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The slope of the tangent of the curve x = t2 + 3t - 8, y = 2t2 - 2t - 5 at the t = 2 is
1. \(\frac{{ - 6}}{7}\)
2. \(\frac{{ 7}}{6}\)
3. \(\frac{{ 6}}{7}\)
4. \(\frac{{ 22}}{7}\)

1 Answer

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Best answer
Correct Answer - Option 3 : \(\frac{{ 6}}{7}\)

Concept:

Parametric Form:

If f(x) and g(x) are the functions in x, then 

\(\rm df(x)\over dg(x)\) = \(\rm \frac{df(x)\over dx}{dg(x)\over dx}\) 

 

Calculation:

Given y = 2t2 - 2t - 5

\(\rm {dy\over dt}\) = 4t - 2

Also x = t2 + 3t - 8

\(\rm dx\over dt\) = 2t + 3 

Now \(\rm dy\over dx\) = \(\rm \frac{dy\over dt}{dx\over dt}\) 

\(\rm dy\over dx\) = \(\rm \frac{4t-2}{2t+3}\)

At t = 2,

\(\rm {dy\over dx}\) = \(\rm \frac{4(2) -2}{2(2)+3}\) = \(\rm \frac{6}{7}\)

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