A curve is defined by the parametric equations x(t)=10t^(3)-5t^(2)+6t and y(t)=-t^(3). Find (dy)/(dx). Answer:

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A curve is defined by the parametric equations  x(t)=10t^(3)-5t^(2)+6t and  y(t)=-t^(3). Find  (dy)/(dx). Answer:

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Find Derivative of x: To find the derivative of y with respect to x (dy/dx) for a curve defined by parametric equations, we need to find dy/dt and dx/dt separately and then divide dy/dt by dx/dt.Find Derivative of y: First, let's find the derivative of x with respect to t, which is dx/dt. Given x(t) = 10t^3 - 5t^2 + 6t, we use the power rule for derivatives: d/dt(t^n) = n*t^(n-1). dx/dt = d/dt(10t^3) - d/dt(5t^2) + d/dt(6t)        = 3*10t^(3-1) - 2*5t^(2-1) + 6*1t^(1-1)        = 30t^2 - 10t + 6Calculate dy/dx: Next, we find the derivative of y with respect to t, which is dy/dt. Given y(t) = -t^3, we again use the power rule for derivatives. dy/dt = d/dt(-t^3)        = -3t^(3-1)        = -3t^2Simplify dy/dx: Now we have dx/dt = 30t^2 - 10t + 6 and dy/dt = -3t^2. To find dy/dx, we divide dy/dt by dx/dt. (dy/dx) = (dy/dt) / (dx/dt)          = (-3t^2) / (30t^2 - 10t + 6)We simplify the expression for dy/dx by dividing both the numerator and the denominator by the common term t^2, assuming t is not equal to zero. (dy/dx) = (-3t^2) / (30t^2 - 10t + 6)          = -3 / (30 - 10/t + 6/t^2)

A curve is defined by the parametric equations $x(t)=10 t^{3}-5 t^{2}+6 t$ and $y(t)=-t^{3}$ . Find $\frac{d y}{d x}$ . $\newline$ Answer:

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A curve is defined by the parametric equations x(t)=10t3−5t2+6t x(t)=10 t^{3}-5 t^{2}+6 t x(t)=10t3−5t2+6t and y(t)=−t3 y(t)=-t^{3} y(t)=−t3. Find dydx \frac{d y}{d x} dxdy​.\newlineAnswer:

A curve is defined by the parametric equations $x(t)=10 t^{3}-5 t^{2}+6 t$ and $y(t)=-t^{3}$ . Find $\frac{d y}{d x}$ . $\newline$ Answer: