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That's an application of the chain rule.

$\noindent It's asking for how the point $M$ moves as $P$ moves. Since we have parametric equations for $x_M$ and $y_M$, namely $x_M = \frac{t}{2}$, $y_M = \frac{1}{2} \left(4t+3\right)$, where $t \in \mathbb{R}$, we can eliminate this parameter to get a Cartesian equation for the locus. $

Re: MATH1231/1241/1251 SOS Thread $\noindent Note $\sec ^{4} x = \sec^{2}x \cdot \sec^{2} x = \left(\tan^{2}x + 1\right)\sec^{2} x$. Since $\sec^{2} x$ is the derivative of $\tan x$, it follows that $\int \sec^{4}x \, \mathrm{d}x = \int \left(\tan^{2} x+1\right)\sec^{2} x \, \mathrm{d}x =...

Re: MATH1231/1241/1251 SOS Thread $\noindent The partial derivative with respect to $y$ is just $2ye^x$.$ $\noindent (Remember, $x$ is just treated as a constant.)$ $\noindent By the way, the above answer was assuming the function was $e^x y^2$. If the function was instead $e^{xy^2}$, then...

I suppose a typical 3U method that completely avoids implicit differentiation (and is thus unnecessarily tedious) is to explicitly find x(t) and y(t) (as functions of t). They are: x(t) = 30 - 50t, and y(t) = 40 - 40t. Then you can find z explicitly in terms of t by writing z(t) =...

Nope, only 4U.

Note z is the distance between the two cars at any time t. Also, x is the distance of the car on the horizontal axis from the intersection, and y is the distance of the car on the vertical axis from the intersection. The rate we are asked to find is dz/dt. By Pythagoras, z^2 = x^2 + y^2. The...

Yeah (pretty sure)

Implicit differentiation is only part of the HSC 4U syllabus.

1) Since y = 2/x, we have y' = -2/x^2. 2) Note y = x/(1-4x). Now we can use the quotient rule to get y' . 3) $\noindent Use implicit differentiation to find $y'$. It gives us $2xy^3 + x^2 \cdot 3y^2 y' + y^2 + x\cdot 2y y' = 0$. Now we can solve this equation for $y'$.$

You may want to check your textbook, lots of these are very basic and there's not much point doing this topic if you haven't learnt the basics yet.

Which parts do you need help with? Do you know how to differentiate/graph/etc. the inverse sine function?

$\noindent Let $y = \cos ^{-1} \left( \frac{1-x^2}{1+x^2}\right)$. To find $y^\prime$, use the chain rule. The derivative of the inner function is $\frac{-2x \left(1+x^2\right) -\left(1-x^2\right)\cdot 2x}{\left(1+x^2\right)^2} = \frac{-4x}{\left(1+x^2\right)^2}$. So $y^\prime =...

$\noindent The interest is paid semi-annually at 3\% per-period. E.g. If you started with \$1, it'd grow to $\$ 1.03$ after 6 months, then $\$ \left(1.03)\times 1.03$ after 1 year, then $\$ 1.03 ^3$ after $18$ months, etc. In other words, one dollar will grow to $R^n$ dollars after $n$...

'Authentic' physics is highly mathematical.

Re: MATH1231/1241/1251 SOS Thread Integral of cos(u) is just sin(u), not (1/u)*sin(u).

Just generally we can always solve these quadratic equations using the quadratic formula (you would generally need to compute the square root of a complex number to use this). Sometimes there are faster ways though. For the last one, have you tried multiplying top and bottom by the conjugate of...

Re: MATH1231/1241/1251 SOS Thread Of course, an obvious and trivial example would be the ODE 0y + y' = 0, i.e. y' = 0.

Re: MATH1231/1241/1251 SOS Thread $\noindent An exact ODE is of the form $I + J \frac{\mathrm{d}y}{\mathrm{d}x} = 0$, where $I$ and $J$ are functions of $x$ and $y$. If this is to be a linear ODE, the $J$ must be a function of $x$ alone, say $g(x)$ (no dependence on $y$). We would also need to...

$\noindent By looking at what the contributions are to the instantaneous rate of increase of the balance, we see that $B^\prime (t) = C + Dt + RB(t)$, with $B(0) = A$ given. This intial value problem can be solved easily since the ODE is first-order and linear.$