MHB Jason's calculus questions

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The discussion focuses on solving calculus problems involving cubic functions and their properties. A cubic function is derived from the conditions that it touches a line at a specific point and has a stationary point at another. The coefficients of the cubic function are determined to be a = 4/3, b = -11/2, c = -3, and d = 15/2. Additionally, the discussion explores stationary points of various functions, including a particle's motion and the instantaneous rate of change of a production output function. The final results highlight the importance of derivatives in finding stationary points and rates of change in different contexts.
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The graph of $\displaystyle \begin{align*} y = a\,x^3 + b\,x^2 + c\,x + d \end{align*}$ touches the line $\displaystyle \begin{align*} 2\,y + 6\,x = 15 \end{align*}$ at the point $\displaystyle \begin{align*} A \left( 0, \frac{15}{2} \right) \end{align*}$ and has a stationary point at $\displaystyle \begin{align*} B\left( 3, -6 \right) \end{align*}$. Find the values of $\displaystyle \begin{align*} a, b, c \end{align*}$ and $\displaystyle \begin{align*} d \end{align*}$.

Since the two functions touch at $\displaystyle \begin{align*} A\left( 0, \frac{15}{2} \right) \end{align*}$ that means that this point lies on the cubic function. Thus

$\displaystyle \begin{align*} \frac{15}{2} &= a\left( 0 \right) ^3 + b\left( 0 \right) ^2 + c\left( 0 \right) + d \\ \frac{15}{2} &= d \end{align*}$

So we can rewrite the cubic as $\displaystyle \begin{align*} y = a\,x^3 + b\,x^2 + c\,x + \frac{15}{2} \end{align*}$.

Also since this is a point where the line just touches the cubic, that means the line is a tangent to the cubic at that point. Thus the gradient of the curve at that point is equal to the gradient of the line.

The gradient of the line is $\displaystyle \begin{align*} -3 \end{align*}$ since the line can be rewritten as $\displaystyle \begin{align*} y = -3\,x + \frac{15}{2} \end{align*}$, thus

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= 3\,a\,x^2 + 2\,b\,x + c \\ -3 &= 3\,a\left( 0 \right) ^2 + 2\,b\left( 0 \right) + c \\ -3 &= c \end{align*}$

So we can rewrite the cubic as $\displaystyle \begin{align*} y = a\,x^3 + b\,x^2 - 3\,x + \frac{15}{2} \end{align*}$.


Since there is a stationary point on the cubic at $\displaystyle \begin{align*} B\left( 3, -6 \right) \end{align*}$, that means that the point lies on the cubic and also the derivative is 0 at that point.

$\displaystyle \begin{align*} -6 &= a\left( 3 \right) ^3 + b\left( 3 \right) ^2 - 3 \left( 3 \right) + \frac{15}{2} \\ -6 &= 27\,a + 9\,b - 9 + \frac{15}{2} \\ -6 &= 27\,a + 9\,b - \frac{3}{2} \\ -\frac{9}{2} &= 9\,a + 3\,b \\ -3 &= 6\,a + 2\,b \end{align*}$

Also

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= 3\,a\,x^2 + 2\,b\,x -3 \\ 0 &= 3\,a\left( 3 \right) ^2 + 2\,b\left( 3 \right) - 3 \\ 3 &= 27\,a + 6\,b \\ 1 &= 9\,a + 2\,b \end{align*}$

Solving these resulting equations simultaneously gives

$\displaystyle \begin{align*} 1 - \left( -3 \right) &= \left( 9\,a + 2\,b \right) - \left( 6\,a + 2\,b \right) \\ 4 &= 3\,a \\ a &= \frac{4}{3} \end{align*}$

and

$\displaystyle \begin{align*} -3 &= 6\left( \frac{4}{3} \right) + 2\,b \\ -3 &= 8 + 2\,b \\ -11 &= 2\,b \\ b &= -\frac{11}{2} \end{align*}$

So the cubic is $\displaystyle \begin{align*} y = \frac{4}{3}\,x^3 - \frac{11}{2}\,x^2 - 3\,x + \frac{15}{2} \end{align*}$.
 
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Find the $\displaystyle \begin{align*} x \end{align*}$ co-ordinates, in terms of $\displaystyle \begin{align*} n \end{align*}$, of the stationary points of the curve with equation $\displaystyle \begin{align*} y = \left( 2\,x - 1 \right) ^n \left( x + 2 \right) \end{align*}$, where $\displaystyle \begin{align*} n \end{align*}$ is a natural number.

Stationary points occur where the derivative is 0, so

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \left( 2\,x - 1 \right) ^n \left( 1 \right) + 2\,n\,\left( 2\,x - 1 \right)^{n-1} \left( x + 2 \right) \\ 0 &= \left( 2\,x - 1 \right) ^{n - 1} \left[ \left( 2\,x - 1 \right) + 2\,n\,\left( x + 2 \right) \right] \\ 0 &= \left( 2\,x - 1 \right) ^{n - 1} \left( 2\,x - 1 + 2\,n\,x + 4\,n \right) \end{align*}$

So

$\displaystyle \begin{align*} \left( 2\,x - 1 \right) ^{n - 1} &= 0 \\ 2\,x - 1 &= 0 \\ 2\,x &= 1 \\ x &= \frac{1}{2} \end{align*}$

and

$\displaystyle \begin{align*} 2\,x - 1 + 2\,n\,x + 4\,n &= 0 \\ \left( 2 + 2\,n \right) x &= 1 - 4\,n \\ x &= \frac{1 - 4\,n }{2 + 2\,n} \end{align*}$
 
Find the co-ordinates of the stationary points of the curve with equation $\displaystyle \begin{align*} y = \frac{x}{x^2 + 1} \end{align*}$.

