Question

40–41. {Use of Tech} Slopes of tangent linesb. Find the slope of the lines tangent to the curve at the origin (when relevant).r = 1 −sin θ

Accepted Answer

Recognize that the given curve is in polar form: $$r = 1 - \sin 	heta. To $$find the slope of the tangent line at a point, we need to convert this polar equation into Cartesian coordinates or use the formula for the slope of a tangent line in polar coordinates. Recall that the slope of the tangent line in Cartesian coordinates is $$\frac{dy}{dx}. $$For polar curves, $$x = r \cos 	heta$$ and $$y = r \sin 	heta. $$Therefore, $$\frac{dy}{dx} = \frac{\frac{dy}{d	heta}}{\frac{dx}{d	heta}}. $$Compute $$\frac{dx}{d	heta}$$ and $$\frac{dy}{d	heta}$$ using the product and chain rules:

$$\frac{dx}{d	heta} = \frac{d}{d	heta}(r \cos 	heta) = \frac{dr}{d	heta} \cos 	heta - r \sin 	heta$$,

$$\frac{dy}{d	heta} = \frac{d}{d	heta}(r \sin 	heta) = \frac{dr}{d	heta} \sin 	heta + r \cos 	heta. $$Find $$\frac{dr}{d	heta} by $$differentiating $$r = 1 - \sin 	heta$$ with respect to $$	heta$$, which gives $$\frac{dr}{d	heta} = -\cos 	heta. $$Evaluate $$\frac{dy}{dx} = \frac{\frac{dy}{d	heta}}{\frac{dx}{d	heta}} at $$the origin. Since the origin corresponds to $$r=0$$, find the value(s) of $$	heta$$ where $$r=0 (i.e$$., solve $$1 - \sin 	heta = 0$$), then substitute those $$	heta$$ values into the expressions for $$\frac{dy}{d	heta}$$ and $$\frac{dx}{d	heta} to $$find the slope(s) of the tangent line(s) at the origin.

40–41. {Use of Tech} Slopes of tangent lines
b. Find the slope of the lines tangent to the curve at the origin (when relevant).
r = 1 −sin θ

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Concetti chiave

Polar Coordinates and Curves

Slope of Tangent Lines in Polar Coordinates

Finding Tangent Lines at the Origin