# Jason's calculus questions

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\right) &= \sqrt{2\,t^2 + 2} \\ x''\left( t \right) &= \frac{\mathrm{d}}{\mathrm{d}t} \left[ \frac{\mathrm{d}}{\mathrm{d}t} \left( \sqrt{2\,t^2 + 2} \right) \right] \\ &= \frac{1}{2\sqrt{2\,t^2 + 2}} \cdot 4\,t \\ &= \frac{2\,t}{\sqrt{2\,t^2 + 2}} \\ &= \frac{2\,t}{\sqrt{2}\sqrt{t
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MHB
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*}.

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*}

## 1. What is "Jason's calculus questions"?

"Jason's calculus questions" is a set of mathematical problems related to calculus, a branch of mathematics that deals with the study of change and motion. These questions are created by Jason, a fictional character, to help students practice and improve their calculus skills.

## 2. Who can benefit from "Jason's calculus questions"?

Anyone who is studying or interested in calculus can benefit from "Jason's calculus questions". It is designed for students at various levels, from beginners to advanced learners, to help them deepen their understanding of calculus concepts and improve their problem-solving skills.

## 3. Are "Jason's calculus questions" suitable for self-study or should they be used in a classroom setting?

"Jason's calculus questions" can be used for both self-study and in a classroom setting. They are designed to be self-explanatory and come with detailed solutions, making them suitable for individual study. However, they can also be used in a classroom as practice problems or as part of a lesson plan.

## 4. How can "Jason's calculus questions" help me prepare for exams or tests?

"Jason's calculus questions" cover a wide range of topics and difficulty levels, making them an excellent resource for exam preparation. By practicing with these questions, you can improve your understanding of calculus concepts and gain confidence in solving similar problems that may appear on exams or tests.

## 5. Are there any additional resources or materials that come with "Jason's calculus questions"?

Yes, "Jason's calculus questions" come with detailed solutions and explanations for each problem. Additionally, there are also resources such as video tutorials and practice quizzes available to help further enhance your understanding of calculus concepts and improve your problem-solving skills.

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