Find the value of the expression (look at the image)

Mathematics

1 answer:

sertanlavr [38]3 years ago

8 0

Answer:

Step-by-step explanation:

Anything raised to the power of zero is one. And 1^-2 is is 1.

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Find dy/dx x=a(cost +sint) , y=a(sint-cost)

MissTica

Answer:

$\begin{aligned} \frac{dy}{dx} &= \frac{\cos(t) + \sin(t)}{\cos(t) - \sin(t)} \end{aligned}$ given that $a \ne 0$ and that $\cos(t) - \sin(t) \ne 0$ .

Step-by-step explanation:

The relation between the $y$ and the $x$ in this question is given by parametric equations (with $t$ as the parameter.)

Make use of the fact that:

$\begin{aligned} \frac{dy}{dx} = \quad \text{$\frac{dy/dt}{dx/dt}$ given that $\frac{dx}{dt} \ne 0$} \end{aligned}$ .

Find $\begin{aligned} \frac{dx}{dt} \end{aligned}$ and $\begin{aligned} \frac{dy}{dt} \end{aligned}$ as follows:

$\begin{aligned} \frac{dx}{dt} &= \frac{d}{dt} [a\, (\cos(t) + \sin(t))] \\ &= a\, (-\sin(t) + \cos(t)) \\ &= a\, (\cos(t) - \sin(t))\end{aligned}$ .

$\begin{aligned} \frac{dx}{dt} \ne 0 \end{aligned}$ as long as $a \ne 0$ and $\cos(t) - \sin(t) \ne 0$ .

$\begin{aligned} \frac{dy}{dt} &= \frac{d}{dt} [a\, (\sin(t) - \cos(t))] \\ &= a\, (\cos(t) - (-\sin(t))) \\ &= a\, (\cos(t) + \sin(t))\end{aligned}$ .

Calculate $\begin{aligned} \frac{dy}{dx} \end{aligned}$ using the fact that $\begin{aligned} \frac{dy}{dx} = \text{$\frac{dy/dt}{dx/dt}$ given that $\frac{dx}{dt} \ne 0$} \end{aligned}$ . Assume that $a \ne 0$ and $\cos(t) - \sin(t) \ne 0$ :

$\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} \\ &= \frac{a\, (\cos(t) + \sin(t))}{a\, (\cos(t) - \sin(t))} \\ &= \frac{\cos(t) + \sin(t)}{\cos(t) - \sin(t)}\end{aligned}$ .