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What is the slope of the tangent to the curve \(\rm x=t^2+3t-8,\ y=2t^2-2t-5\) at \(t = 2\)?
1. 7/6
2. 6/7
3. 1
4. 5/6
5. None of these

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Correct Answer - Option 2 : 6/7


Slope of tangent to curve: 

  • The first derivative of f(x).
  • Plug x value of the indicated point into f '(x) to find the slope at x


\(\rm x=t^2+3t-8,\ y=2t^2-2t-5\)

\( \begin{aligned} &\rm \frac{d x}{d t}=2 t+3 \text { and } \frac{d y}{d t}=4 t-2\\\ &\rm \frac{\frac{d y}{dt}}{\frac{d x}{d t}}=\frac{4 t-2}{2 t+3}\\ &\rm\left.\frac{d y}{dx}\right|_{t=2}=\frac{4 (2)-2}{2 (2)+3}\\ &=\frac{6}{7} \end{aligned}\)

Hence, option (2) is correct.

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