The curve represented by x = t^5 + 5t^3 + 20t | vedclass

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(C) Given $x = t^5 + 5t^3 + 20t + 7$ and $y = 4t^3 - 3t^2 - 18t + 3$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = 5t^4 + 15t^2 + 20

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$(-1, 3/2)$

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(C) Given $x = t^5 + 5t^3 + 20t + 7$ and $y = 4t^3 - 3t^2 - 18t + 3$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = 5t^4 + 15t^2 + 20 = 5(t^4 + 3t^2 + 4)$.Note that $t^4 + 3t^2 + 4$ is always positive for all real $t$ (since its discriminant $D = 3^2 - 4(1)(4) = 9 - 16 = -7 $\frac{dy}{dt} = 12t^2 - 6t - 18 = 6(2t^2 - t - 3)$.Now,the derivative of the curve is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6(2t^2 - t - 3)}{5(t^4 + 3t^2 + 4)}$.The curve is decreasing when $\frac{dy}{dx} Since the denominator $5(t^4 + 3t^2 + 4)$ is always positive,we only need the numerator to be negative:$6(2t^2 - t - 3) $2t^2 - t - 3 Factorizing the quadratic expression: $(2t - 3)(t + 1) The roots are $t = -1$ and $t = 3/2$.Testing the intervals,the expression is negative for $t \in (-1, 3/2)$.

The curve represented by $x = t^5 + 5t^3 + 20t + 7$ and $y = 4t^3 - 3t^2 - 18t + 3$ is decreasing in the interval

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