Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
The equations with tangents at (0,0) are [tex]y = \frac{4}{7} x[/tex] and
[tex]y = -\frac{4}{7} x[/tex].
In this question,
The curves are x = 7 cos(t), y = 4 sin(t) cos(t)
Two tangents at (0, 0)
In this case, the parametric derivative of x and y are expressed in terms of t.
The first derivative dy/dx can be expressed as
[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]
Now, dy/dt is obtained by differentiate y with respect to t,
[tex]\frac{dy}{dt}= 4[cos(t)(cos(t))+sin(t)(-sin(t))][/tex]
⇒ [tex]\frac{dy}{dt}= 4[cos^{2} (t)-sin^{2} (t)][/tex]
Now, dx/dt is obtained by differentiate x with respect to t,
[tex]\frac{dx}{dt} =7(-sin(t))[/tex]
⇒ [tex]\frac{dx}{dt} =-7sin(t)[/tex]
Thus, [tex]\frac{dy}{dx}=\frac{4[cos^{2}(t)-sin^{2}(t ) ]}{-7sin(t)}[/tex]
At (0,0) x = 0 and y = 0, Then
0 = 7 cos(t)
0 = 4 sin(t) cos(t)
and
cos(t) = 0
sin(t) cos(t) = 0
There are two values between -π and π which satisfy these equations simultaneously are
t = π/2
t = -π/2
The equation of a straight line given a point and its slope is
y-y₀ = m(x-x₀)
The two tangents lies at (0,0), so the equation becomes
y = mx
Then the two straight lines will be
y = m₁x and
y = m₂x
For t = π/2,
[tex]m_1=\frac{dy}{dx}=\frac{4[cos^{2}(\frac{\pi }{2} )-sin^{2}(\frac{\pi }{2} ) ]}{-7sin(\frac{\pi }{2} )}[/tex]
⇒ [tex]m_1=-\frac{4[0-1]}{7(1)}[/tex]
⇒ [tex]m_1=\frac{4}{7}[/tex]
For t = -π/2,
[tex]m_2=\frac{dy}{dx}=\frac{4[cos^{2}(-\frac{\pi }{2} )-sin^{2}(-\frac{\pi }{2} ) ]}{-7sin(-\frac{\pi }{2} )}[/tex]
⇒ [tex]m_2=-\frac{4[0-1]}{-7(1)}[/tex]
⇒ [tex]m_2=-\frac{4}{7}[/tex]
Thus the equations with tangents at (0,0) are [tex]y = \frac{4}{7} x[/tex] and
[tex]y = -\frac{4}{7} x[/tex].
Learn more about equation of curve here
https://brainly.com/question/9959935
#SPJ4
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
