Calculus Practice Paper 1

Question 1 §

Question 1ai §

$${\int }_0^{\frac{\pi }{2}}{\sin }^3(x)\cos (x)dx$$

Using the substitution $u=\sin x$, we get $dx=\frac{du}{\cos x}$, thus the expression becomes…

$${\int }_0^1{u}^{3}du={\left[\frac{1}{4}{u}^4\right]}_0^1=\frac{1}{4}-0=\frac{1}{4}$$

Question 1aii §

$${\int }_0^{\frac{\pi }{2}}{\cos }^2(x)dx$$

If we use the identity $\cos 2\theta \equiv 2{\cos }^2\theta -1$, rearranged to be ${\cos }^2\theta \equiv \frac{1+\cos 2\theta }{2}$, then we can get the expression…

$${\int }_0^{\frac{\pi }{2}}\frac{1}{2}+\frac{1}{2}\cos (2\theta )dx={\left[\frac{1}{2}x+\frac{1}{4}\sin (2\theta )\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{4}-0=\frac{\pi }{4}$$

Question 1bi §

$$\begin{array}{rl}{\int }_0^3{\int }_1^2{x}^2{y}^{2}dxdy& ={\int }_0^3{\left[\frac{1}{3}{x}^3{y}^2\right]}_1^{2}dy\\ & ={\int }_0^3\left(\frac{8}{3}{y}^2-\frac{1}{3}{y}^2\right)dy\\ & ={\int }_0^3\frac{7}{3}{y}^{2}dy\\ & ={\left[\frac{7}{9}{y}^3\right]}_0^3\\ & =\frac{27\cdot 7}{9}-0\\ & =21\end{array}$$

Question 1bii §

$$\begin{array}{rl}{\int }_0^1{\int }_x^{\sqrt{x}}ydydx& ={\int }_0^1{\left[\frac{1}{2}{y}^2\right]}_x^{\sqrt{x}}dx\\ & ={\int }_0^1\left(\frac{1}{2}x-\frac{1}{2}{x}^2\right)dx\\ & =\frac{1}{2}-\frac{1}{2}-0-0\\ & =0\end{array}$$

Question 1c §

$$f(x,y)=x^4-x^2+y^2$$

First, finding the partial derivatives in terms of $x$ and $y$…

$$\frac{\delta f}{\delta x}=4x^3-2x,\frac{\delta f}{\delta y}=2y$$

These are both equal to 0 when the function is stationary, therefore we have the system of simultaneous equations:

$$\{\begin{array}{l}x(4x^2-2)=0\\ 2y=0\end{array}$$

Therefore, the stationary points are $(0,0)$, $(\frac{1}{2},0)$, and $(-\frac{1}{2},0)$.

To now determine the nature of these points, we’ll calculate the second partial derivatives and then calculate $D$:

$$D=\left(\frac{{\partial }^{2}f}{\partial x^2}\right)\left(\frac{{\partial }^{2}f}{\partial y^2}\right)-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^2$$

These second partial derivatives are as follows:

$$\frac{{\partial }^{2}f}{\partial x^2}=12x^2-2,\frac{{\partial }^{2}f}{\partial y^2}=2,\frac{{\partial }^{2}f}{\partial x\partial y}=0$$

$$\begin{array}{rl}D& =\left(\frac{{\partial }^{2}f}{\partial x^2}\right)\left(\frac{{\partial }^{2}f}{\partial y^2}\right)-{\left(\frac{{\partial }^{2}f}{\partial x\partial y}\right)}^2\\ & =(12x^2-2)(2)-(0)\\ & =24x^2-4\end{array}$$

Using this determinant $D$, we can find the nature of the points based on the following:

• If $D>0$ and $\frac{{\partial }^{2}f}{\partial x^2}>0$, then $(x_0,y_0)$ is a local minimum.
• If $D>0$ and $\frac{{\partial }^{2}f}{\partial x^2}<0$, then $(x_0,y_0)$ is a local maximum.
• If $D<0$, then $(x_0,y_0)$ is a saddle point.
• If $D=0$, the test is inconclusive.

Hence,

$$\begin{array}{cc}\text{Saddle point:}& (0,0)\\ \text{Local minima:}& \left(\frac{1}{2},0\right),\left(-\frac{1}{2},0\right)\end{array}$$

Question 2 §

Question 2ai §

Given the parametric curve where $x=t^2-1$ and $y=t^3-3t$,

$$\frac{dx}{dt}=2t,\frac{dy}{dt}=3t^2-3$$

$$\frac{dy}{dx}=\frac{dy}{dt}\cdot {\left(\frac{dx}{dt}\right)}^{-1}=\frac{3t^2-3}{2t}$$

If we repeat the process once more, then…

$$\frac{d^{2}x}{d{t}^2}=2,\frac{d^{2}y}{d{t}^2}=6t$$

$$\frac{d^{2}y}{d{x}^2}=\frac{d^{2}y}{d{t}^2}\cdot {\left(\frac{d^{2}x}{d{t}^2}\right)}^{-1}=\frac{6t}{2}=3t$$

Question 2aii §

Converting the polar equation $r=4\cos \theta$ to Cartesian then becomes…

$$r^2=4r\cos \theta ⟺x^2+y^2=4x⟺x^2-4x+y^2=0$$

$$(x-2{)}^2+y^2=2^2$$

Hence the centre of circle is $(2,0)$ with radius $2$.

