Lena772
(a) find an equation of the tangent line to
the graph of at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your
results.
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Lena772
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function: y= cos 3x
point: (pi/4, -sqrt2/2)
ganeshie8
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where wer u stuck ?
ganeshie8
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i saw u doing a similar problem the other day ? :)
Decart
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take the derivative
Lena772
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I just get confused when trig is added in
Decart
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do you know what the derivative is
Lena772
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dy/dx
Decart
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-3sin3x
Decart
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right
Lena772
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yes
ganeshie8
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so, slope = m = dy/dx = -3sin3x
ganeshie8
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since u want the slope at
point: (pi/4, -sqrt2/2)
evaluate slope at x = pi/4
ganeshie8
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dy/dx at x = pi/4 :-
-3sin(3*pi/4)
= ?
Lena772
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-0.123
Lena772
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@ganeshie8
ganeshie8
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that doesnt look right
you should get :-
dy/dx at x = pi/4 :-
-3sin(3*pi/4)
= \(\large \frac{-3}{\sqrt{2}}\)
ganeshie8
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now that u knw slope, m = \(\large \frac{-3}{\sqrt{2}}\)
point: (pi/4, -sqrt2/2)
wats the equation of tangent in point slope form ?
Lena772
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y-(-sqrt2/3)=(-3/sqrt2)(x-(pi/4)
Lena772
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?
Lena772
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@hartnn
ganeshie8
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thats right !
Lena772
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So that's for a? What would the graph look like for b? Do I just put that equation into Geogebra?
Lena772
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@ganeshie8
ganeshie8
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yup
graph all 3 below :-
1) function: y= cos 3x
2) point: (pi/4, -sqrt2/2)
3) tangent : y-(-sqrt2/\(\color{red}{2}\))=(-3/sqrt2)(x-(pi/4)
ganeshie8
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u should get exactly like the one attached