B. Find the equation of the tangent line. Derived from a function parametrically. 2) x = √t² + 1 y = t4 t = √3

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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段階的に解決し、 人工知能を使用せず、 優れた仕事を行います
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B. Find the equation of the tangent line.
Derived from a function parametrically.
2) x = vt2+1
3) x = Cos x
y=14
y = sin x
t=v3
8=
π
-6
Transcribed Image Text:段階的に解決し、 人工知能を使用せず、 優れた仕事を行います ご支援ありがとうございました SOLVE STEP BY STEP IN DIGITAL FORMAT DON'T USE AI DON'T USE AI DON'T USE AI DON'T USE AI B. Find the equation of the tangent line. Derived from a function parametrically. 2) x = vt2+1 3) x = Cos x y=14 y = sin x t=v3 8= π -6
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According to Bratleby Guidelines we are supposed to answer only fast question. So repost other parts in the next question.

Given that  x equals square root of t squared plus 1 end root space space space a n d space space y equals t to the power of 4 space space space semicolon space t equals square root of 3 space space

when t equals square root of 3 space comma space space x equals square root of 3 plus 1 end root equals square root of 4 equals 2 space space space a n d space space space y equals left parenthesis square root of 3 space right parenthesis to the power of 4 equals 3 squared equals 9

So we have to find the tangent line of the parametric curve at the point left parenthesis 2 comma 9 right parenthesis 

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