The slope of the tangent to the curve x = t^2 | vedclass

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(B) Given the parametric equations of the curve are $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$. First,we find the value of $t$ at the point $(2, -1)$.

Vedclass · Accepted Answer

$6/7$

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(B) Given the parametric equations of the curve are $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$. First,we find the value of $t$ at the point $(2, -1)$. Substituting $x = 2$ into the equation for $x$: $2 = t^2 + 3t - 8 \implies t^2 + 3t - 10 = 0 \implies (t+5)(t-2) = 0$. Thus,$t = 2$ or $t = -5$. Checking $t = 2$ in the equation for $y$: $y = 2(2)^2 - 2(2) - 5 = 8 - 4 - 5 = -1$. This matches the given point. Now,calculate the derivatives with respect to $t$: $\frac{dx}{dt} = 2t + 3$ and $\frac{dy}{dt} = 4t - 2$. The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}$. At $t = 2$,the slope is $\frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

The slope of the tangent to the curve $x = t^2 + 3t - 8$,$y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is

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