Wize University Calculus 2 Textbook > Parametric & Polar Curves

Calculus w/ Parametric Curves

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Tangents of Parametric Curves (Derivatives)
Consider a parametric curve with equations (x(t), y(t))\left(x\left(t\right),\ y\left(t\right)\right)(x(t), y(t)) for a≤t≤ba\le t\le ba≤t≤b.
First Derivative
(a.k.a. slope of the tangent to the curve)
dydx=y′(t)x′(t)\frac{dy}{dx}=\frac{y'\left(t\right)}{x'\left(t\right)}dxdy​=x′(t)y′(t)​ if x′(t)≠0x'\left(t\right)\ne0x′(t)=0
Find the derivatives of y and x in terms of t
Divide the two expressions
Second Derivative
d2ydx2=ddx(dydx)=ddt(dydx)x′(t)\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{x'\left(t\right)}dx2d2y​=dxd​(dxdy​)=x′(t)dtd​(dxdy​)​
Find the derivative of the first derivative dydx\frac{dy}{dx}dxdy​ in terms of t
Find the derivative of x in terms of t
Divide the two expressions

The parametric curve has 
A Horizontal Tangent when y′(t)=0y'\left(t\right)=0y′(t)=0 and x′(t)≠0x'\left(t\right)\ne0x′(t)=0
A Vertical Tangent when x′(t)=0x'\left(t\right)=0x′(t)=0 and y′(t)≠0y'\left(t\right)\ne0y′(t)=0
*If both y′(t)=0 y'\left(t\right)=0\ y′(t)=0  and x′(t)=0x'\left(t\right)=0x′(t)=0, use L'Hospital's rule

Practice Question
If x and y are described with parametric equations x=sin⁡t+cos⁡tx = \sin t + \cos t x=sint+costand y=tan⁡ty = \tan ty=tant, find   ⁣dy  ⁣dx\frac{\de{y}}{ \de{x}}  dx dy​at t=π/3 t = \pi/3t=π/3.
  

21−3\frac{2}{1-\sqrt{3}}1−3​2​

21+23\frac{2}{1+2\sqrt{3}}1+23​2​

42−3\frac{4}{2-\sqrt{3}}2−3​4​

81−3\frac{8}{1-\sqrt{3}}1−3​8​

81+3\frac{8}{1+\sqrt{3}}1+3​8​

I don't know

Practice: Tangents
Given the parametric curves x=t2+1,  y=t3−3t,  t∈I ⁣R.x = t^2 + 1, \space \space y = t^3 - 3t,\space \space  t \in \R. x=t2+1,  y=t3−3t,  t∈IR.,
a.) Find the equation of the tangent line to the curve at the point (x, y)=(5, 2)\left(x,\ y\right)=\left(5,\ 2\right)(x, y)=(5, 2)
b.) Identify all points where the tangent is horizontal
c.) Identify all points where the tangent is vertical

a.) y−2=94(x−5)y-2=\frac{9}{4}\left(x-5\right)y−2=49​(x−5)
b.) (2, 2)\left(2,\ 2\right)(2, 2) and (2, −2)\left(2,\ -2\right)(2, −2)
c.) (1, 0)\left(1,\ 0\right)(1, 0) and (−1, 0)\left(-1,\ 0\right)(−1, 0)

a.) y−2=94(x−5)y-2=\frac{9}{4}\left(x-5\right)y−2=49​(x−5)
b.) (2, 2)\left(2,\ 2\right)(2, 2) and (2, −2)\left(2,\ -2\right)(2, −2)
c.) (1, 0)\left(1,\ 0\right)(1, 0)

a.) y−2=14(x−5)y-2=\frac{1}{4}\left(x-5\right)y−2=41​(x−5)
b.) (2, 2)\left(2,\ 2\right)(2, 2)
c.) (1, 0)\left(1,\ 0\right)(1, 0)

a.) y−2=14(x−5)y-2=\frac{1}{4}\left(x-5\right)y−2=41​(x−5)
b.) (2, 2)\left(2,\ 2\right)(2, 2) and (2, −2)\left(2,\ -2\right)(2, −2)
c.) (1, 0)\left(1,\ 0\right)(1, 0) and (−1, 0)\left(-1,\ 0\right)(−1, 0)

a.) y−2=15(x−5)y-2=\frac{1}{5}\left(x-5\right)y−2=51​(x−5)
b.) (0, 0)\left(0,\ 0\right)(0, 0)
c.) (0, 1)\left(0,\ 1\right)(0, 1)