Question 2b §

Finding the first four terms in the Taylor series of $\sqrt{x}$ about $x=1$,

Using the formula,

$$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{\prime \prime }(a)}{2!}(x-a{)}^2+\frac{f^{\prime \prime \prime }(a)}{3!}(x-a{)}^3+\cdots$$

We get the result:

$$f(x)=1+\frac{1}{2}{x}^{-\frac{1}{2}}(x-1)-\frac{1}{4}{x}^{-\frac{3}{2}}(x-1{)}^2+\frac{3}{8}{x}^{-\frac{5}{2}}(x-1{)}^3-\cdots$$

Question 2c §

To calculate the volume under a surface $z=xy$ and above the region in the $xy$ plane bounded $y=x^2$ and $y=\sqrt{2-x^2}$ on the interval $0\le x\le 1$.

Formulating this into a mathematical expression…

\begin{align*} V&=\int_{0}^{1}\int^\sqrt{2-x^{2}}_{x^{2}} (xy) dydx\\ &= \int_{0}^{1}\left[\frac{1}{2}xy^{2}\right]_{x^{2}}^\sqrt{2-x^{2}}dx\\ &= \int_{0}^{1}\left(\frac{1}{2}x(2-x^{2}) - \frac{1}{2}x^{5}\right)dx\\ &= \int_{0}^{1}x- \frac{1}{2}x^{3}- \frac{1}{2}x^{5}dx\\ &= \left[\frac{1}{2}x^{2}- \frac{1}{8}x^{4}- \frac{1}{12}x^{6}\right]_{0}^{1}\\ &= \frac{1}{2}- \frac{1}{8}- \frac{1}{12}\\ &= \frac{12}{24}- \frac{3}{24} - \frac{2}{24}\\ &= \frac{7}{24} \end{align*}

Question 3 §

Question 3ai §

$$\begin{array}{rl}(1-i)(1+i)& =1^2-i^2\\ & =1+1\\ & =2+0i\end{array}$$

Question 3aii §

$$\begin{array}{rl}\frac{1-i}{1+i}& =\frac{1-i}{1+i}\cdot \frac{1-i}{1-i}\\ & =\frac{(1-i{)}^2}{2}\\ & =\frac{1^2-i^2-2i}{2}\\ & =1-i\end{array}$$

Question 3aiii §

$$\begin{array}{rl}{\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)}^{20}& ={\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)}^{20}\\ & =\cos 5\pi +i\sin 5\pi \\ & =-1+0i\end{array}$$

Question 3b §

$$x^{2}z+y^{2}z+2x{y}^2-z^3=0$$

We can use this to calculate the two partial derivatives, $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$.

$$2x\frac{\partial z}{\partial x}+y^2\frac{\partial z}{\partial x}+2y^2-3z^2\frac{\partial z}{\partial x}=0⟺\frac{\partial z}{\partial x}=-\frac{2y^2}{2x+y^2-3z^2}$$

$$x^2\frac{\partial z}{\partial y}+2y\frac{\partial z}{\partial y}4xy-3z^2\frac{\partial z}{\partial y}=0⟺\frac{\partial z}{\partial y}=\frac{1}{x^2+2y-3z^2}$$

Question 3c §

$$\begin{array}{rl}{z}^4& =\frac{\sqrt{3}}{2}+\frac{1}{2}i\\ & =\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\end{array}$$

$$\begin{array}{rl}⟺z& =\exp {\left(\frac{\pi }{6}+2n\pi \right)}^{\frac{1}{4}}\\ & =\exp \left(\frac{\frac{\pi }{6}+2n\pi }{4}\right)\\ & =e^{\frac{\pi }{24}},e^{\frac{13\pi }{24}},e^{\frac{25\pi }{24}},e^{\frac{37\pi }{24}}\end{array}$$

Question 4 §

Question 4a §

$$f(t)=t\ln t-t⟺f^{\prime }(t)=\ln t$$

$$g(t)=-\frac{t}{\sqrt{2\pi }}\exp \left(-\frac{t^2}{2}\right)⟺g^{\prime }(t)=\frac{t^2}{\sqrt{2}\pi }\exp \left(-\frac{t^2}{2}\right)$$

$$\begin{array}{rl}h(t)=\frac{t-1}{t^2+1}⟺h^{\prime }(t)& =\frac{(1)(t^2+1)-(t-1)(2t)}{t^2+1}\\ & =\frac{-t^2+2t+1}{t^2+1}\end{array}$$

Question 4b §

$$\begin{array}{rl}\underset{u\to 3}{\lim }\frac{u^2-u-6}{u-3}& =\underset{u\to 3}{\lim }\frac{2u-1}{1}\\ & =5\text{ by L’Hoptial’s Rule}\end{array}$$

$$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{x}{\ln (x+1)}& =\frac{x}{\frac{1}{x+1}}\\ & =0\end{array}$$

Question 4c §

$$f(x)=x^2+\frac{2}{x}$$

Question 4ci §

To find the vertical asymptote of the function, $f(x)$, we must evaluate where the function tends to infinity: this is at $x=0$.

Question 4cii §

To find the local extremum of the function, we must first find its derivative and solve it when equal to zero,

$$f^{\prime }(x)=2x-\frac{2}{x^2}:f^{\prime }(x)=0⟺x=1⟺y=3$$

To find the nature of this point, we’ll evaluate the second derivative at that point,

$$f^{\prime \prime }(x)=2+\frac{4}{x^3}:f^{\prime \prime }(1)=6>0\therefore \text{local minimum at }(1,3)$$

Question 4ciii §

The function is increasing when the first derivative is greater than zero, and is decreasing when the first derivative is less than zero, hence…

$$f^{\prime }(x)=2x-\frac{2}{x^2}:\{\begin{array}{l}{f}^{\prime }(x)<0⟺x<1\\ f^{\prime }(x)>0⟺x>1\end{array}$$

Therefore, the function is decreasing on $x<0$ and $0<x<1$, and increasing $x>1$.