17Calculus - Vector Lines-Planes Application - Normals & Tangent Planes
17Calculus
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We will be adding more discussion here soon. In the meantime, enjoy these practice problems.
Practice
Find the equation of the tangent plane and the symmetric equations of the normal line to the surface \( 2(x-2)^2 + (y-1)^2 + (x-3)^2 = 10 \) at the point \( (3,3,5) \).
Problem Statement |
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Find the equation of the tangent plane and the symmetric equations of the normal line to the surface \( 2(x-2)^2 + (y-1)^2 + (x-3)^2 = 10 \) at the point \( (3,3,5) \).
Solution |
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1482 video
video by Krista King Math |
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Determine the equation of the line that passes through the point \( (1,-1,1) \) and is normal to the plane \( 2x + 3y - z = 4 \).
Problem Statement |
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Determine the equation of the line that passes through the point \( (1,-1,1) \) and is normal to the plane \( 2x + 3y - z = 4 \).
Solution |
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1767 video
video by Dr Chris Tisdell |
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Determine a normal vector and the equation of the tangent plane to the surface \( z = x^2 + 2y^2 \) at the point \( A(2,-1,6) \).
Problem Statement |
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Determine a normal vector and the equation of the tangent plane to the surface \( z = x^2 + 2y^2 \) at the point \( A(2,-1,6) \).
Solution |
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1828 video
video by Dr Chris Tisdell |
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Find the tangent plane to the surface \( x=u^2, y=u-v^2, z=v^2 \) for \( u,v \geq 0 \) at the point \((1,0,1)\).
Problem Statement |
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Find the tangent plane to the surface \( x=u^2, y=u-v^2, z=v^2 \) for \( u,v \geq 0 \) at the point \((1,0,1)\).
Solution |
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2529 video
video by Michael Hutchings |
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Really UNDERSTAND Calculus
Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
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Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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