A parametric curve x = f ( t ) x = f(t) x = f ( t ) y = g ( t ) y = g(t) y = g ( t ) x x x y y y
d y d x = d y / d t d x / d t wherever d x d t ≠ 0 \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \qquad \text{wherever } \frac{dx}{dt} \ne 0 d x d y = d x / d t d y / d t wherever d t d x = 0 Over a short time Δ t \Delta t Δ t d x d t Δ t \tfrac{dx}{dt}\,\Delta t d t d x Δ t d y d t Δ t \tfrac{dy}{dt}\,\Delta t d t d y Δ t Δ t \Delta t Δ t d y d t = d y d x ⋅ d x d t \tfrac{dy}{dt} = \tfrac{dy}{dx} \cdot \tfrac{dx}{dt} d t d y = d x d y ⋅ d t d x d x d t \tfrac{dx}{dt} d t d x The slope compares how fast y y y x x x
One arch of the cycloid x = t − sin t, y = 1 − cos t
One arch of a cycloid, the path of a point on a rolling wheel of radius 1, rising from the origin to a height of 2 at x = π and coming back down at x = 2π. A dot marks the point at t = π/2, near (0.57, 1), where a dashed tangent line of slope 1 touches the curve.
2 4 6 1 2 x y
t = π/2
One arch of the cycloid x = t − sin t, y = 1 − cos t
Example 1 — A tangent line to the cycloid
Find the tangent line to x = t − sin t x = t - \sin t x = t − sin t y = 1 − cos t y = 1 - \cos t y = 1 − cos t t = π 2 t = \tfrac{\pi}{2} t = 2 π
d x d t = 1 − cos t = 1 \tfrac{dx}{dt} = 1 - \cos t = 1 d t d x = 1 − cos t = 1 d y d t = sin t = 1 \tfrac{dy}{dt} = \sin t = 1 d t d y = sin t = 1 t = π 2 t = \tfrac{\pi}{2} t = 2 π 1 1 = 1 \tfrac{1}{1} = 1 1 1 = 1 ( π 2 − 1 , 1 ) \left(\tfrac{\pi}{2} - 1, 1\right) ( 2 π − 1 , 1 )
y − 1 = x − ( π 2 − 1 ) y - 1 = x - \left(\frac{\pi}{2} - 1\right) y − 1 = x − ( 2 π − 1 ) The ratio shows where the tangent lines turn flat or upright:
Horizontal where d y d t = 0 \tfrac{dy}{dt} = 0 d t d y = 0 d x d t ≠ 0 \tfrac{dx}{dt} \ne 0 d t d x = 0 Vertical where d x d t = 0 \tfrac{dx}{dt} = 0 d t d x = 0 d y d t ≠ 0 \tfrac{dy}{dt} \ne 0 d t d y = 0
Where both derivatives are 0 0 0 cusp , as the cycloid does wherever it touches the ground.
The second derivative is the rate of change of the slope with respect to x x x t t t
d 2 y d x 2 = d d t ( d y d x ) / d x d t \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \bigg/ \frac{dx}{dt} d x 2 d 2 y = d t d ( d x d y ) / d t d x It is not d 2 y / d t 2 d 2 x / d t 2 \tfrac{d^2y/dt^2}{d^2x/dt^2} d 2 x / d t 2 d 2 y / d t 2
A short step has length Δ x 2 + Δ y 2 ≈ ( d x / d t ) 2 + ( d y / d t ) 2 Δ t \sqrt{\Delta x^2 + \Delta y^2} \approx \sqrt{(dx/dt)^2 + (dy/dt)^2}\;\Delta t Δ x 2 + Δ y 2 ≈ ( d x / d t ) 2 + ( d y / d t ) 2 Δ t t t t a a a b b b
L = ∫ a b ( d x d t ) 2 + ( d y d t ) 2 d t L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt L = ∫ a b ( d t d x ) 2 + ( d t d y ) 2 d t Example 2 — Horizontal and vertical tangents
Find the horizontal and vertical tangents of x = t 2 x = t^2 x = t 2 y = t 3 − 3 t y = t^3 - 3t y = t 3 − 3 t
The curve x = t², y = t³ − 3t
A curve that comes in from the upper right, loops around, crosses itself at (3, 0) and leaves toward the lower right. Dots mark horizontal tangents at (1, 2) and (1, −2), with short dashed horizontal lines, and the vertical tangent at the origin, where the curve touches the y-axis.
