Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
