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Find the slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at t = 2. |
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Answer» Slope of the tangent =\(\frac{dy}{dx}\) = \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)= \(\frac{4t-2}{2t+3}\) = \(\frac{dy}{dx}\) at t = 2 = \((\frac{4t-2}{2t+3})_{at\,t = 2}\) = \(\frac{4(2)-2}{2(2)+3}\) = \(\frac{6}{7}\) |
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