Question

Math

Posted 5 months ago

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1. The position of a particle moving in the $x y$-plane is given by the parametric equations $x(t)=\frac{6 t}{t+1}$ and $y(t)=\frac{-8}{t^{2}+4}$. What is the slope of the line tangent to the path of the particle at the point where $t=2$ ?
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) $\frac{4}{3}$
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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

To find the slope of the tangent line to the path of the particle at a specific point, we need to find the derivatives of $x(t)$ and $y(t)$ with respect to $t$ and then evaluate them at $t=2$

step 2

The derivative of $x(t)$ with respect to $t$ is given by $\frac{dx}{dt} = \frac{6}{(t + 1)^2}$

step 3

The derivative of $y(t)$ with respect to $t$ is given by $\frac{dy}{dt} = \frac{16t}{(t^2 + 4)^2}$

step 4

Evaluate $\frac{dx}{dt}$ at $t=2$ to get $\frac{dx}{dt}\bigg|_{t=2} = \frac{6}{(2 + 1)^2} = \frac{6}{9} = \frac{2}{3}$

step 5

Evaluate $\frac{dy}{dt}$ at $t=2$ to get $\frac{dy}{dt}\bigg|_{t=2} = \frac{16 \cdot 2}{(2^2 + 4)^2} = \frac{32}{(4 + 4)^2} = \frac{32}{64} = \frac{1}{2}$

step 6

The slope of the tangent line at the point where $t=2$ is the ratio of the derivatives, which is $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ evaluated at $t=2$

step 7

Calculate the slope: $\frac{\frac{1}{2}}{\frac{2}{3}} = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$

Answer

$\frac{3}{4}$

Key Concept

Slope of the tangent line in parametric equations

Explanation

The slope of the tangent line to a curve defined by parametric equations at a point is the ratio of the derivatives of the $y$-coordinate and the $x$-coordinate with respect to the parameter at that point.

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