Jason's calculus questions

[Prove It]

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 $\displaystyle \begin{align*} y = \frac{4}{3}\,x^3 - \frac{11}{2}\,x^2 - 3\,x + \frac{15}{2} \end{align*}$.

 

[Prove It]

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

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

 

[Prove It]

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 $\displaystyle \begin{align*} \left( 1, \frac{1}{2} \right) \end{align*}$.

 

[Prove It]

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

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

 

[Prove It]

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 $\displaystyle \begin{align*} m \end{align*}$.

$\displaystyle \begin{align*} \frac{\mathrm{d}m}{\mathrm{d}t} &= -0.2\cdot 2\,\mathrm{e}^{-0.2\,t} \\ \frac{\mathrm{d}m}{\mathrm{d}t} &= -0.2\,m \end{align*}$