Question

If x=3t☻-1,y=t♥-t,then dy/dx is equal to

Answer 1

Answer on Question# 54574– Mathematics – Differential Equations

**Question:**

If $x = 3t \bullet -1, y = t \nabla - t$, then $\frac{dy}{dx}$ is equal to...

**Answer:**

**Definition of parametric differentiation:** if $x = x(t)$ and $y = y(t)$ then

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \quad \text{provided} \quad \frac{dx}{dt} \neq 0.$$

1) If according to the statement of the problem we have

$$\left\{ \begin{array}{l} x(t) = 3t^{-1}, \\ y(t) = t^{-t}. \end{array} \right.$$

then using (1) we obtain

$$\frac{dx}{dt} = 3 \cdot (-1)t^{-2} = -\frac{3}{t^2};$$ $$\ln(y) = -t \ln(t) \Rightarrow \frac{1}{y} \frac{dy}{dt} = -\ln(t) - \frac{t}{t} \Rightarrow \frac{dy}{dt} = -y (\ln(t) + 1) = -t^{-t} (\ln(t) + 1);$$ $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-t^{-t} (\ln(t) + 1)}{-\frac{3}{t^2}} = \frac{1}{3} t^{2-t} (\ln(t) + 1);$$ $$\frac{dy}{dx} = \frac{1}{3} t^{2-t} (\ln(t) + 1).$$

2) If according to the statement of the problem we have

$$\left\{ \begin{array}{l} x(t) = 3t^m - 1, \\ y(t) = t^n - t, \end{array} \right.$$

then we receive

$$\frac{dx}{dt} = 3mt^{m-1}; \quad \frac{dy}{dt} = nt^{n-1} - 1;$$ $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{nt^{n-1} - 1}{3mt^{m-1}}.$$

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