1 2 3 4 5 -3 -2 -1 1 2 3 x y
t = −1
t = 1
t = 0
The curve x = t², y = t³ − 3t
d y d t = 3 t 2 − 3 = 0 \tfrac{dy}{dt} = 3t^2 - 3 = 0 d t d y = 3 t 2 − 3 = 0 t = ± 1 t = \pm 1 t = ± 1 d x d t = 2 t ≠ 0 \tfrac{dx}{dt} = 2t \ne 0 d t d x = 2 t = 0 ( 1 , − 2 ) (1, -2) ( 1 , − 2 ) ( 1 , 2 ) (1, 2) ( 1 , 2 ) d x d t = 2 t = 0 \tfrac{dx}{dt} = 2t = 0 d t d x = 2 t = 0 t = 0 t = 0 t = 0 d y d t = − 3 ≠ 0 \tfrac{dy}{dt} = -3 \ne 0 d t d y = − 3 = 0
Example 3 — The second derivative
For the same curve, find d 2 y d x 2 \tfrac{d^2y}{dx^2} d x 2 d 2 y t = 1 t = 1 t = 1
d y d x = 3 t 2 − 3 2 t \tfrac{dy}{dx} = \tfrac{3t^2 - 3}{2t} d x d y = 2 t 3 t 2 − 3
d d t ( 3 t 2 − 3 2 t ) = 6 t ⋅ 2 t − ( 3 t 2 − 3 ) ⋅ 2 4 t 2 = 3 t 2 + 3 2 t 2 \frac{d}{dt}\left(\frac{3t^2 - 3}{2t}\right) = \frac{6t \cdot 2t - (3t^2 - 3) \cdot 2}{4t^2} = \frac{3t^2 + 3}{2t^2} d t d ( 2 t 3 t 2 − 3 ) = 4 t 2 6 t ⋅ 2 t − ( 3 t 2 − 3 ) ⋅ 2 = 2 t 2 3 t 2 + 3 Divide by d x d t = 2 t \tfrac{dx}{dt} = 2t d t d x = 2 t d 2 y d x 2 = 3 t 2 + 3 4 t 3 \tfrac{d^2y}{dx^2} = \tfrac{3t^2 + 3}{4t^3} d x 2 d 2 y = 4 t 3 3 t 2 + 3 t = 1 t = 1 t = 1 6 4 = 1.5 > 0 \tfrac{6}{4} = 1.5 > 0 4 6 = 1.5 > 0
Example 4 — The length of one arch
Find the length of one arch of the cycloid, 0 ≤ t ≤ 2 π 0 \le t \le 2\pi 0 ≤ t ≤ 2 π
( d x d t ) 2 + ( d y d t ) 2 = ( 1 − cos t ) 2 + sin 2 t = 2 − 2 cos t = 4 sin 2 t 2 \left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2 = (1 - \cos t)^2 + \sin^2 t = 2 - 2\cos t = 4\sin^2\tfrac{t}{2} ( d t d x ) 2 + ( d t d y ) 2 = ( 1 − cos t ) 2 + sin 2 t = 2 − 2 cos t = 4 sin 2 2 t 1 − cos t = 2 sin 2 t 2 1 - \cos t = 2\sin^2\tfrac{t}{2} 1 − cos t = 2 sin 2 2 t [ 0 , 2 π ] [0, 2\pi] [ 0 , 2 π ] sin t 2 ≥ 0 \sin\tfrac{t}{2} \ge 0 sin 2 t ≥ 0
L = ∫ 0 2 π 2 sin t 2 d t = [ − 4 cos t 2 ] 0 2 π = 4 + 4 = 8 L = \int_0^{2\pi} 2\sin\frac{t}{2}\,dt = \Big[-4\cos\frac{t}{2}\Big]_0^{2\pi} = 4 + 4 = 8 L = ∫ 0 2 π 2 sin 2 t d t = [ − 4 cos 2 t ] 0 2 π = 4 + 4 = 8 A wheel of radius 1 1 1 2 π ≈ 6.28 2\pi \approx 6.28 2 π ≈ 6.28 8 8 8
Example 5 — Checking with a circle
Use the formula to find the circumference of x = r cos t x = r\cos t x = r cos t y = r sin t y = r\sin t y = r sin t
r 2 sin 2 t + r 2 cos 2 t = r \sqrt{r^2\sin^2 t + r^2\cos^2 t} = r r 2 sin 2 t + r 2 cos 2 t = r L = ∫ 0 2 π r d t = 2 π r L = \int_0^{2\pi} r\,dt = 2\pi r L = ∫ 0 2 π r d t = 2 π r
Common mistake
Dividing the derivatives upside down. The slope is d y / d t d x / d t \tfrac{dy/dt}{dx/dt} d x / d t d y / d t x = t 2 x = t^2 x = t 2 y = t 3 y = t^3 y = t 3 3 t 2 2 t = 3 t 2 \tfrac{3t^2}{2t} = \tfrac{3t}{2} 2 t 3 t 2 = 2 3 t 2 t 3 t 2 \tfrac{2t}{3t^2} 3 t 2 2 t