Stationary points occur where the derivative is 0, so

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1\left( x^2 + 1 \right) - x \left( 2\,x \right)}{\left( x^2 + 1 \right) ^2} \\ 0 &= \frac{x^2 + 1 - 2\,x^2}{\left( x^2 + 1 \right)^2} \\ 0 &= \frac{1 - x^2}{\left( x^2 + 1 \right) ^2} \\ 0 &= 1 - x^2 \\ x^2 &= 1 \\ x &= \pm 1 \end{align*}$

When $\displaystyle \begin{align*} x = -1 \end{align*}$

$\displaystyle \begin{align*} y &= \frac{-1}{\left( -1 \right) ^2 + 1 } \\ &= \frac{-1}{1 + 1} \\ &= -\frac{1}{2} \end{align*}$

and when $\displaystyle \begin{align*} x = 1 \end{align*}$

$\displaystyle \begin{align*} y &= \frac{1}{1^2 + 1} \\ &= \frac{1}{1 + 1} \\ &= \frac{1}{2} \end{align*}$

Thus the stationary points are $\displaystyle \begin{align*} \left( -1, -\frac{1}{2} \right) \end{align*}$ and $\displaystyle \begin{align*} \left( 1, \frac{1}{2} \right) \end{align*}$.
 
A particle moves in a straight line such that its position, $\displaystyle \begin{align*} x \end{align*}$ cm, relative to a point $\displaystyle \begin{align*} O \end{align*}$, at time $\displaystyle \begin{align*} t \end{align*}$ seconds is given by the equation $\displaystyle \begin{align*} x\left( t \right) = 8+ 2\,t - t^2 \end{align*}$. Find:

a) its initial position
b) its initial velocity
c) when and where the velocity is zero
d) its acceleration at time $\displaystyle \begin{align*} t \end{align*}$.

a) Initially $\displaystyle \begin{align*} t = 0 \end{align*}$ so

$\displaystyle \begin{align*} x \left( 0 \right) &= 8 + 2\left( 0 \right) - 0^2 \\ &= 8 \end{align*}$

b) The velocity is the derivative of position, so

$\displaystyle \begin{align*} v\left( t \right) &= 2 - 2\,t \\ v \left( 0 \right) &= 2 - 2 \left( 0 \right) \\ &= 2 \end{align*}$

c)
$\displaystyle \begin{align*} 0 &= 2 - 2\,t \\ 2\,t &= 2 \\ t &= 1 \\ \\ x\left( 1 \right) &= 8 + 2\left( 1 \right) - 1^2 \\ &= 8 + 2 - 1 \\ &= 9 \end{align*}$

d) Acceleration is the derivative of velocity, so

$\displaystyle \begin{align*} a\left( t \right) &= -2 \end{align*}$
 
A particle is moving in a straight line such that its position, $\displaystyle \begin{align*} x \end{align*}$ cm, relative to a point $\displaystyle \begin{align*} O \end{align*}$ at time $\displaystyle \begin{align*} t \end{align*}$ seconds, is given by $\displaystyle \begin{align*} x\left( t \right) = \sqrt{2\,t^2 + 2} \end{align*}$. Find the acceleration as a function of $\displaystyle \begin{align*} t \end{align*}$.

Acceleration is the second derivative of position, so

$\displaystyle \begin{align*} x\left( t \right) &= \left( 2\,t^2 + 2 \right) ^{\frac{1}{2}} \\ \\ v\left( t \right) &= 4\,t \left( \frac{1}{2} \right) \left( 2\,t^2 + 2 \right) ^{-\frac{1}{2}} \\ &= 2\,t \, \left( 2\,t^2 + 2 \right) ^{-\frac{1}{2}} \\ \\ a\left( t \right) &= 2\,\left( 2\,t^2+ 2 \right) ^{-\frac{1}{2}} + 2\,t \left( -\frac{1}{2} \right) \left( 2\,t^2 + 2 \right) ^{-\frac{3}{2}} \\ &= 2\,\left( 2\,t^2 + 2 \right) ^{-\frac{1}{2}} - t \,\left( 2\,t^2 + 2 \right) ^{-\frac{3}{2}} \end{align*}$

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A manufacturing company has a daily output on day $\displaystyle \begin{align*} t \end{align*}$ of a production run given by $\displaystyle \begin{align*} y = 6000\,\left( 1 - \mathrm{e}^{-0.5\,t} \right) \end{align*}$, where the first day of the production run is $\displaystyle \begin{align*} t = 0 \end{align*}$. Find the instantaneous rate of change of output $\displaystyle \begin{align*} y \end{align*}$ with respect to $\displaystyle \begin{align*} t \end{align*}$ on the 10th day.

The 10th day is when $\displaystyle \begin{align*} t = 9 \end{align*}$.

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} &= 6000\,\left( -0.5\,\mathrm{e}^{-0.5\,t} \right) \\ &= -3000\,\mathrm{e}^{-0.5\,t} \\ &= -3000\,\mathrm{e}^{-0.5 \cdot 9} \\ &= -3000\,\mathrm{e}^{-4.5} \end{align*}$

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The mass, $\displaystyle \begin{align*} m \end{align*}$ kg, of radioactive lead remaining in a sample $\displaystyle \begin{align*} t \end{align*}$ hours after observation began is given by $\displaystyle \begin{align*} m = 2\,\mathrm{e}^{-0.2\,t} \end{align*}$. Express the rate of decay as a function of $\displaystyle \begin{align*} m \end{align*}$.

$\displaystyle \begin{align*} \frac{\mathrm{d}m}{\mathrm{d}t} &= -0.2\cdot 2\,\mathrm{e}^{-0.2\,t} \\ \frac{\mathrm{d}m}{\mathrm{d}t} &= -0.2\,m \end{align*}$
 
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Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

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