35. Consider the curve in the \( xy \)-plane represented by \( x = \dfrac{2}{t} \) and \( y = \ln t \) for \( t > 0 \). The slope of the line tangent to the curve at the point where \( x = 1 \) is






Answer is: option1

\( -1 \)

Solution:

\( x = \dfrac{2}{t} \)
\( y = \ln t \)

\( \dfrac{dx}{dt} = -\dfrac{2}{t^{2}} \)
\( \dfrac{dy}{dt} = \dfrac{1}{t} \)

Then \( \dfrac{dy}{dx} = \left(\dfrac{1}{t}\right)\left(-\dfrac{t^{2}}{2}\right) = -\dfrac{t}{2} \)

When \( x = 1 \), \( t = 2 \). Then slope \( \dfrac{dy}{dx} = -\dfrac{2}{2} = -1 \)

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