zero and the numerator is also zero.

The usual horizontal-or-vertical test does not decide the tangent.

Look at the limit of \(dy/dx\) as \(t\) approaches that value.

Answer: the test does not decide it

5
\(x=2\cos t\) and \(y=3\sin t\). Find \(dy/dx\) at \(t=\pi/6\).
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\(\dfrac{dy}{dx}=\dfrac{3\cos t}{-2\sin t}\)

At \(\pi/6\): \(\dfrac{3\cdot \sqrt{3}/2}{-2\cdot 1/2}\).

\(\dfrac{3\sqrt{3}/2}{-1}\)

\(-\dfrac{3\sqrt{3}}{2}\)

Answer: \(-\dfrac{3\sqrt{3}}{2}\)

Level 3

Hard

A tangent line at a horizontal point, or a slope that still depends on t.

Example
Write the equation of the horizontal tangent to \(x=t^2\), \(y=t^3-3t\) at \(t=1\).

The point is \((1,-2)\).

The slope is 0.

\(y=-2\)

Answer: \(y=-2\)

1
\(x=t+\dfrac{1}{t}\) and \(y=t-\dfrac{1}{t}\) for \(t>0\). Find \(dy/dx\) at \(t=2\).
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\(\dfrac{dx}{dt}=1-\dfrac{1}{t^2}\)

\(\dfrac{dy}{dt}=1+\dfrac{1}{t^2}\)

At \(t=2\): \(\dfrac{1+1/4}{1-1/4}\).

\(\dfrac{5/4}{3/4}\)

\(\dfrac{5}{3}\)

Answer: \(\dfrac{5}{3}\)

2
A curve is given by \(x=\sin(2t)\) and \(y=\cos t\). Find the slope at \(t=\pi/2\).
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\(\dfrac{dx}{dt}=2\cos(2t)\)

\(\dfrac{dy}{dt}=-\sin t\)

At \(\pi/2\): \(dx/dt=2\cos\pi=-2\) and \(dy/dt=-1\).

\(\dfrac{dy}{dx}=\dfrac{-1}{-2}\)

\(\dfrac{1}{2}\)

Answer: \(\dfrac{1}{2}\)

3
Which statement produces a vertical tangent?
A. \(dx/dt=0\) and \(dy/dt\neq 0\)
B. \(dy/dt=0\) and \(dx/dt\neq 0\)
C. both derivatives zero
D. \(dy/dx=0\)
Show Answer

A vertical tangent is an undefined slope with a real point still being traced.

That is a zero in the denominator and a nonzero numerator.

Answer: A. \(dx/dt=0\) and \(dy/dt\neq 0\)

4
\(x=t^2-4t\) and \(y=t^2\). Find \(dy/dx\) when the curve crosses the y-axis.
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The y-axis means \(x=0\): \(t(t-4)=0\), so \(t=0\) or \(t=4\).

\(\dfrac{dy}{dx}=\dfrac{2t}{2t-4}=\dfrac{t}{t-2}\)

At \(t=0\), the slope is 0. The point is \((0,0)\).

At \(t=4\), the slope is \(4/2=2\). The point is \((0,16)\).

Answer: slope 0 at \((0,0)\) and slope 2 at \((0,16)\)

5
Explain why \(dy/dx=dy/dt\) is not the slope of \(x=t^3\), \(y=t^2\) at \(t=2\). Give the correct slope.
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The denominator \(dx/dt=3t^2\) is not 1.

At \(t=2\): \(\dfrac{2t}{3t^2}=\dfrac{2}{3t}\).

\(\dfrac{2}{6}=\dfrac{1}{3}\)

Answer: \(\dfrac{1}{3}\)

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