I don't know

Practice Question
Given the cycloid x=2(θ−sin⁡θ)x=2\left(\theta-\sin\theta\right)x=2(θ−sinθ) and y=2(1−cos⁡θ)y=2\left(1-\cos\theta\right)y=2(1−cosθ), find all points where the slope of the tangent is horizontal or vertical

Horizontal tangent at points: (2nπ, 4)\left(2n\pi,\ 4\right)(2nπ, 4)
Vertical tangent at points: (2nπ, 0)\left(2n\pi,\ 0\right)(2nπ, 0)

Horizontal tangent at points: (4nπ, 4)\left(4n\pi,\ 4\right)(4nπ, 4)
Vertical tangent at points: (2nπ, 0)\left(2n\pi,\ 0\right)(2nπ, 0)

Horizontal tangent at points: (2nπ, 4)\left(2n\pi,\ 4\right)(2nπ, 4)
Vertical tangent at points: (4nπ, 0)\left(4n\pi,\ 0\right)(4nπ, 0)

Horizontal tangent at points: ((2n+1)π, 4)\left(\left(2n+1\right)\pi,\ 4\right)((2n+1)π, 4)
Vertical tangent at points: (4nπ, 0)\left(4n\pi,\ 0\right)(4nπ, 0)

Horizontal tangent at points: (2(2n+1)π, 4)\left(2\left(2n+1\right)\pi,\ 4\right)(2(2n+1)π, 4)
Vertical tangent at points: (4nπ, 0)\left(4n\pi,\ 0\right)(4nπ, 0)

I don't know

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Applications of Parametric Curves
Consider the parametric curve defined by {x(t), y(t): t∈[a,b]}\left\{x\left(t\right),\ y\left(t\right):\ t\in\left[a,b\right]\right\}{x(t), y(t): t∈[a,b]} where x′(t)x'\left(t\right)x′(t) and y′(t)y'\left(t\right)y′(t) are continuous on [a,b]\left[a,b\right][a,b], and the curve is traversed exactly once on [a,b]\left[a,b\right][a,b].
Arc Length
L=∫ab[x′(t)]2 + [y′(t)]2  ⁣dtL={\displaystyle\int_a^b}\sqrt{
\left[x'(t)\right]^2\ + \ 
\left[y'(t)\right]^2}\de{t}L=∫ab​[x′(t)]2 + [y′(t)]2​ dt

Area under the Curve
A=∫aby(t)x′(t)dtA=
{\displaystyle \int_a^b}
y\left(t\right)x'\left(t\right)dtA=∫ab​y(t)x′(t)dt

Surface Area
Obtained by revolving the curve about the x-axis (or al ine parallel to the x-axis):
S=∫ab2πy(t)[x′(t)]2+[y′(t)]2dtS=
{\displaystyle \int_a^b}
2\pi y\left(t\right)\sqrt{\left[x'\left(t\right)\right]^2+\left[y'\left(t\right)\right]^2}dtS=∫ab​2πy(t)[x′(t)]2+[y′(t)]2​dt
 

Practice Question
Write down the integral representing the length of the curve parameterized by x=t3+t+1,y=t2−1x = t^3+t+1, y = t^2 - 1x=t3+t+1,y=t2−1, from the point (1, -1) to the point (3, 0). 
  

