Questions: Determine el valor de dy/dt de la pendiente de la recta tangente para la curva x=t^4+10t, y=t^10-4t^2+t en el punto t=1. (A) 0 (B) -3/2 (C) 2/3 (D) 3/2

To determine the value of \(\frac{dy}{dt}\) for the tangent line to the curve given by \(x = t^4 + 10t\) and \(y = t^{10} - 4t^2 + t\) at the point \(t = 1\), we need to:

1. Compute the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
2. Evaluate these derivatives at \(t = 1\).
3. Use the chain rule to find the slope of the tangent line, which is \(\frac{dy/dt}{dx/dt}\) at \(t = 1\).

Definimos las funciones paramétricas de la curva como: \[ x(t) = t^4 + 10t \] \[ y(t) = t^{10} - 4t^2 + t \]

Calculamos las derivadas de \(x\) y \(y\) con respecto a \(t\): \[ \frac{dx}{dt} = 4t^3 + 10 \] \[ \frac{dy}{dt} = 10t^9 - 8t + 1 \]

Evaluamos las derivadas en el punto \(t = 1\): \[ \frac{dx}{dt}\bigg|_{t=1} = 4(1)^3 + 10 = 14 \] \[ \frac{dy}{dt}\bigg|_{t=1} = 10(1)^9 - 8(1) + 1 = 3 \]

La pendiente de la recta tangente en el punto \(t = 1\) se calcula como: \[ \text{slope} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3}{14} \]

La pendiente de la recta tangente para la curva en el punto \(t = 1\) es: \[ \boxed{\frac{3}{14}} \]