# Tutor profile: Jonny S.

## Questions

### Subject: Calculus

Given a function $$f(x,y) = \begin{cases} (x^2 + y^2) \sin(\frac{1}{x^2 + y^2})& \text{if $(x,y) \neq (0,0)$} \\ C & \text{if $(x,y) = (0,0)$} \\ \end{cases} $$ $$(a)$$ Find the value of $$C$$ so that $$f(x,y)$$ is continuous at $$(0,0)$$ $$(b)$$ Find $$\delta_xf, \delta_yf$$ at the point $$(x,y) \neq (0,0)$$

For part $$(a)$$, First, we need a definition as follows: A function $$f(x,y)$$ is defined as continuous at the point $$(a,b)$$ if $$\lim_{(x,y) \to (a,b)} f(x,y) = f(a,b)$$ Following this definition, we need to take the limit of $$f(x,y)$$ as $$(x,y) \rightarrow (0,0)$$ This gives $$\lim_{(x,y) \to (0,0)} (x^2 + y^2) \sin(\frac{1}{x^2 + y^2})$$ We use polar coordinates $$x = r\cos\theta$$, $$y =r\sin\theta$$ to simplify the limit above, giving $$\lim_{r \to 0} r^2 \sin\frac{1}{r^2}$$ Using the Squeeze Theorem, we can evaluate this limit. Since $$ -1 < \sin\theta <1 \hspace {0.1cm} \forall \hspace{0.1cm} \theta$$, we can set up the following inequality: $$-r^2 \leq r^2 \sin\frac{1}{r^2} \leq r^2$$ Taking the limit as $$r \rightarrow 0$$ gives $$0 \leq r^2 \sin\frac{1}{r^2} \leq 0$$ So by Squeeze Theorem, $$\lim_{(x,y) \to (0,0)} (x^2 + y^2) \sin(\frac{1}{x^2 + y^2}) = 0$$ By the definition above, it follows that $$C=0$$. For part $$(b)$$, Taking the partial derivatives directly and simplifying, $$\delta_xf = 2x [ \sin(\frac{1}{x^2+y^2}) - \frac{1}{x^2+y^2}\cos(\frac{1}{x^2+y^2})]$$ $$\delta_yf = 2y [ \sin(\frac{1}{x^2+y^2}) - \frac{1}{x^2+y^2}\cos(\frac{1}{x^2+y^2})]$$

### Subject: Physics

Find the magnetic field at the origin of a wire parametrized by $$s = ae^{b\phi}, 0 < \phi < 2\pi N$$ using Biot-Savart law.

Biot-Savart Law states $$B = \frac{\mu_0I}{4\pi} \int_{}\frac{\vec{dl} \times (\vec{r}-\vec{r'})}{|\vec{r}-\vec{r'}|^3}$$ In this example, the parametrization $$s = ae^{b\phi}$$ spirals outward from the origin (Graph this using a caluclator to get an idea), where the current $$I$$ flows in the same direction. $$\vec{r}$$ is defined as the location we are trying to solve for the field at. So it follows $$\vec{r} = 0$$. $$\vec{r'}$$ is defined as the location of the source. So it follows $$\vec{r} = ae^{b\phi} \hat{s}$$. $$\vec{dl}$$ is defined as the infinitesimal path along the wire. So we have $$\vec{dl} = ds\hat{s} + s d\phi \hat{\phi} $$ Substituting these values into Biot-Savart's Law, we have $$B = \frac{\mu_0I}{4\pi} \int_{}\frac{ds \hat{s} \times (-ae^{b\phi}\hat{s}) + sd\phi\hat{\phi}\times(-ae^{b\phi}\hat{s})}{(ae^{b\phi})^3}$$ Note the first term in the integral is equal to $$0$$ as $$\hat{s} \times \hat{s} = 0$$. This reduces the integral to $$B = \frac{\mu_0I}{4\pi} \int_{}\frac{ae^{b\phi} \cdot ae^{b\phi}(\hat{\phi} \times (-\hat{s}))d\phi}{(ae^{b\phi})^3}$$ Reducing, $$B = \frac{\mu_0I}{4\pi} \int_{}\frac{d\phi}{ae^{b\phi}} \hat{z}$$ $$B = \frac{\mu_0I}{4\pi} \frac{1}{a} \hat{z} \int_{}\frac{d\phi}{e^{b\phi}}$$ Solving the integral, $$B = \left. \frac{\mu_0I}{4\pi} \frac{1}{ab} \hat{z} \frac{1}{e^{b\phi}} \right|_{0}^{2 \pi N}$$ Evaluating gives the answer: $$B = \frac{\mu_0I}{4\pi} (1-\frac{1}{e^{b 2 \pi N}}) \hat{z}$$

### Subject: Differential Equations

Solve the following IVP: $$\frac{dy}{dx} = \frac{3x^2 + 4x - 4}{2y - 4} \hspace{1 cm} y(1) = 3$$

The differential equation above is separable. Rewrite the equation as: $$(2y - 4) \hspace{0.1cm} dy = 3x^2 + 4x - 4 \hspace{0.1cm} dx$$ Integrate on both sides $$\int_{} 2y-4 \hspace{0.1cm} dy = \int_{} 3x^2 + 4x - 4 \hspace{0.1cm} dx$$ $$y^2 - 4y = x^3 + 2x^2 - 4x + C $$ Substitute the initial condition $$y(1) = 3$$ to determine the value of $$C$$, which gives $$(3)^2 - 4(3) = (1)^3 + 2(1)^2 - 4(1) + C \implies C=-2$$ The implicit solution is then $$y^2 - 4y = x^3 + 2x^2 - 4x - 2 $$

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