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Calculus with Parametric Curves

Calculus 2 · Chapter 9: Parametric Equations & Polar CoordinatesSection 9.2 · lesson 46 of 5410 min · free

← Parametric CurvesNext lesson: Arc Length and Surface Area in Parametric Form →

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In this lesson

Slopes and concavity without eliminating the parameter.

  • Compute dy/dx as a ratio of t-derivatives
  • Locate horizontal and vertical tangents
  • Apply the second-derivative formula correctly
  • Avoid the quotient-of-second-derivatives trap

Transcript

Everything said in the video, section by section (1,481 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 9.2

Calculus with Parametric Curves

Section nine point one showed how to describe a curve with a parameter. Now we want calculus on those curves. Slopes. Tangent lines. Concavity. And the obvious approach — eliminate the parameter first, then differentiate — is often messy. And sometimes impossible. So there's a better way. The chain rule lets you differentiate without ever eliminating t.

The First Derivative

A Ratio of t-Derivatives

Here's the idea. The chain rule says d y d t equals d y d x, times d x d t. Solve that for d y d x. You get d y d x equals d y d t, over d x d t. Provided d x d t is not zero. So: differentiate each equation separately with respect to t. Then divide. The parameter never has to be eliminated at all. Let's do one. A curve is given by x equals t squared plus three t, and y equals two t cubed minus six t. Find the slope of the tangent at t equals two. Step one. Differentiate each equation with respect to t. d x d t is two t plus three. And d y d t is six t squared minus six. Step two. Form the quotient. d y d x equals six t squared minus six, over two t plus three. Step three. Substitute t equals two, evaluating numerator and denominator separately. The numerator is six times four, minus six. Twenty-four minus six. Eighteen. The denominator is four plus three. Seven. So the slope is eighteen sevenths. Step four. Locate the point. Substitute t equals two into the original equations. x is four plus six, which is ten. And y is sixteen minus twelve, which is four. So at the point ten comma four, the curve has slope eighteen sevenths. And note the denominator two t plus three is nonzero at t equals two, so the theorem applies.

Tangent Lines

When the Numerator or Denominator Vanishes

Now, when does a curve have a horizontal or vertical tangent? Think about the quotient. A quotient equals zero when its numerator is zero. And it fails to exist when its denominator is zero. Apply that to d y d x equals d y d t over d x d t, and you get two conditions. Horizontal tangent. Occurs at values of t where d y d t equals zero, and d x d t is not zero. The slope there is zero. Vertical tangent. Occurs where d x d t equals zero, and d y d t is not zero. The slope is undefined. And a third case. Indeterminate. If both derivatives vanish at the same t, no conclusion follows from the formula alone. The point may be a cusp, or a corner, or even a perfectly smooth point. It requires separate investigation. So those and clauses are essential. Requiring that only one of the two derivatives vanish is exactly what makes the test valid. When both vanish, the formula gives zero over zero. And the geometry must be examined on its own terms.

Worked Example 2

Locating Every Tangent

Let's find every tangent on a specific curve. x equals t squared minus one. y equals t cubed minus three t. Compute both derivatives. d x d t is two t. And d y d t is three t squared minus three. First, horizontal tangents. Set d y d t equal to zero. Three t squared minus three equals zero. Factor out the three, then use the difference of squares. Three times t minus one, times t plus one, equals zero. So t equals one, or t equals minus one. Now confirm d x d t is not zero at those values. At t equals one, d x d t is two. At t equals minus one, it's minus two. Both qualify. Compute the points. At t equals one: x is one minus one, zero. And y is one minus three, minus two. So zero comma minus two. At t equals minus one: x is again zero. And y is minus one plus three, which is two. So zero comma two. Both horizontal tangents occur at x equals zero. At two different heights. Now vertical tangents. Set d x d t equal to zero. Two t equals zero, so t equals zero. Confirm d y d t is not zero there. Three times zero, minus three, is minus three. Not zero. So this is a genuine vertical tangent. And the point is x equals minus one, y equals zero. Now here's something worth noticing. The same x-value, zero, yields two distinct points. So this curve cannot possibly be the graph of a function of x. A reminder that parametric curves are free to intersect themselves.

Figure 9.2.1

The Curve and Its Tangents

Look at the curve. The two gold points, at zero comma two and zero comma minus two, carry the horizontal tangents. And the point at minus one comma zero carries the vertical one. You can see the curve loop back on itself, crossing near the right side. And notice how the same vertical line at x equals zero passes through two points of the curve. That's exactly what makes this impossible as a function of x. And exactly what parametric equations exist to permit.

