Any vector can be defined by its projections onto the coordinate axes or by its magnitude and direction. For example, the velocity vector shows the rate of change of an object's coordinates when using the first method of definition, and in the second case, it determines how fast and in what direction the object is moving.

Thus, the acceleration vector, which shows the change in velocity over time, must contain information about the rate of change of the velocity magnitude (speed) and the rate of change of the velocity direction. In this problem, we will try to demonstrate this and write equations that describe it.

The following formulas can be useful for you:
- Definition of the scalar product (dot product) of vectors $\vec A$ and $\vec B$, the angle between which is equal to $\alpha$:
$$\left(\vec A \cdot \vec B\right) = \left(\vec B \cdot \vec A\right) = \left|\vec A\right|\cdot\left|\vec B\right|\cdot\cos\alpha$$- Expression for the dot product of these vectors in terms of their coordinates:
$$\left(\vec A \cdot \vec B\right) = A_xB_x + A_yB_y$$- Linearity of the scalar product:
$$\left(\vec A \cdot \left(\vec B+\vec C\right)\right) = \left(\vec A \cdot \vec B\right) + \left(\vec A \cdot \vec C\right)$$- Designations of derivatives (notation):
$$\frac{df(t)}{dt} = f^\prime(t) = \dot f(t)$$- Derivative properties:
$$\left(f(x) + g(x)\right)^\prime = f^\prime(x) + g^\prime(x)$$$$\left(f(x)g(x)\right)^\prime = f(x) g^\prime(x) + f^\prime(x)g(x)$$$$\left(\frac{f(x)}{g(x)}\right)^\prime = \frac{f^\prime(x)g(x) - f(x)g^\prime(x)}{\left(g(x)\right)^2}$$$$\left(f\left(g(x)\right)\right)^\prime = f^\prime\left(g(x)\right)\cdot g^\prime(x)$$- Derivatives of some functions:
$$\frac{d x^n}{dx} = nx^{n-1}$$$$\frac{d e^x}{dx} = e^x$$$$\frac{d \ln x}{dx} = \frac{1}{x}$$$$\frac{d \sin x}{dx} = \cos x$$$$\frac{d \cos x}{dx} = -\sin x$$

Let's introduce the notation for velocity and acceleration:
$$\vec v = \left(v_x, v_y\right)$$$$\vec a = \left(a_x, a_y\right) = \left(\dot v_x,\dot v_y\right)$$

Part A. Transverse and longitudinal acceleration

First, let's prove the auxiliary formulas.

A1  ^0.50 Express $\left|\vec v\right|^2$ in terms of $v_x$ and $v_y$.

Let's decompose the acceleration vector into a sum of two vectors, one of which is parallel to the velocity, and the other is perpendicular to it.
$$\vec a = \vec a_\parallel + \vec a_\perp,\qquad \vec a_\parallel \parallel \vec v,\qquad \vec a_\perp \perp \vec v$$

A2  ^0.50 Express $\left|\vec a\right|$ in terms of $\left|\vec a_\parallel\right|$ and $\left|\vec a_\perp\right|$.

A3  ^1.00 Prove that $\left(\vec v \cdot \vec a\right) = \left(\vec v\cdot \vec a_\parallel\right)$ and $\left|\left(\vec v \cdot \vec a\right)\right| = \left|\vec v\right|\cdot \left|\vec a_\parallel\right|$.

A4  ^1.00 Express $\left|\vec a_\perp\right|$ in terms of $\left|\vec a\right|$, $\left|\vec v\right|$ and $\left(\vec v \cdot \vec a\right)$.

A5  ^0.50 Prove that
$$\frac{d\left(f(t)\right)^2}{dt} = Af(t)\dot f(t),$$where $A$ is a constant, and find the value of $A$. If you are unable to complete this step, assume $A = 2$.

A6  ^1.00 Using the results of A1 and A5, express $\frac{d\left(\vec v\cdot\vec v\right)}{dt}$ in terms of $\left(\vec v\cdot\vec a\right)$.

Part B. Transverse and longitudinal components of acceleration

Now let's proceed to studying the magnitude and direction of velocity.

Let us denote by $v$ and $\varphi$ the magnitude of velocity and the angle it forms with the axis $x$, respectively.

B1  ^1.00 Express $v_x$ and $v_y$ in terms of $v$ and $\varphi$.

B2  ^1.00 Express $\frac{d\left(\vec v\cdot\vec v\right)}{dt}$ in terms of $v$ and $\dot v$.

B3  ^1.00 Express $a_x$ and $a_y$ in terms of $v$, $\varphi$, $\dot v$ and $\dot\varphi$.

B4  ^1.00 Express $\left|\vec a_\perp\right|$ in terms of $v$, $\varphi$, $\dot v$ and $\left|\dot\varphi\right|$.

B6  ^0.50 Express $\dot v$ in terms of $v$ and $\left(\vec v\cdot\vec a\right)$.

B7  ^1.00 Express $\left|\dot\varphi\right|$ in terms of $v$, $\left(\vec v\cdot\vec a\right)$ and $\left|\vec a\right|$.