The position of a particle moving in the xy-plane is given by the parametric equations x(t)=(6t)/(t+1) and y(t)=(-8)/(t^(2)+4). What is the slope of the line tangent to the path of the particle at the point where t=2 ?

Find Derivative of x(t): First, we need to find the derivatives of x(t) and y(t) to determine the slope of the tangent line at t=2. Start with x(t) = (6t)/(t+1). Using the quotient rule, the derivative x'(t) is given by: x'(t) = [(6(t+1) - 6t) / (t+1)^2] = [6 / (t+1)^2].Calculate x'(2): Next, calculate x'(2) using the derivative formula: x'(2) = 6 / (2+1)^2 = 6 / 9 = 2/3.Find Derivative of y(t): Now, find the derivative of y(t) = (-8)/(t^2+4). Using the derivative of a quotient, y'(t) is: y'(t) = [0*(t^2+4) - (-8)*2t] / (t^2+4)^2 = (16t) / (t^2+4)^2.Calculate y'(2): Calculate y'(2) using the derivative formula: y'(2) = 16*2 / (2^2+4)^2 = 32 / 36 = 8/9.Calculate Slope at t=2: The slope of the tangent line at t=2 is the ratio of dy/dx, which is y'(2)/x'(2): Slope = 8/9 / 2/3 = (8/9)*(3/2) = 4/3.

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The position of a particle moving in the
xy-plane is given by the parametric equations
x(t)=(6t)/(t+1) and
y(t)=(-8)/(t^(2)+4). What is the slope of the line tangent to the path of the particle at the point where
t=2 ?

The position of a particle moving in the $\newline$ $xy$ -plane is given by the parametric equations $\newline$ $x(t)=\frac{6t}{t+1}$ and $\newline$ $y(t)=\frac{-8}{t^{2}+4}$ . What is the slope of the line tangent to the path of the particle at the point where $\newline$ $t=2$ ?

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Calculus

Relate position, velocity, speed, and acceleration using derivatives

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Q. The position of a particle moving in the

\newline

xy

-plane is given by the parametric equations

\newline

x(t)=\frac{6t}{t+1}

and

\newline

y(t)=\frac{-8}{t^{2}+4}

. What is the slope of the line tangent to the path of the particle at the point where

\newline

t=2

Find Derivative of $x(t)$ : First, we need to find the derivatives of $x(t)$ and $y(t)$ to determine the slope of the tangent line at $t=2$ . Start with $x(t) = \frac{6t}{t+1}$ . Using the quotient rule, the derivative $x'(t)$ is given by: $\newline$ $x'(t) = \left[\frac{6(t+1) - 6t}{(t+1)^2}\right] = \left[\frac{6}{(t+1)^2}\right]$ .
Calculate $x'(2)$ : Next, calculate $x'(2)$ using the derivative formula: $\newline$ $x'(2) = \frac{6}{(2+1)^2} = \frac{6}{9} = \frac{2}{3}.$
Find Derivative of $y(t)$ : Now, find the derivative of $y(t) = \frac{-8}{t^2+4}$ . Using the derivative of a quotient, $y'(t)$ is: $\newline$ $y'(t) = \frac{0\cdot(t^2+4) - (-8)\cdot 2t}{(t^2+4)^2} = \frac{16t}{(t^2+4)^2}$ .
Calculate $y'(2)$ : Calculate $y'(2)$ using the derivative formula: $\newline$ $y'(2) = \frac{16\cdot 2}{(2^2+4)^2} = \frac{32}{36} = \frac{8}{9}$ .
Calculate Slope at $t=2$ : The slope of the tangent line at $t=2$ is the ratio of $\frac{dy}{dx}$ , which is $\frac{y'(2)}{x'(2)}$ : $\newline$ Slope = $\frac{8}{9} / \frac{2}{3} = \left(\frac{8}{9}\right)\left(\frac{3}{2}\right) = \frac{4}{3}$ .