Common mistake
Dividing second derivatives. d 2 y d x 2 \tfrac{d^2y}{dx^2} d x 2 d 2 y d 2 y / d t 2 d 2 x / d t 2 \tfrac{d^2y/dt^2}{d^2x/dt^2} d 2 x / d t 2 d 2 y / d t 2 d y d x \tfrac{dy}{dx} d x d y t t t d x d t \tfrac{dx}{dt} d t d x
Common mistake
Counting a curve twice in arc length. The formula measures the path as t t t a a a b b b
For x = t 2 x = t^2 x = t 2 y = t 3 y = t^3 y = t 3 d y d x \tfrac{dy}{dx} d x d y t = 2 t = 2 t = 2
Answer
3 3 3
Full solution
d y d x = 3 t 2 2 t = 3 t 2 \tfrac{dy}{dx} = \tfrac{3t^2}{2t} = \tfrac{3t}{2} d x d y = 2 t 3 t 2 = 2 3 t 3 3 3 t = 2 t = 2 t = 2
For x = 3 cos t x = 3\cos t x = 3 cos t y = 3 sin t y = 3\sin t y = 3 sin t t = π 4 t = \tfrac{\pi}{4} t = 4 π
Answer
− 1 -1 − 1
Full solution
d y d x = 3 cos t − 3 sin t = − cot t \tfrac{dy}{dx} = \tfrac{3\cos t}{-3\sin t} = -\cot t d x d y = − 3 s i n t 3 c o s t = − cot t − 1 -1 − 1 t = π 4 t = \tfrac{\pi}{4} t = 4 π
Find the tangent line to x = t + 1 x = t + 1 x = t + 1 y = t 2 y = t^2 y = t 2 t = 2 t = 2 t = 2
Answer
y − 4 = 4 ( x − 3 ) y - 4 = 4(x - 3) y − 4 = 4 ( x − 3 )
Full solution
The point is ( 3 , 4 ) (3, 4) ( 3 , 4 ) d y d x = 2 t 1 = 4 \tfrac{dy}{dx} = \tfrac{2t}{1} = 4 d x d y = 1 2 t = 4 t = 2 t = 2 t = 2
Find the horizontal and vertical tangents of x = t 3 − 3 t x = t^3 - 3t x = t 3 − 3 t y = t 2 y = t^2 y = t 2
Answer
Horizontal at ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( − 2 , 1 ) (-2, 1) ( − 2 , 1 ) ( 2 , 1 ) (2, 1) ( 2 , 1 )
Full solution
d y d t = 2 t = 0 \tfrac{dy}{dt} = 2t = 0 d t d y = 2 t = 0 t = 0 t = 0 t = 0 d x d t = − 3 ≠ 0 \tfrac{dx}{dt} = -3 \ne 0 d t d x = − 3 = 0 d x d t = 3 t 2 − 3 = 0 \tfrac{dx}{dt} = 3t^2 - 3 = 0 d t d x = 3 t 2 − 3 = 0 t = ± 1 t = \pm 1 t = ± 1 d y d t = ± 2 ≠ 0 \tfrac{dy}{dt} = \pm 2 \ne 0 d t d y = ± 2 = 0 ( − 2 , 1 ) (-2, 1) ( − 2 , 1 ) ( 2 , 1 ) (2, 1) ( 2 , 1 )
For x = t 2 x = t^2 x = t 2 y = t 3 y = t^3 y = t 3 d 2 y d x 2 \tfrac{d^2y}{dx^2} d x 2 d 2 y t = 1 t = 1 t = 1
Answer
3 4 \tfrac{3}{4} 4 3
Full solution
d y d x = 3 t 2 \tfrac{dy}{dx} = \tfrac{3t}{2} d x d y = 2 3 t t t t 3 2 \tfrac{3}{2} 2 3 d x d t = 2 t \tfrac{dx}{dt} = 2t d t d x = 2 t 3 4 t \tfrac{3}{4t} 4 t 3 3 4 \tfrac{3}{4} 4 3 t = 1 t = 1 t = 1
Find the length of x = t 2 x = t^2 x = t 2 y = 2 3 t 3 y = \tfrac{2}{3}t^3 y = 3 2 t 3 0 ≤ t ≤ 1 0 \le t \le 1 0 ≤ t ≤ 1
Answer
2 3 ( 2 2 − 1 ) ≈ 1.22 \tfrac{2}{3}\left(2\sqrt{2} - 1\right) \approx 1.22 3 2 ( 2 2 − 1 ) ≈ 1.22