∫−109t4+10t2+1  ⁣dt{\displaystyle\int}_{-1}^0\sqrt{9t^4+10t^2+1}\de{t}∫−10​9t4+10t2+1​ dt

∫019t4+10t2+1  ⁣dt{\displaystyle\int}^1_0\sqrt{9t^4+10t^2+1}\de{t}∫01​9t4+10t2+1​ dt

∫139t4+10t2+1  ⁣dt{\displaystyle\int}_{1}^3\sqrt{9t^4+10t^2+1}\de{t}∫13​9t4+10t2+1​ dt

∫−103t2+2t+1  ⁣dt{\displaystyle\int}_{-1}^0\sqrt{3t^2+2t+1}\de{t}∫−10​3t2+2t+1​ dt

∫013t2+2t+1  ⁣dt{\displaystyle\int}_{0}^1\sqrt{3t^2+2t+1}\de{t}∫01​3t2+2t+1​ dt

I don't know

Practice Question
Find the area under the cruve {x=t2, y=cos⁡t: t∈[0,π2]}\left\{x=t^2,\ y=\cos t:\ t\in\left[0,\frac{\pi}{2}\right]\right\}{x=t2, y=cost: t∈[0,2π​]}.

Practice: Surface Area
Find the area of the surface obtained by revolving the curve (x(t), y(t))=(2−t3, 3t)\left(x\left(t\right),\ y\left(t\right)\right)=\left(2-t^3,\ 3t\right)(x(t), y(t))=(2−t3, 3t) on t∈[0,2]t\in\left[0,2\right]t∈[0,2] about the x-axis.

3π2[417+ln⁡(4+17)]\frac{3\pi}{2}\left[4\sqrt{17}+\ln\left(4+\sqrt{17}\right)\right]23π​[417​+ln(4+17​)]

9π2[417+ln⁡(4+17)]\frac{9\pi}{2}\left[4\sqrt{17}+\ln\left(4+\sqrt{17}\right)\right]29π​[417​+ln(4+17​)]

9π[417+ln⁡(4+17)]9\pi\left[4\sqrt{17}+\ln\left(4+\sqrt{17}\right)\right]9π[417​+ln(4+17​)]

9π[417−ln⁡(4+17)]9\pi\left[4\sqrt{17}-\ln\left(4+\sqrt{17}\right)\right]9π[417​−ln(4+17​)]

I don't know

Practice Question
Find the integral that represents the area of the surface of revolution obtained by rotating the curve defined by {x=6ln⁡t, y=3t+3t: t∈[1,e]}\left\{x=6\ln t,\ y=3t+\frac{3}{t}:\ t\in\left[1,e\right]\right\}{x=6lnt, y=3t+t3​: t∈[1,e]} about the x-axis.

∫1e2π(3t+3t)9t4+18t2+9 dt{\displaystyle \int_1^e}
2\pi(3t+\frac{3}{t})

\sqrt{\frac{9}{t^4}+\frac{18}{t^2}+9}\ dt∫1e​2π(3t+t3​)t49​+t218​+9​ dt

∫1e2π(3t+3t)6t+3−3t2 dt{\displaystyle \int_1^e}
2\pi(3t+\frac{3}{t})

\sqrt{\frac{6}{t}+3-\frac{3}{t^2}}\ dt∫1e​2π(3t+t3​)t6​+3−t23​​ dt

∫1e2π(3t+3t)6t−3+3t2 dt{\displaystyle \int_1^e}
2\pi(3t+\frac{3}{t})

\sqrt{\frac{6}{t}-3+\frac{3}{t^2}}\ dt∫1e​2π(3t+t3​)t6​−3+t23​​ dt

∫1e2π(6ln⁡t)9t4+18t2+9 dt{\displaystyle \int_1^e}
2\pi(6\ln t)

\sqrt{\frac{9}{t^4}+\frac{18}{t^2}+9}\ dt∫1e​2π(6lnt)t49​+t218​+9​ dt

∫1e2π(6ln⁡t)54t2−9t4−9 dt{\displaystyle \int_1^e}
2\pi(6\ln t)

\sqrt{\frac{54}{t^2}-\frac{9}{t^4}-9}\ dt∫1e​2π(6lnt)t254​−t49​−9​ dt

I don't know

Extra Practice

Practice: Area Under the Parametric Curve

Find the area under the curve {x=t2, y=cos⁡t: t∈[0,π2]}\left\{x=t^2,\ y=\cos t:\ t\in\left[0,\frac{\pi}{2}\right]\right\}{x=t2, y=cost: t∈[0,2π​]}.

