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Sagot :
The slope of the line tangent to the polar curve [tex]\rm r =3sin\theta[/tex] is 1.79.
What is the slope?
A surface on which one end or side is at a higher level than another; a rising or falling surface.
The slope of the line tangent to the polar curve;
[tex]\rm r =3sin\theta[/tex]
The slope of the tangent is given by the following formula;
[tex]\rm \text{Slope of tangent line} = \dfrac{dy}{dx} =\dfrac{ (\dfrac{dr}{d\theta}sin\theta + cos\theta)}{(\dfrac{dr}{d\theta}cos\theta-rsin\theta)}[/tex]
Differentiate the given polar curve
[tex]\rm r =3sin\theta\\\\\dfrac{dr}{d\theta}=3cos\theta\\\\[/tex]
Substitute all the values in the formula
[tex]\rm \text{Slope of tangent line} = \dfrac{dy}{dx} =\dfrac{ (\dfrac{dr}{d\theta}sin\theta + rcos\theta)}{(\dfrac{dr}{d\theta}cos\theta-rsin\theta)}\\\\\rm \text{Slope of tangent line} = \dfrac{dy}{dx} =\dfrac{ (3cos\theta \times sin\theta +3 sin \theta \times cos\theta)}{(3cos\theta \times cos\theta-3 sin \theta \times sin\theta)}\\\\\rm \text{Slope of tangent line} = \dfrac{dy}{dx} =\dfrac{3 (cos\theta sin\theta + sin \theta cos\theta)}{3(cos\theta cos\theta- sin \theta sin\theta)}\\\\[/tex][tex]\rm \text{Slope of tangent line} = \dfrac{dy}{dx} =\dfrac{ 2sin\theta cos\theta}{(cos^2\theta -sin^2 \theta)}[/tex]
Substitute the value of [tex]\theta[/tex] in the equation
[tex]\rm \text{Slope of tangent line} = \dfrac{dy}{dx} =\dfrac{ 2sin \dfrac{\pi }{6} cos\dfrac{\pi }{6} }{(cos^2\dfrac{\pi }{6} -sin^2 \dfrac{\pi }{6} )}\\\\\text{Slope of tangent line} = \dfrac{dy}{dx} =\dfrac{2 \times 0.5 \times 0.86}{0.86^2 -0.5^2}\\\\\text{Slope of tangent line} = \dfrac{dy}{dx} =\dfrac{0.86}{0.73- 0.25}\\\\\text{Slope of tangent line} = \dfrac{dy}{dx} =\dfrac{0.86}{0.48}\\\\\text{Slope of tangent line} = \dfrac{dy}{dx} =1.79[/tex]
Hence, the slope of the line tangent to the polar curve [tex]\rm r =3sin\theta[/tex] is 1.79.
Learn more about slopes here;
https://brainly.com/question/1593553
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