Second Derivatives of Parametric Equations
Introduced| English | Meaning |
|---|---|
| second derivative/ˈsekənd dɪˈrɪvətɪv/ | The derivative of a function's first derivative, describing how its gradient changes. |
| concavity/kənˈkævɪti/ | The direction in which a curve bends, indicated by how its gradient changes. |
How does the tangent slope change?
- A moving point can trace the same path in either direction. To study the path's bending as a graph of $y$ against $x$, measure how its slope changes with $x$, rather than with time alone.
- The second derivative 二阶导数 $d^2y/dx^2$ measures this change and gives concavity 凹凸性 on a smooth local graph. It is not, by itself, the geometric curvature of a path.
Apply the chain rule again
- Write the first slope as $m(t)=\dfrac{dy/dt}{dx/dt}$, where $dx/dt\ne0$. Then $dm/dt=(dm/dx)(dx/dt)$, so $\dfrac{d^2y}{dx^2}=\dfrac{dm/dt}{dx/dt}$.
- Differentiate the slope in $t$, then divide by $dx/dt$ again. Check $dx/dt\ne0$ at every point where you use this formula; simplification does not remove that restriction.
The parametric second derivative $\dfrac{d^2y}{dx^2}$ equals...
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Differentiate $\tfrac{dy}{dx}$ in $t$, then divide by $\tfrac{dx}{dt}$.
Order the steps to find the parametric second derivative.
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Slope, differentiate in $t$, divide by $\tfrac{dx}{dt}$.
The regular second-derivative formula requires $dx/dt\ne0$.
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Both the first slope formula and the final division require a nonzero $dx/dt$.
Work the two divisions
- For $x=t^2$, $y=t^3$ and $t\ne0$, the first slope is $m(t)=3t^2/(2t)=3t/2$. Its derivative in $t$ is $dm/dt=3/2$.
- Divide once more: $\dfrac{d^2y}{dx^2}=\dfrac{3/2}{2t}=\dfrac3{4t}$. At $t=1$, this is $3/4$. The intermediate value $3/2$ is not the second derivative in $x$.
For $x=t^2$, $y=t^3$ ($\tfrac{dy}{dx}=\tfrac{3t}{2}$), $\dfrac{d^2y}{dx^2}=$
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$\frac{d}{dt}(\tfrac{3t}{2})=\tfrac32$; divide by $2t$: $\tfrac{3}{4t}$.
After differentiating $\tfrac{dy}{dx}$ in $t$, you divide by $\dfrac{dx}{dt}$, the ____ denominator as the first derivative.
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Both steps divide by $\tfrac{dx}{dt}$.
For $x=t^2$, $y=t^3$, find $d^2y/dx^2$ at $t=1$.
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$3/(4t)=3/4$.
Read the branches separately
- For this curve, $3/(4t)>0$ when $t>0$: the upper branch is concave up. For $t<0$, the lower branch is concave down. Both branches have $x=t^2>0$.
- At $t=0$, $dx/dt=0$ and the displayed formula is unavailable. The two branches meet at a cusp; a sign change across this parameter value does not automatically establish an inflection point of one smooth graph $y=f(x)$.
Concavity on a parametric path
At t=1 for x=t²,y=t³, compare slope 3/2, its t-derivative 3/2, and second x-derivative 3/4. Compare reversed travel, zero second derivatives and a singular parameter value.
The sign of the parametric second derivative tells you the curve's...
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Positive → concave up, negative → concave down.
For the same curve, find $d^2y/dx^2$ at $t=-1$.
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$3/[4(-1)]=-3/4$, so the lower branch is concave down.
Compare the tempting wrong answers
- The ratio $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ is not a general formula for $d^2y/dx^2$. In the example it gives $6t/2=3t$, which is $3$ at $t=1$, instead of $3/4$.
- Differentiating $m(t)$ without the final division is also wrong in general. Keep the three quantities distinct: $dy/dx$, $dm/dt$ and $d^2y/dx^2$.
The parametric second derivative equals $\dfrac{d^2y/dt^2}{d^2x/dt^2}$.
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That naive ratio is wrong.
For $x=t^2$, $y=t^3$, what is $d(dy/dx)/dt$ when $t\ne0$?
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The first slope is $3t/2$, whose t-derivative is $3/2$. A further division by $2t$ is still needed.
The incorrect ratio $(d^2y/dt^2)/(d^2x/dt^2)$ gives what value for $x=t^2$, $y=t^3$ at $t=1$?
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It gives $6/2=3$; the actual second x-derivative is $3/4$.
Simplify before choosing a rule
- If the first slope simplifies to $3t/2$, differentiate that directly. You need the quotient rule only when the remaining expression requires it, not simply because the slope began as a ratio.
