# What is the equation of the tangent line of #r=tan^2theta - sintheta# at #theta=pi/4#?

0.1197y-0.2143x-0.0103=0.

I have given my best answer. For this purpose, I had made self-corrections,

periodically.

The two infinitely long vertical loops touch here.

graph{(x^2(x^2+y^2)-y^2sqrt(x^2+y^2)-x^2y)((x-.207)^2+(y-.207)^2-.02)=0 [-10, 10, -5, 5]}

and is given by,

Here, the point of contact of the tangent P is

So, the equation to the tangent is

rsin(theta-0.8740)=0.2929sin(0.7854-0.8740)#

rsin(theta^o-50.08^o)=-0.0259#

Expanding and converting to Cartesian form, this is

0.6417y-0.7889x+0.0259=0.

The Socratic twin graph is for the Cartesian frame.

See the graph below.

graph{(x^2(x^2+y^2)-y^2sqrt(x^2+y^2)-x^2y)(0.6417y-0.7889x+0.0259)((x--.207)^2+(y-.207)^2-.001)=0 [-139.4, 139.1, -66.7, 72.5]}

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To find the equation of the tangent line of ( r = \tan^2(\theta) - \sin(\theta) ) at ( \theta = \frac{\pi}{4} ), we first find the derivative of ( r ) with respect to ( \theta ), then evaluate it at ( \theta = \frac{\pi}{4} ).

The derivative ( \frac{dr}{d\theta} ) is given by ( \frac{d}{d\theta}(\tan^2(\theta) - \sin(\theta)) ).

Differentiating ( \tan^2(\theta) ) with respect to ( \theta ) using the chain rule yields ( 2\tan(\theta) \sec^2(\theta) ).

Differentiating ( -\sin(\theta) ) with respect to ( \theta ) gives ( -\cos(\theta) ).

So, ( \frac{dr}{d\theta} = 2\tan(\theta)\sec^2(\theta) - \cos(\theta) ).

Evaluate ( \frac{dr}{d\theta} ) at ( \theta = \frac{\pi}{4} ) to get ( \frac{dr}{d\theta} = 2\tan(\frac{\pi}{4})\sec^2(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = 2(1)(2) - \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} ).

Now, we need the slope of the tangent line, which is ( \frac{dr}{d\theta} ) evaluated at ( \theta = \frac{\pi}{4} ), so ( \frac{3\sqrt{2}}{2} ).

To find the point on the curve where the tangent line touches, plug ( \theta = \frac{\pi}{4} ) into ( r = \tan^2(\theta) - \sin(\theta) ) to get ( r = 1 - \frac{\sqrt{2}}{2} ).

Using the point-slope form of a line ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is a point on the line and ( m ) is the slope, and substituting ( (x_1, y_1) = (\frac{\pi}{4}, 1 - \frac{\sqrt{2}}{2}) ) and ( m = \frac{3\sqrt{2}}{2} ), the equation of the tangent line is ( r - (1 - \frac{\sqrt{2}}{2}) = \frac{3\sqrt{2}}{2}(\theta - \frac{\pi}{4}) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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