The curve given by the parametric equations #x=16 - t^2#, #y= t^3 - 1 t#
is symmetric about the x-axis. At which x value is the tangent to this curve horizontal?

Question

Emma Aldridge · Accepted Answer

#x=15 2/3#

Slope of the tangent at a point on the cuurve, is the value of first derivative at that point.

Hence the tangent will be hrizontal when first derivative #(dy)/(dx)# is zero.

#(dy)/(dx)# of a parametric equation is given by #((dy)/(dt))/((dx)/(dt))#

as #(dy)/(dt)=3t^2-1# and #(dx)/(dt)=-2t#

#(dy)/(dx)=-(3t^2-1)/(2t)# and this is zero when #3t^2=1# ad #t!=0#

i.e. #t=+-1/sqrt3#

i.e. #x=16-1/3=47/3=15 2/3#

Note that for same value of #x#, there are two values of #y#, where tangent will be hrizontal, depending on sign of #t#. graph{(16-x)(15-x)^2-y^2=0 [1.34, 21.34, -5.12, 4.88]}

Christopher Allers · Answer

undefined To find the x-value at which the tangent to the curve is horizontal, we need to determine when the derivative of y with respect to x is zero.

Given the parametric equations $x = 16 - t^2$ and $y = t^3 - 1$, we can eliminate the parameter $t$ to express $y$ solely in terms of $x$. Then, differentiate $y$ with respect to $x$ and set the derivative equal to zero to find the critical points.

1. Eliminate the parameter $t$ by solving the equation $x = 16 - t^2$ for $t$:
$$t = \pm \sqrt{16 - x}$$

2. Substitute the expression for $t$ into the equation for $y$:
$$y = (\sqrt{16 - x})^3 - 1$$

3. Simplify the expression for $y$:
$$y = (16 - x)^{3/2} - 1$$

4. Differentiate $y$ with respect to $x$ to find the derivative:
$$\frac{dy}{dx} = -\frac{3}{2}(16 - x)^{1/2}$$

5. Set the derivative equal to zero and solve for $x$:
$$-\frac{3}{2}(16 - x)^{1/2} = 0$$
$$16 - x = 0$$
$$x = 16$$

So, the x-value at which the tangent to the curve is horizontal is $x = 16$.

The curve given by the parametric equations #x=16 - t^2#, #y= t^3 - 1 t# is symmetric about the x-axis. At which x value is the tangent to this curve horizontal?