Practice: Area Under the Parametric Curve

Practice Question
Find the area under the cruve {x=t2, y=cos⁡t: t∈[0,π2]}\left\{x=t^2,\ y=\cos t:\ t\in\left[0,\frac{\pi}{2}\right]\right\}{x=t2, y=cost: t∈[0,2π​]}.

Practice: Area Under the Parametric Curve

Practice Question
Write down the integral representing the area under the curve parameterized by x=2tan⁡tx = 2\tan tx=2tant, y=2ty = 2ty=2t, for 0≤t≤π/40 \le t \le \pi/40≤t≤π/4. 
  

Practice: Area Under the Parametric Curve

Practice Question
Find the area under the cruve {x=t2, y=cos⁡t: t∈[0,π2]}\left\{x=t^2,\ y=\cos t:\ t\in\left[0,\frac{\pi}{2}\right]\right\}{x=t2, y=cost: t∈[0,2π​]}.

⭐ TRICKY! Find, if possible, the maximum and minimum values of the curvature of the graph of y = ex  and the points at which they occur. 

Wize University Calculus 2 Textbook > Parametric & Polar Curves

Calculus w/ Parametric Curves

Popular Courses

Tangents of Parametric Curves (Derivatives)

First Derivative

Second Derivative

Practice Question

Practice: Tangents

Practice Question

Applications of Parametric Curves

Arc Length

$L={\displaystyle\int_a^b}\sqrt{ \left[x'(t)\right]^2\ + \ \left[y'(t)\right]^2}\de{t}$

Area under the Curve

$A= {\displaystyle \int_a^b} y\left(t\right)x'\left(t\right)dt$

Surface Area

$S= {\displaystyle \int_a^b} 2\pi y\left(t\right)\sqrt{\left[x'\left(t\right)\right]^2+\left[y'\left(t\right)\right]^2}dt$

Practice Question

Practice Question

Practice: Surface Area

Practice Question

Extra Practice

Practice: Area Under the Parametric Curve

Practice: Area Under the Parametric Curve

Practice: Area Under the Parametric Curve

Practice: Area Under the Parametric Curve

Wize University Calculus 2 Textbook > Parametric & Polar Curves

Calculus w/ Parametric Curves

Popular Courses

Tangents of Parametric Curves (Derivatives)

First Derivative

Second Derivative

Practice Question

Practice: Tangents

Practice Question

Applications of Parametric Curves

Arc Length

L=∫ab[x′(t)]2 + [y′(t)]2 ⁣dtL={\displaystyle\int_a^b}\sqrt{ \left[x'(t)\right]^2\ + \ \left[y'(t)\right]^2}\de{t}L=∫ab​[x′(t)]2 + [y′(t)]2​ dt

Area under the Curve

A=∫aby(t)x′(t)dtA= {\displaystyle \int_a^b} y\left(t\right)x'\left(t\right)dtA=∫ab​y(t)x′(t)dt

Surface Area

S=∫ab2πy(t)[x′(t)]2+[y′(t)]2dtS= {\displaystyle \int_a^b} 2\pi y\left(t\right)\sqrt{\left[x'\left(t\right)\right]^2+\left[y'\left(t\right)\right]^2}dtS=∫ab​2πy(t)[x′(t)]2+[y′(t)]2​dt

Practice Question

Practice Question

Practice: Surface Area

Practice Question

Extra Practice

Practice: Area Under the Parametric Curve

Practice: Area Under the Parametric Curve

Practice: Area Under the Parametric Curve

Practice: Area Under the Parametric Curve

$L={\displaystyle\int_a^b}\sqrt{ \left[x'(t)\right]^2\ + \ \left[y'(t)\right]^2}\de{t}$

$A= {\displaystyle \int_a^b} y\left(t\right)x'\left(t\right)dt$

$S= {\displaystyle \int_a^b} 2\pi y\left(t\right)\sqrt{\left[x'\left(t\right)\right]^2+\left[y'\left(t\right)\right]^2}dt$