The Second Derivative

A Trap Worth Naming

Concavity requires the second derivative. And here a common error has to be confronted immediately. The second derivative is not the quotient of the two second derivatives. Writing d two y d t squared, over d two x d t squared, is incorrect. And it will produce wrong answers. The dividing factor is d x d t. Not d two x d t squared. The correct reasoning treats the second derivative as the derivative of d y d x, with respect to x. And since d y d x is a function of t, differentiating it with respect to x requires the first theorem a second time — applied to d y d x, in place of y. So: the second derivative equals d d t of d y d x, all divided by d x d t. In words. First form d y d x as a function of t. Then differentiate that expression with respect to t. Then divide by d x d t once more. The curve is concave up where that's positive, and concave down where it's negative.

Worked Example 3

Applying the Formula, Step by Step

Let's apply it. For x equals t squared, y equals t cubed minus three t, find the second derivative and evaluate at t equals two. Step one. Compute the first derivatives. d x d t is two t. d y d t is three t squared minus three. Step two. Form d y d x. That's three t squared minus three, over two t. And here's a useful move — split the fraction into two simpler terms. Three t squared over two t is three halves t. And minus three over two t is minus three halves, t to the minus one. Step three. Differentiate that with respect to t. The derivative of three halves t is three halves. For the second term, the power rule gives the derivative of t to the minus one is minus t to the minus two. So the derivative of minus three halves t inverse is plus three halves t to the minus two. Combine over the common denominator two t squared. You get three t squared plus three, over two t squared. Step four. Divide by d x d t, which is two t. As the theorem requires. Dividing by two t is the same as multiplying by one over two t. So the second derivative is three t squared plus three, over four t cubed. Step five. Evaluate at t equals two. The numerator is three times four, plus three. Fifteen. The denominator is four times eight. Thirty-two. Fifteen thirty-seconds. Positive. So the curve is concave up at t equals two. Now for contrast — what would the incorrect formula have given? d two y d t squared is six t. And d two x d t squared is two. So their quotient is three t. At t equals two, that's six. A completely different number. And wrong.

Key Takeaways

What to Carry Forward

Five takeaways. The slope is d y d t over d x d t, wherever d x d t is nonzero. No elimination of the parameter is required. Horizontal tangents need d y d t equal to zero, with d x d t nonzero. Vertical tangents need the reverse. And when both vanish, no conclusion follows — the point requires separate investigation. The second derivative is the t-derivative of d y d x, divided by d x d t. Never the quotient of the two second derivatives. Its sign gives concavity. Positive is concave up. Negative is concave down. And a parametric curve may cross itself. So one x-value can carry several points, and several tangents. Something no graph of a function can do. Next up is section nine point three — Arc Length and Surface Area for parametric curves.

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FAQ

Questions students ask us

About Calculus with Parametric Curves

How do you find dy/dx for parametric equations?
To find dy/dx for a parametric curve, differentiate each equation with respect to t and divide: dy/dx = \frac{dy/dt}{dx/dt}, provided dx/dt \neq 0. This follows from the chain rule, so you never need to eliminate the parameter. For example, if x = t^2 + 3t and y = 2t^3 - 6t, then dy/dx = \frac{6t^2 - 6}{2t + 3}. That equals 18/7 at t = 2, which is the point (10, 4).
How do you find horizontal and vertical tangent lines on a parametric curve?
A parametric curve has a horizontal tangent where dy/dt = 0 and dx/dt \neq 0, and a vertical tangent where dx/dt = 0 and dy/dt \neq 0. For x = t^2 - 1, y = t^3 - 3t, horizontal tangents occur at t = \pm 1, giving (0, -2) and (0, 2). A vertical tangent occurs at t = 0, giving (-1, 0). If both derivatives are zero, the test is inconclusive and the point needs separate investigation.
What is the formula for the second derivative of a parametric curve?
The second derivative of a parametric curve is \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(dy/dx)}{dx/dt}. First form dy/dx as a function of t, then differentiate it with respect to t, then divide by dx/dt. For x = t^2, y = t^3 - 3t, this gives \frac{3t^2 + 3}{4t^3}, which equals 15/32 at t = 2. Since that value is positive, the curve is concave up there.
Why can't you divide d²y/dt² by d²x/dt² to get the second derivative?
Dividing d^2y/dt^2 by d^2x/dt^2 gives the wrong answer because the correct formula divides the t-derivative of dy/dx by dx/dt, not by d^2x/dt^2. For x = t^2, y = t^3 - 3t, the wrong formula gives 6t/2 = 3t, or 6 at t = 2. The correct second derivative there is 15/32. Always find dy/dx first, then differentiate it with respect to t and divide by dx/dt.

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