- With $x'=dx/dt$ and $y'=dy/dt$, the equivalent formula is $\dfrac{d^2y}{dx^2}=\dfrac{x'y''-y'x''}{(x')^3}$, for $x'\ne0$. It follows from the quotient rule and the final division by $x'$.
You must use the quotient rule even after the first slope simplifies to a linear expression.
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You can differentiate the simplified expression directly, while keeping its domain restriction.
A decreasing x reverses the time-based sign
- For $x=-t$, $y=t^2$, the slope is $m(t)=-2t$, so $dm/dt=-2$. But $dx/dt=-1$, giving $d^2y/dx^2=(-2)/(-1)=2>0$.
- The curve is $y=x^2$, which is concave up. The slope decreases as $t$ increases because the point moves toward decreasing $x$. The sign of $dm/dt$ alone does not decide concavity.
For $x=-t$, $y=t^2$, find $d^2y/dx^2$.
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The slope is $-2t$. Divide its derivative $-2$ by $dx/dt=-1$ to obtain 2.
If $dm/dt<0$, a parametric curve must be concave down.
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When $dx/dt<0$, division reverses the sign. For $x=-t,y=t^2$, the curve is concave up.
Work a nonconstant denominator
- For $x=t^3+t$, $y=t^2$, $dx/dt=3t^2+1>0$ and $m(t)=2t/(3t^2+1)$. The quotient rule gives $dm/dt=2(1-3t^2)/(3t^2+1)^2$.
- Dividing again gives $d^2y/dx^2=2(1-3t^2)/(3t^2+1)^3$. At $t=1$, this is $-1/16$, so the curve is locally concave down at $(2,1)$.
For $x=t^3+t$, $y=t^2$, find the second x-derivative at $t=1$.
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$2(1-3t^2)/(3t^2+1)^3=-4/64=-1/16$.
Why does $x=t^3+t$ cause no zero-denominator restriction for real t?
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$dx/dt=3t^2+1\ge1$, so it is always positive.
A zero second derivative needs a sign check
- For $x=t$, $y=t^3$, the second derivative is $6t$. It changes from negative to positive through $t=0$, so the origin is an inflection point.
- For $x=t$, $y=t^4$, the second derivative is $12t^2$. It is zero at the origin but positive on both sides: there is no change of concavity there. Zero alone is not enough.
For $x=t$, $y=t^4$, the origin is an inflection point because $d^2y/dx^2=0$ there.
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$12t^2$ is positive on both sides of 0, so concavity does not change.
For $x=t$, $y=t^3$, what establishes an inflection at the origin?
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$d^2y/dx^2=6t$ changes from negative to positive through a regular graph point.
Keep singular points separate
- If $dx/dt=0$, do not insert that value into a denominator or cancel it away and claim the regular formula applies there. The geometry may require a separate argument.
- For $x=t^3$, $y=t^3$, the regular formula is unavailable at $t=0$, yet the path is $y=x$. That straight line has second derivative $0$. Failure of the parameter formula does not prove failure of a graph derivative.
If $dx/dt=0$, the graph can never have a second derivative at the corresponding point.
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A stationary parametrization can describe a straight line: $x=t^3,y=t^3$ gives $y=x$, whose second derivative is 0.
The path $x=t^3$, $y=t^3$ is $y=x$. What is its second derivative as a graph, including at the origin?
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A straight-line graph has constant first slope 1, so its second derivative is 0. This conclusion uses the path, not direct division at t=0.
Try fresh concavity problems
- For $x=t^3+2t$, $y=t^2$, derive $dy/dx$ and $d^2y/dx^2$, then evaluate the second derivative at $t=1$ and state the concavity.
- For $x=-2t$, $y=t^3$, find $dm/dt$ and $d^2y/dx^2$ at $t=-1$. Explain why their signs differ. Then explain why $x=t$, $y=t^4$ has no inflection at $t=0$.
For $x=t^3+2t$, $y=t^2$, derive the first and second x-derivatives. Evaluate the second at t=1 and state the concavity.
Compare your chain-rule steps, restrictions and sign reasoning with the model.
For $x=-2t$, $y=t^3$, find $dm/dt$ and $d^2y/dx^2$ at t=-1. Explain their signs.
Compare your chain-rule steps, restrictions and sign reasoning with the model.
Follow the derivative with respect to x
- Find the first slope, differentiate it in $t$, and divide by $dx/dt$ again. Keep both the denominator restriction and the distinction between the intermediate derivative and the final result.
- Interpret the final sign on a smooth local graph. Check a change in concavity before claiming an inflection, and investigate singular parameter values separately.
For $x=t^2$, $y=t^3$, where does the regular formula $3/(4t)$ apply?
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$dx/dt=2t$, so t=0 must be excluded.
Compare the origin for $x=t,y=t^3$ and for $x=t,y=t^4$. Why does a zero second derivative not settle both cases?
Compare your chain-rule steps, restrictions and sign reasoning with the model.