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

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(B) Given the parametric equations of the curve: $x = t^{2} + 3t - 8 \quad (1)$ $y = 2t^{2} - 2t - 5 \quad (2)$ At the point $(2, -1)$,we first find t

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$\frac{6}{7}$

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(B) Given the parametric equations of the curve: $x = t^{2} + 3t - 8 \quad (1)$ $y = 2t^{2} - 2t - 5 \quad (2)$ At the point $(2, -1)$,we first find the value of $t$. From equation $(2)$: $-1 = 2t^{2} - 2t - 5 \Rightarrow 2t^{2} - 2t - 4 = 0 \Rightarrow t^{2} - t - 2 = 0$ $(t - 2)(t + 1) = 0 \Rightarrow t = 2$ or $t = -1$. From equation $(1)$: $2 = t^{2} + 3t - 8 \Rightarrow t^{2} + 3t - 10 = 0$ $(t + 5)(t - 2) = 0 \Rightarrow t = 2$ or $t = -5$. The common value for $t$ is $t = 2$. Now,we find the derivatives with respect to $t$: $\frac{dx}{dt} = 2t + 3$ $\frac{dy}{dt} = 4t - 2$ The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}$. At $t = 2$: $\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

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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