Consider the curve given by the parametric equations x = t(t^{2} - 48), y = 4(t^{2} - 48)| a)...

Question:

Consider the curve given by the parametric equations

{eq}x = t(t^{2} - 48), y = 4(t^{2} - 48)| {/eq}

a) Determine the point on the curve where the tangent is horizontal.

t=

b) Determine the points {eq}t_{1}, t_{2} {/eq} where the tangent is vertical and {eq}t_{1} < t_{2}. {/eq}

{eq}t_{1}= {/eq}

{eq}t_{2}= {/eq}

Slope of Parametric Curve:

A parametric curve is described by the set of equations

{eq}x=x(t) \\ y=y(t). {/eq}

The slope to the parametric curve is given by the formula

{eq}\displaystyle m = \frac{ y'(t) }{ x'(t) }. {/eq}

Answer and Explanation:

We are given the parametric curve

{eq}x = t(t^{2} - 48) \\ y = 4(t^{2} - 48). {/eq}

The slope to the curve is calculated by means of the formula

{eq}\displaystyle m = \frac{y'(t)}{x'(t)} = \frac{8t }{3t^2-48 }. {/eq}

a) The point on the curve where the tangent is horizontal is found setting the slope of the curve to zero

{eq}\displaystyle m = 0 \Rightarrow t = 0 \Rightarrow \\ x=0, \; y = -192. {/eq}

b) The points on the curve where the tangent is vertical are found setting the slope of the curve to infinity

{eq}\displaystyle m = \infty \Rightarrow 3t^2-48 = 0 \Rightarrow t^2 = 12 \Rightarrow \\ \displaystyle t = \pm \sqrt{12} \Rightarrow \\ \displaystyle t_1 = -\sqrt{12} \Rightarrow x =36\sqrt{12} , \; y = -144 \\ \displaystyle t_2 = \sqrt{12} \Rightarrow x =-36\sqrt{12} , \; y = -144. {/eq}


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Slopes and Tangents on a Graph

from Math 104: Calculus

Chapter 1 / Lesson 8
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