Full solution
( 2 t ) 2 + ( 2 t 2 ) 2 = 2 t 1 + t 2 \sqrt{(2t)^2 + (2t^2)^2} = 2t\sqrt{1 + t^2} ( 2 t ) 2 + ( 2 t 2 ) 2 = 2 t 1 + t 2 u = 1 + t 2 u = 1 + t^2 u = 1 + t 2 ∫ 0 1 2 t 1 + t 2 d t = 2 3 [ ( 1 + t 2 ) 3 / 2 ] 0 1 = 2 3 ( 2 3 / 2 − 1 ) \int_0^1 2t\sqrt{1 + t^2}\,dt = \tfrac{2}{3}\big[(1 + t^2)^{3/2}\big]_0^1 = \tfrac{2}{3}\left(2^{3/2} - 1\right) ∫ 0 1 2 t 1 + t 2 d t = 3 2 [ ( 1 + t 2 ) 3/2 ] 0 1 = 3 2 ( 2 3/2 − 1 )
Find the length of x = 3 cos t x = 3\cos t x = 3 cos t y = 3 sin t y = 3\sin t y = 3 sin t 0 ≤ t ≤ π 0 \le t \le \pi 0 ≤ t ≤ π
Answer
3 π 3\pi 3 π
Full solution
The integrand is 9 sin 2 t + 9 cos 2 t = 3 \sqrt{9\sin^2 t + 9\cos^2 t} = 3 9 sin 2 t + 9 cos 2 t = 3 ∫ 0 π 3 d t = 3 π \int_0^{\pi} 3\,dt = 3\pi ∫ 0 π 3 d t = 3 π 3 3 3
Find the length of x = e t cos t x = e^t\cos t x = e t cos t y = e t sin t y = e^t\sin t y = e t sin t 0 ≤ t ≤ 1 0 \le t \le 1 0 ≤ t ≤ 1
Answer
2 ( e − 1 ) ≈ 2.43 \sqrt{2}(e - 1) \approx 2.43 2 ( e − 1 ) ≈ 2.43
Full solution
d x d t = e t ( cos t − sin t ) \tfrac{dx}{dt} = e^t(\cos t - \sin t) d t d x = e t ( cos t − sin t ) d y d t = e t ( sin t + cos t ) \tfrac{dy}{dt} = e^t(\sin t + \cos t) d t d y = e t ( sin t + cos t ) e 2 t ( ( cos t − sin t ) 2 + ( sin t + cos t ) 2 ) = 2 e 2 t e^{2t}\left((\cos t - \sin t)^2 + (\sin t + \cos t)^2\right) = 2e^{2t} e 2 t ( ( cos t − sin t ) 2 + ( sin t + cos t ) 2 ) = 2 e 2 t L = ∫ 0 1 2 e t d t = 2 ( e − 1 ) L = \int_0^1 \sqrt{2}\,e^t\,dt = \sqrt{2}(e - 1) L = ∫ 0 1 2 e t d t = 2 ( e − 1 )
The curve x = t 2 x = t^2 x = t 2 y = t 3 − 3 t y = t^3 - 3t y = t 3 − 3 t ( 3 , 0 ) (3, 0) ( 3 , 0 )
Answer
3 \sqrt{3} 3 − 3 -\sqrt{3} − 3
Full solution
( 3 , 0 ) (3, 0) ( 3 , 0 ) t = ± 3 t = \pm\sqrt{3} t = ± 3 d y d x = 3 t 2 − 3 2 t = 6 ± 2 3 = ± 3 \tfrac{dy}{dx} = \tfrac{3t^2 - 3}{2t} = \tfrac{6}{\pm 2\sqrt{3}} = \pm\sqrt{3} d x d y = 2 t 3 t 2 − 3 = ± 2 3 6 = ± 3
A student finds the slope of x = t 2 x = t^2 x = t 2 y = t 3 y = t^3 y = t 3 t = 2 t = 2 t = 2 d x / d t d y / d t = 4 12 = 1 3 \tfrac{dx/dt}{dy/dt} = \tfrac{4}{12} = \tfrac{1}{3} d y / d t d x / d t = 12 4 = 3 1
Hint
Slope is rise over run. Which derivative measures the rise?
Answer
The student divided upside down. The slope is 12 4 = 3 \tfrac{12}{4} = 3 4 12 = 3
Full solution
d y d x = d y / d t d x / d t = 3 t 2 2 t \tfrac{dy}{dx} = \tfrac{dy/dt}{dx/dt} = \tfrac{3t^2}{2t} d x d y = d x / d t d y / d t = 2 t 3 t 2 12 4 = 3 \tfrac{12}{4} = 3 4 12 = 3 t = 2 t = 2 t = 2 1 3 \tfrac{1}{3} 3 1 d x d y \tfrac{dx}{dy} d y d x