The curve is given by rho=at³, phi=bt². Find the angle
The curve is given by $\rho=at^{3},\ \varphi=bt^{2}$. Find the angle $\alpha$ between the polar axis and the tangent at a general point.
Explanation of Method
The tangent-addition formula appears naturally from the ratio of two polar derivatives.
Step 1: The inclination as $\varphi+\psi$
The tangent's inclination to the polar axis is $\alpha=\varphi+\psi$, where $\tan\psi$ is given by the polar-with-parameter formula $\tan\psi=\dfrac{\rho\dot\varphi}{\dot\rho}$ (dots denoting $d/dt$).
Step 2: Compute $\tan\psi$
\[ \dot\rho=3at^{2}, \qquad \dot\varphi=2bt, \qquad \tan\psi = \frac{at^{3}\cdot2bt}{3at^{2}} = \frac{2bt^{2}}{3}. \tag{1} \]
Step 3: Verification
\[a=1.3,\ b=0.7,\ t=0.9\text{: } \tan\alpha \text{ from the formula matches the numerically computed } dy/dx\ \checkmark\]
Result
Final answer:
\[\boxed{\,\alpha = \varphi+\psi = bt^{2} + \arctan\frac{2bt^{2}}{3}\,}\]
Remark
Remark. Writing the slope $dy/dx$ directly as a ratio of Cartesian derivatives and dividing through by $\cos(bt^2)$ turns it into exactly the tangent-addition pattern $\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ with $A=bt^2$, $B=\arctan\dfrac{2bt^2}3$, giving the same result (1) from the Cartesian side.