The line tangent to the curve at [tex](x(0),y(0))[/tex] has slope
[tex]\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{\theta=0}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}\bigg|_{\theta=0}=\dfrac{\cos\theta-4\sin4\theta}{-\sin\theta+4\cos4\theta}\bigg|_{\theta=0}=\dfrac14[/tex]
Then the tangent line has equation
[tex]y-y(0)=\dfrac14(x-x(0))[/tex]
[tex]y-1=\dfrac14(x-1)[/tex]
[tex]\boxed{y=\dfrac{x+3}4}[/tex]
The equation of the tangent to the curve is [tex](x,y) = (1,1) + \theta\cdot (4,1)[/tex].
Given the parametric equations of a curve, the parametric equation of the line is presented below:
[tex](x,y) = (x(0), y(0)) + \theta \cdot (\Delta x, \Delta y)[/tex] (1)
[tex]m = \frac{\frac{dy}{d\theta} }{\frac{dx}{d\theta} } = \frac{\Delta y}{\Delta x }[/tex] (2)
Where:
The first derivatives of the curve are described below:
[tex]\frac{dx}{d\theta} = -\sin \theta + 4\cdot \cos 4\theta[/tex] (3)
[tex]\frac{dy}{d\theta} = \cos \theta -4\cdot \sin 4\theta[/tex] (4)
If we know that [tex]\theta = 0[/tex], then we have the following expression:
By (3) and (4):
[tex]\frac{dx}{d\theta} = -\sin 0 + 4\cdot \cos (4\cdot 0)[/tex]
[tex]\frac{dx}{d\theta} = 4[/tex]
[tex]\frac{dy}{d\theta} = \cos 0 - 4\cdot \sin (4\cdot 0)[/tex]
[tex]\frac{dy}{d\theta} = 1[/tex]
[tex]m = \frac{1}{4}[/tex]
[tex]\Delta x = 4, \Delta y = 1[/tex]
[tex]x(0) = \cos 0 + \sin (4\cdot 0)[/tex]
[tex]x(0) = 1[/tex]
[tex]y(0) = \sin 0 + \cos (4\cdot 0)[/tex]
[tex]y(0) = 1[/tex]
Hence, the equation of the tangent to the curve is:
[tex](x,y) = (1,1) + \theta\cdot (4,1)[/tex]
We kindly invite to check this question on tangent lines: https://brainly.com/question/15585522
