At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
The slope of the tangent line to the curve at (8, 2) is given by the derivative [tex]\frac{dy}{dx}[/tex] at that point. By the chain rule,
[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
Differentiate the given parametric equations with respect to [tex]t[/tex] :
[tex]x = 4t \implies \dfrac{dx}{dt} = 4[/tex]
[tex]y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}[/tex]
Then
[tex]\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}[/tex]
We have [tex]x=8[/tex] and [tex]y=2[/tex] when [tex]t=2[/tex], so the slope at the given point is [tex]\frac{dy}{dx} = -\frac14[/tex].
The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation
[tex]y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}[/tex]
Alternatively, we can eliminate the parameter and express [tex]y[/tex] explicitly in terms of [tex]x[/tex] :
[tex]x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x[/tex]
Then the slope of the tangent line is
[tex]\dfrac{dy}{dx} = -\dfrac{16}{x^2}[/tex]
At [tex]x = 8[/tex], the slope is again [tex]-\frac{16}{64}=-\frac14[/tex], so the normal has slope +4, and so on.
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
