Respuesta :
Answer:
[tex]\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )[/tex].
Step-by-step explanation:
We need to find all the points on the curve x=4cos(t),y=4sin(t) that have the slope of 1/2.
[tex]x=4cos (t)[/tex]
[tex]\dfrac{dx}{dt}=-4sin (t)[/tex]
[tex]y=4sin (t)[/tex]
[tex]\dfrac{dy}{dt}=4cos (t)[/tex]
Now,
[tex]\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}[/tex]
[tex]\dfrac{dy}{dx}=4cos (t)\times \dfrac{1}{-4sin (t)}[/tex]
[tex]\dfrac{dy}{dx}=-\cot t[/tex]
So, slope of the curve is [tex]-\cot t[/tex].
[tex]-\cot t=\dfrac{1}{2}[/tex]
[tex]-\tan t=2[/tex]
[tex]\tan t=-2[/tex] ...(1)
Using [tex]\sec^2t=1+\tan^2t[/tex], we get
[tex]\sec^2t=1+(-2)^2[/tex]
[tex]\sec^2t=1+4[/tex]
[tex]\sec t=\pm \sqrt{5}[/tex]
[tex]\cos t=\pm \dfrac{1}{\sqrt{5}}[/tex]
Now,
[tex]\sin^2t=1-cos^2t[/tex]
[tex]\sin t=\pm \sqrt{1-\dfrac{1}{5}}[/tex]
[tex]\sin t=\pm \sqrt{\dfrac{4}{5}}[/tex]
[tex]\sin t=\pm \dfrac{2}{\sqrt{5}}[/tex]
It equation (1), tan(t) is negative. So, sin and cos have different signs.
If [tex]\sin t= \dfrac{2}{\sqrt{5}}[/tex], then [tex]\cos t=- \dfrac{1}{\sqrt{5}}[/tex].
[tex]x=4cos (t)=-\dfrac{4}{\sqrt{5}}[/tex]
[tex]y=4sin (t)=\dfrac{8}{\sqrt{5}}[/tex]
If [tex]\sin t=- \dfrac{2}{\sqrt{5}}[/tex], then [tex]\cos t= \dfrac{1}{\sqrt{5}}[/tex].
[tex]x=4cos (t)=\dfrac{4}{\sqrt{5}}[/tex]
[tex]y=4sin (t)=-\dfrac{8}{\sqrt{5}}[/tex]
Therefore, the two points are [tex]\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )[/tex].
