A parametric curve is defined by x = t², y = t³. A student computes the slope dy/dx as t³/t² = t. What error did they make, and what is the correct answer?
ANo error — dy/dx = y/x = t³/t² = t is correct
BThey used y/x instead of (dy/dt)/(dx/dt); the correct slope is (3t²)/(2t) = 3t/2
CThey forgot the chain rule entirely; the slope is dy/dt = 3t²
DThey should have eliminated the parameter first and differentiated y = x^(3/2) directly
The slope of a parametric curve is dy/dx = (dy/dt)/(dx/dt), not y/x. Here, dy/dt = 3t² and dx/dt = 2t, giving dy/dx = 3t²/(2t) = 3t/2. The error in option A confuses the coordinates (y, x) with the derivatives (dy/dt, dx/dt) — a very common mistake when the parameter t appears in both x and y as simple powers. The correct formula comes from the chain rule: dy/dx = (dy/dt) · (dt/dx) = (dy/dt)/(dx/dt).
Question 2 Multiple Choice
For the curve x = t², y = t³ − 3t, a student wants to find where horizontal and vertical tangents occur. They set dx/dt = 2t = 0, getting t = 0, and call this a horizontal tangent. What is wrong?
CThe student should eliminate t first and then find where dy/dx = 0 in terms of x
Ddx/dt = 0 has no geometric meaning for parametric curves
This is the most common confusion in parametric calculus. A horizontal tangent requires dy/dx = 0, which means the numerator (dy/dt) equals zero while the denominator (dx/dt) is nonzero. For this curve, dy/dt = 3t² − 3 = 0 gives t = ±1 → horizontal tangents. A vertical tangent requires dy/dx to be undefined, meaning the denominator (dx/dt) equals zero while the numerator is nonzero. dx/dt = 2t = 0 gives t = 0 → vertical tangent. The student swapped the conditions.
Question 3 True / False
The formula d²y/dx² = [d/dt(dy/dx)] / (dx/dt) is derived by applying the same chain rule logic that gives dy/dx = (dy/dt)/(dx/dt), but treating dy/dx as the new 'y' quantity.
TTrue
FFalse
Answer: True
This is exactly the logic. The first derivative dy/dx is itself a function of t. To find how it changes with respect to x (i.e., its derivative with respect to x), apply the same chain rule: d(dy/dx)/dx = [d(dy/dx)/dt] / (dx/dt). This is why the second derivative formula works — it is the chain rule applied twice, not some separate rule. The formula is NOT d²y/dt² divided by d²x/dt², which would be a ratio of second derivatives in t with no geometric meaning.
Question 4 True / False
If dx/dt = 0 at a point on a parametric curve, the curve has a horizontal tangent there.
TTrue
FFalse
Answer: False
dx/dt = 0 means the curve is momentarily not moving in the x-direction — it creates a vertical tangent (the curve moves purely vertically at that instant, making the slope undefined). A horizontal tangent occurs when dy/dt = 0 (no vertical motion) while dx/dt ≠ 0. Mixing these two conditions is listed as a core misconception for this topic because the conditions feel symmetric but their geometric consequences are opposite.
Question 5 Short Answer
Explain why the second derivative of a parametric curve is NOT computed as d²y/dt² divided by d²x/dt².
Think about your answer, then reveal below.
Model answer: d²y/dx² measures how the slope dy/dx changes as you move along the x-axis — it is a geometric quantity about the curve's concavity. d²y/dt² and d²x/dt² measure how fast y and x accelerate as the parameter t changes — these are properties of the parameterization, not the curve's geometry. Dividing them gives a ratio that changes when you reparameterize the same curve (e.g., using t² instead of t), so it cannot represent a geometric property. The correct formula applies the chain rule to d(dy/dx)/dx, treating dy/dx as a new function and converting dt to dx.
A useful sanity check: the second derivative d²y/dx² must be independent of the choice of parameterization (it is a property of the curve, not of how you describe it). The ratio d²y/dt² / d²x/dt² fails this test — it changes if you reparameterize. The correct formula [d/dt(dy/dx)] / (dx/dt) passes the test because (dy/dx) is already parameterization-independent, and dividing its t-derivative by dx/dt correctly converts back to a derivative with respect to x.