Suppose a curve is traced by the parametric equations

x=1sin(t)
y=18-3cos^2(t)-4sin(t)
At what point (x,y) on this curve is the tangent line horizontal?

Well, well, well, looks like we've got ourselves a curve that wants to play tricks on us! So, let's see if we can find the point where the tangent line gets horizontal.

First, we need to find the derivative of the function y with respect to t. So grab your calculus cap and let's go:

x = 1sin(t) ---(1)
y = 18 - 3cos^2(t) - 4sin(t) ---(2)

Taking the derivative of equation (2) with respect to t, we get:
dy/dt = 2(3sin(t)cos(t)) - 4cos(t)

Now, for the tangent line to be horizontal, the derivative dy/dt should be zero. After all, horizontal lines tend to be a bit lazy and don't move up or down much.

So, we set dy/dt = 0 and solve for t:
2(3sin(t)cos(t)) - 4cos(t) = 0

Let's divide everything by 2cos(t):
3sin(t) - 2 = 0

Now, we just need to find the values of t that satisfy this equation. Oh wait, we have some trigonometry to help us out here!

sin(t) = 2/3

This sounds like a trig party! So, find the fancy inverse sine (also known as arcsin) of both sides:
t = arcsin(2/3)

Now, we can plug this value of t back into equations (1) and (2) to get the corresponding (x, y) coordinates. But hey, don't forget to bring your calculator to this party!

x = 1sin(arcsin(2/3))
y = 18 - 3cos^2(arcsin(2/3)) - 4sin(arcsin(2/3))

Once you've done the calculations, you'll have the point (x, y) on the curve where the tangent line is horizontal. Enjoy your horizontal adventures on the curve, and remember to bring your sense of humor along the way!

To find the point on the curve where the tangent line is horizontal, we need to find the value of t that satisfies the condition dy/dt = 0.

Given:
x = sin(t)
y = 18 - 3cos^2(t) - 4sin(t)

To find dy/dt, we need to find the derivative of y with respect to t:

dy/dt = d/dt (18 - 3cos^2(t) - 4sin(t))

The derivative of a constant (18) with respect to t is zero, so we only need to consider the derivative of -3cos^2(t) - 4sin(t):

dy/dt = -6cos(t)(-sin(t)) - 4cos(t)

Simplifying this expression, we get:

dy/dt = 6cos(t)sin(t) - 4cos(t)

Now, we set dy/dt = 0 and solve for t:

0 = 6cos(t)sin(t) - 4cos(t)

Factoring out cos(t), we get:

0 = cos(t)(6sin(t) - 4)

Setting each factor equal to zero, we have:

cos(t) = 0 OR 6sin(t) - 4 = 0

From the first equation, we know that cos(t) = 0 when t = π/2 + kπ, where k is an integer.

From the second equation, we solve for sin(t):

6sin(t) = 4
sin(t) = 4/6
sin(t) = 2/3

Using the unit circle, we know that sin(t) = 2/3 when t = π/3 or t = 2π/3.

Therefore, we have three potential values for t where the tangent line is horizontal:

t = π/2 + kπ, t = π/3, and t = 2π/3.

To find the corresponding (x, y) points on the curve for each value of t, substitute the values of t into the given parametric equations:

For t = π/2 + kπ:
x = sin(t)
y = 18 - 3cos^2(t) - 4sin(t)

For t = π/3:
x = sin(t)
y = 18 - 3cos^2(t) - 4sin(t)

For t = 2π/3:
x = sin(t)
y = 18 - 3cos^2(t) - 4sin(t)

Evaluate each pair of (x, y) points to get the complete solution.

To find the point on the curve where the tangent line is horizontal, we need to find the value of t that satisfies the condition.

First, let's find the derivative of y with respect to t, dy/dt:
dy/dt = d/dt(18 - 3cos^2(t) - 4sin(t))
dy/dt = 0 + 6cos(t)sin(t) - 4cos(t)

To find when the tangent line is horizontal, we need to find the value of t for which dy/dt = 0:
0 = 6cos(t)sin(t) - 4cos(t)

Factor out cos(t) from the equation:
0 = cos(t)(6sin(t) - 4)

Now we have two possibilities:
1. cos(t) = 0
2. 6sin(t) - 4 = 0

Let's solve each possibility:

1. When cos(t) = 0:
This occurs when t = π/2 or t = 3π/2. Substitute these values of t into the parametric equations to find the corresponding point (x, y).

When t = π/2:
x = 1sin(π/2) = 1
y = 18 - 3cos^2(π/2) - 4sin(π/2) = 18 - 3(0) - 4(1) = 14

So the point (x, y) when t = π/2 is (1, 14).

When t = 3π/2:
x = 1sin(3π/2) = 1(-1) = -1
y = 18 - 3cos^2(3π/2) - 4sin(3π/2) = 18 - 3(0) - 4(-1) = 14

So the point (x, y) when t = 3π/2 is (-1, 14).

2. When 6sin(t) - 4 = 0:
Solving this equation gives us sin(t) = 4/6 = 2/3.
Using the unit circle or a calculator, we can find the two values of t where this is true. Let's call them t₁ and t₂.

When sin(t₁) = 2/3:
t₁ ≈ 0.84 (in radians)

When sin(t₂) = 2/3:
t₂ ≈ 2.46 (in radians)

Substitute these values of t into the parametric equations to find the corresponding points (x, y).

When t = t₁ ≈ 0.84:
x = 1sin(0.84) ≈ 0.75
y = 18 - 3cos^2(0.84) - 4sin(0.84) ≈ 12.46

So the point (x, y) when t = t₁ ≈ 0.84 is (0.75, 12.46).

When t = t₂ ≈ 2.46:
x = 1sin(2.46) ≈ 0.93
y = 18 - 3cos^2(2.46) - 4sin(2.46) ≈ 12.46

So the point (x, y) when t = t₂ ≈ 2.46 is (0.93, 12.46).

To summarize:
The points (x, y) on the curve where the tangent line is horizontal are (1, 14), (-1, 14), (0.75, 12.46), and (0.93, 12.46).

When dy/dx = (dy/dt)/(dx/dt) = 0

Do the calculation and first find the value of t. Use that co calculate x and y