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

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(D) The given curve is defined by $x=t^{2}+3t-8$ and $y=2t^{2}-2t-5$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = 2t+3$ and $\fr

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$\frac{6}{7}$

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(D) The given curve is defined by $x=t^{2}+3t-8$ and $y=2t^{2}-2t-5$.First,we find 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 $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t-2}{2t+3}$.To find the value of $t$ at the point $(2, -1)$,we set $x=2$ and $y=-1$:For $x=2$: $t^{2}+3t-8=2 \Rightarrow t^{2}+3t-10=0 \Rightarrow (t-2)(t+5)=0$,so $t=2$ or $t=-5$.For $y=-1$: $2t^{2}-2t-5=-1 \Rightarrow 2t^{2}-2t-4=0 \Rightarrow 2(t^{2}-t-2)=0 \Rightarrow 2(t-2)(t+1)=0$,so $t=2$ or $t=-1$.The common value of $t$ is $t=2$.Substituting $t=2$ into the expression for the slope:$\left. \frac{dy}{dx} \right|_{t=2} = \frac{4(2)-2}{2(2)+3} = \frac{8-2}{4+3} = \frac{6}{7}$.Thus,the slope of the tangent is $\frac{6}{7}$.

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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