Calculus problem #2154

Equation of the given curve is $x=cos^3 \theta, y=sin^3 \theta$

Now, $x=cos^3 \theta$ Differentiating both sides w.r.t $x$ , $\frac{dx}{d \theta} = -3sin \theta cos^2 \theta$ And, $y=sin^3 \theta$ Differentiating both sides w.r.t $x$ , $\frac{dy}{d \theta} = 3cos \theta sin^2 \theta$ So, $\frac{dy}{dx} = \frac{ \frac{dy}{d \theta}}{ \frac{dx}{d \theta}}$ or, $\frac{dy}{dx} = - \frac{3cos \theta sin^2 \theta}{-3sin \theta cos^2 \theta}$ or, $\frac{dy}{dx} = -tan \theta$

This is the slope of the tangent to the curve at any arbitrary point $(x, y)$ .

When $\theta = \frac{2 \pi}{3}$ , Slope of tangent = $-tan \theta=-tan \frac{2 \pi}{3}=-(- \sqrt{3})= \sqrt{3}$

So, the equation of the tangent at $(x, y)$ using the point-slope form is given by $y-sin^3 \theta= \sqrt{3}(x-cos^3 \theta)$ Upon simplification, we get $y= \sqrt{3}x+sin^3 \theta- \sqrt{3}cos^3 \theta$ But, $\theta = \frac{2 \pi}{3}$ . So, the equation of tangent becomes $y= \sqrt{3}x+sin^3 \frac{2 \pi}{3}- \sqrt{3}cos^3 \frac{2 \pi}{3}$

Therefore, $a= \sqrt{3}$ and $b=sin^3 \frac{2 \pi}{3}- \sqrt{3}cos^3 \frac{2 \pi}{3}$ $b$ can be further simplified as $b= \frac{3 \sqrt{3}}{8}+ \frac{ \sqrt{3}}{8}= \frac{ \sqrt{3}+3 \sqrt{3}}{8}= \frac{4 \sqrt{3}}{8}= \frac{ \sqrt{3}}{2}$

So, finally we get $a= \sqrt{3}$ and $b=\frac{ \sqrt{3}}{2}$

Therefore, $a \times b =\sqrt{3} \times \frac{ \sqrt{3}}{2}= \frac{3}{2} = \boxed{1.5}$

First, we notice that at $\theta = 2\pi/3$ we have $$(x,y)=\left(-\frac{1}{8},\frac{3\sqrt{3}}{8}\right).$$ To find the tangent line, we must find its slope, or $\frac{dy}{dx}$ . $$\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{3 \,\sin^2 \theta \, \cos \theta}{-3 \, \cos^2 \theta \, \sin \theta}=-\tan \theta.$$ So, at $\theta=2\pi/3$ , we have $$\frac{dy}{dx} = \sqrt{3}.$$ Using the point-slope formula, the equation for the tangent line is $$y- \frac{3 \sqrt{3}}{8} = \sqrt{3} \left(x+\frac{1}{8}\right).$$ Rearranging, we get $$y=\sqrt{3}x+\frac{\sqrt{3}}{2},$$ so $a=\sqrt{3}$ and $b=\frac{\sqrt{3}}{2}$ and their product is $$a \times b = \frac{3}{2} = \boxed{1.5}.$$

the gradient of the normal to the point at $\Theta =(2/3)\Pi$ is given by : $gradient=a=\frac{(y_{2}-y_{1})}{(x_{2}-x_{1})}$ with points : $( x_{1},y_{1})=(cos^3(\frac{2\Pi }{3}),sin^3(\frac{2\Pi }{3}))=(\frac{-1}{8},\frac{3\sqrt3}{8})$ and : $(x_{2},y_{2})=(cos^3(0),sin^3(0))=(1,0)$ therefore, $a=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{\frac{3\sqrt3}{8}-0}{\frac{-1}{8}-1}=\frac{-\sqrt3}{3}$ Now , the gradient of the tangent at the point $\Theta =(2/3)\Pi$ is : $a_{tangent}=\frac{-1}{a_{normal}}=\frac{-1}{\frac{-\sqrt3}{3}}=\sqrt3$ Now we have that $y=ax+b$ which is : $sin^3(\Theta )=\sqrt3 cos^3(\Theta )+b$ therefore , $b=sin^3(\Theta )-\sqrt cos^3(\Theta )$ now putting $\Theta =\frac{2\Pi }{3}$ gives : $b=sin^3(\frac{2\Pi }{3})-\sqrt3 cos^3(\frac{2\Pi }{3})=\frac{\sqrt3}{2}$ $a*b=\sqrt3*\frac{\sqrt3}{2}=\frac{3}{2}=1.5$