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Question
Math
Posted 5 months ago
1. The position of a particle moving in the xyx y-plane is given by the parametric equations x(t)=6tt+1x(t)=\frac{6 t}{t+1} and y(t)=8t2+4y(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=2t=2 ?
(A) 12\frac{1}{2}
(B) 23\frac{2}{3}
(C) 34\frac{3}{4}
(D) 43\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)x(t) and y(t)y(t) with respect to tt and then evaluate them at t=2t=2
step 2
The derivative of x(t)x(t) with respect to tt is given by dxdt=6(t+1)2\frac{dx}{dt} = \frac{6}{(t + 1)^2}
step 3
The derivative of y(t)y(t) with respect to tt is given by dydt=16t(t2+4)2\frac{dy}{dt} = \frac{16t}{(t^2 + 4)^2}
step 4
Evaluate dxdt\frac{dx}{dt} at t=2t=2 to get dxdtt=2=6(2+1)2=69=23\frac{dx}{dt}\bigg|_{t=2} = \frac{6}{(2 + 1)^2} = \frac{6}{9} = \frac{2}{3}
step 5
Evaluate dydt\frac{dy}{dt} at t=2t=2 to get dydtt=2=162(22+4)2=32(4+4)2=3264=12\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=2t=2 is the ratio of the derivatives, which is dydtdxdt\frac{\frac{dy}{dt}}{\frac{dx}{dt}} evaluated at t=2t=2
step 7
Calculate the slope: 1223=1232=34\frac{\frac{1}{2}}{\frac{2}{3}} = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}
Answer
34\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 yy-coordinate and the xx-coordinate with respect to the parameter at that point.

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