Practicing Success
Find the slope of the normal to the curve x=a\( { cos}^{3 }\theta \), y=\(a { sin}^{3 }\theta \) at $\theta=\frac{\pi}{4}$ |
1/2 1 2/3 3/2 |
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Slope of the tamgent of the curve is dy/dx $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ $sin^3\theta = \frac{3sin\theta - sin3\theta}{4}$ $ \frac{dy}{d\theta} = a(\frac{3cos\theta - 3cos3\theta}{4})$ $cos^3\theta = \frac{3cos\theta + cos3\theta}{4}$ $ \frac{dx}{d\theta} =a( -\frac{3sin\theta - 3sin3\theta}{4})$ $\frac{dy}{dx} = \frac{3cos\theta - 3cos3\theta}{3sin\theta - 3sin3\theta}$ At $\theta = \frac{\pi}{4} $ $\text{Slope of the normal is }-\frac{dx}{dy} = -\frac{3sin\theta - 3sin3\theta}{3cos\theta - 3cos3\theta}=1$
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