Question

Notice that the curve given by the parametric equations

x = 16-t²

y = t³- 4t

is symmetric about the x-axis. (If gives us the point (x,y), then − will give (x,−y)). At which x value is the tangent to this curve horizontal and which t value is the tangent to this curve vertical?

Answer 1

The x value at which the tangent to the curve is horizontal is 44/3. The t value at which the tangent to the curve is vertical is t = 0.

Step-by-step explanation:

The given parametric equations are:

x = 16 - t²

y = t³ - 4t

To determine the x value at which the tangent to the curve is horizontal, we need to find the values of t when the slope dy/dt is equal to 0.

We can find the slope dy/dt by differentiating the equation for y with respect to t.

dy/dt = 3t² - 4

Setting dy/dt equal to 0 and solving for t:

0 = 3t² - 4

3t² = 4

t² = 4/3

t = ±√(4/3)

For the tangent to be horizontal, the y-coordinate of the corresponding point on the curve should be the same. Plugging in the value of t into the equation for y:

y = (√(4/3))³ - 4(√(4/3))

y = (√(4/3))³ - 4(√(4/3))

y = 8/3 - 4√(4/3)

Therefore, the x value at which the tangent to the curve is horizontal is 16 - (√(4/3))² = 16 - 4/3 = 44/3.

To find the t value at which the tangent to the curve is vertical, we need to find the value of t when the derivative dx/dt is equal to 0. Differentiating the equation for x with respect to t:

dx/dt = -2t

Setting dx/dt equal to 0 and solving for t:

0 = -2t

t = 0

Therefore, the t value at which the tangent to the curve is vertical is t = 0.

Answer 2

To find the x value at which the tangent to the curve is horizontal, we set dy/dx = 0 to find the t value. To find the x value for the vertical tangent, we set dx/dt = 0 to find the t value. The t value for the horizontal tangent is t = √(4/3) and the x value for the vertical tangent is x = 16.

Step-by-step explanation:

To find the x value at which the tangent to the curve is horizontal, we need to find the t value at this point. The slope of the tangent to the curve is given by the derivative of y with respect to x, which is dy/dx = (dy/dt)/(dx/dt). So, we need to find the values of t for which dy/dx = 0. To find the t value at which the tangent to the curve is vertical, we need to find the x value at this point. The slope of the tangent to the curve is given by the derivative of x with respect to t, which is dx/dt. So, we need to find the values of t for which dx/dt = 0.

Given the parametric equations x = 16 - t² and y = t³ - 4t, we can find dx/dt and dy/dt. Then we set dy/dx = 0 to find the t value for the horizontal tangent, and we set dx/dt = 0 to find the x value for the vertical tangent.

First, we find dx/dt: dx/dt = d/dt (16 - t²) = -2t

Then, we find dy/dt: dy/dt = d/dt (t³ - 4t) = 3t² - 4

Next, we find dy/dx: dy/dx = (dy/dt)/(dx/dt) = (3t² - 4)/(-2t)

To find the t value for the horizontal tangent, we set dy/dx = 0 and solve for t: (3t² - 4)/(-2t) = 0

Simplifying the equation, we multiply both sides by -2t to eliminate the denominator: 3t² - 4 = 0

Using the quadratic formula, we solve for t: t = ±√(4/3)

Since the curve is symmetric about the x-axis, we are only interested in the positive value for t, so the t value for the horizontal tangent is t = √(4/3)

To find the x value for the vertical tangent, we set dx/dt = 0 and solve for x: -2t = 0

Since the equation -2t = 0 is true when t = 0, the vertical tangent occurs at the x value x = 16 - t² = 16 - 0² = 16