What is the equation of the tangent line of #r=cos^2(theta-pi) +sin^2(theta-3pi)-theta# at #theta=(-13pi)/4#?

Answer 1

#y = frac{8+13pi}{13pi}x + frac{(4+13pi)^2}{26sqrt2pi}#

#~~ (1.196)x + 17.406#

First, simplify #r#.

#r = cos^2(theta-pi) + sin^2(theta-3pi) - theta#

#= cos^2(theta-pi) + sin^2(theta-pi) - theta#

#= 1 - theta#

When #theta = -{13pi}/4#, #r = 1 + {13pi}/4#.

In cartesian coordinates, that would be

#x = rcostheta#

#= (1+{13pi}/4) * (-1/sqrt2)#
#~~ -7.927#

#y = rsintheta#

#= (1+{13pi}/4) * (1/sqrt2)#
#~~ 7.927#

The tangent line has to pass through the point #(-7.927,7.927)#.

The gradient of the tangent line is equal to #frac{"d"y}{"d"x}# at #theta = -{13pi}/4#. To find #frac{"d"y}{"d"x}#, we use the chain rule and the product rule.

#frac{"d"y}{"d"x} = frac{frac{"d"y}{"d" theta}}{frac{"d"x}{"d" theta}}#

#= frac{frac{"d"}{"d" theta}(rsintheta)}{frac{"d"}{"d" theta}(rcostheta)}#

#= frac{sinthetafrac{"d"}{"d" theta}(r)+rfrac{"d"}{"d" theta}(sintheta)}{costhetafrac{"d"}{"d" theta}(r)+rfrac{"d"}{"d" theta}(costheta)}#

#= frac{sinthetafrac{"d"r}{"d" theta}+rcostheta}{costhetafrac{"d"r}{"d" theta}-rsintheta}#

Next, we find #frac{"d"r}{"d"theta}#.

#frac{"d"r}{"d"theta} = frac{"d"}{"d" theta}(1-theta)#

#= -1#

Therefore,

#frac{"d"y}{"d"x} = frac{sintheta(-1)+(1 - theta)costheta}{costheta(-1)-(1 - theta)sintheta}#

#= frac{-sintheta+costheta-thetacostheta}{-costheta-sintheta+thetasintheta}#.

So, now we can find #frac{"d"y}{"d"x}# at #theta = -{13pi}/4#.

#frac{"d"y}{"d"x}|_{theta = -{13pi}/4} #

#= frac{-sin(-{13pi}/4)+cos(-{13pi}/4)-(-{13pi}/4)cos(-{13pi}/4)}{-cos(-{13pi}/4)-sin(-{13pi}/4)+(-{13pi}/4)sin(-{13pi}/4)}#

#= frac{sin({13pi}/4)+cos({13pi}/4)+({13pi}/4)cos({13pi}/4)}{-cos({13pi}/4)+sin({13pi}/4)+({13pi}/4)sin({13pi}/4)}#

#= frac{(-1/sqrt2)+(-1/sqrt2)+({13pi}/4)(-1/sqrt2)}{-(-1/sqrt2)+(-1/sqrt2)+({13pi}/4)(-1/sqrt2)}#

#= frac{1+1+({13pi}/4)(1)}{-(1)+(1)+({13pi}/4)(1)}#

#= frac{2+{13pi}/4}{{13pi}/4}#

#= frac{8}{13pi}+1#

#~~= 1.196#

So now that we have the gradient of the line and one point that it passes through, we can write the equation of the line in point-slope form.

#y - ({4+13pi}/{4sqrt2}) = frac{8+13pi}{13pi}(x - (-{4+13pi}/{4sqrt2}))#

Now we rearrange the terms to get the slope-intercept form.

#y = frac{8+13pi}{13pi}x + (frac{8+13pi}{13pi}+1) {4+13pi}/{4sqrt2}#

#= frac{8+13pi}{13pi}x + frac{(4+13pi)^2}{26sqrt2pi}#

#~~ (1.196)x + 17.406#

Below is a graph for additional reference

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Answer 2

The equation of the tangent line of ( r = \cos^2(\theta - \pi) + \sin^2(\theta - 3\pi) - \theta ) at ( \theta = \frac{-13\pi}{4} ) is ( r = -\frac{1}{2}\theta - \frac{3\sqrt{2}}{2} ).

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Answer from HIX Tutor

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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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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