# Jason's calculus questions

#### Prove It

##### Well-known member
MHB Math Helper
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

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MHB Math Helper
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

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MHB Math Helper
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

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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

##### Well-known member
MHB Math Helper
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*}