Question

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter.

x = 6 + ln(t), y = t^{2} +
10, (6, 11)

y=?

Answer #1

Find an equation of the tangent to the curve at the given point
by both eliminating the parameter and without eliminating the
parameter. x = 4 + ln(t), y = t2 + 4, (4, 5)

Find the exact length of the curve. x = 8 + 9t2, y = 3 + 6t3, 0
≤ t ≤ 5
Find an equation of the tangent to the curve at the given point
by both eliminating the parameter and without eliminating the
parameter. x = 6 + ln(t), y = t2 + 1, (6, 2) y =
Find dy/dx. x = t 3 + t , y = 3 + t
Find the distance traveled by a particle...

Find an equation of the tangent to the curve at the point
corresponding to the given value of the parameter.
x = cos(θ) +
sin(4θ), y =
sin(θ) +
cos(4θ); θ = 0

find an equation for the line tangent to the given curve at the
point defined by the given value of t.
x sin t + 2x =t, t sin t - 2t =y, t=pi

Find an equation of the tangent line to the curve at the given
point.
1.) y= sqrt(5x+ 9), at x= 10.
2.) y= cos(x) + cos^3(x), at x=π/6.

Find an equation of the tangent line to the curve at the given
point. A) y = 6x + 3 cos x, P = (0, 3) B)y = 8 x cos x P = \(pi ,
-8 pi)
B)Find an equation of the tangent line to the curve at the given
point.
y = 8 x cos x
C)
If H(θ) = θ cos θ, find H'(θ) and H''(θ).
find H'(
θ)
and H"(θ)

Find the equation of the line tangent to the parametric curve x
= t2 − 2t3 y = t2 when t = 2.

Find an equation for the line tangent to the curve at the point
defined by the given value of t. Also, find the value of
StartFraction d squared y Over dx squared EndFractiond2ydx2
at this point.
x=secant^2 t−1
,
y=cos t;
t=negative StartFraction pi Over 3 EndFraction−π/3
Write the equation of the tangent line.
y=?xplus+?
(Type exact answers, using radicals as needed.)
What is the value of
StartFraction d squared y Over dx squared EndFractiond2ydx2
at this point?
StartFraction d squared...

Find an equation for the tangent of the curve x=2t-lnt, y=t+lnt
at the point corresponding to t=1

a) Find the equation of the tangent line to the curve x= 2sin2t,
y= 3sint at the point where the same.
b) Find the points on the curve x= t^2-t+2, y=t^3-3t where the
tangent is horizontal.

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