Exit Exam-Part 3: September 14, 2005

Problem 1. Find the derivatives for the following functions. Assume $a$, and $b$ are constants.

(i)

$$f(t) = 2t e^{(a\;t)} - \frac{1}{\sqrt{t}}$$

(ii)

$$y = 2^{\sin(x)} \cos( b \; x)$$

Answer. For (i) we have

$$f'(t) = 2 e^{a t} + 2 t a e^{at} - \left(-\frac{1}{2}\right) t^{-3/2}$$

since $$\frac{1}{\sqrt{t}} = t^{-1/2}$$. So

$$f'(t) = (2 + 2 t a) e^{at} + \frac{1}{2}t^{-3/2}\;.$$

For (ii), we start by noting that

$$\bigg(2^{\sin(x)}\bigg)' = \bigg(e^{\sin(x) \ln(2)}\bigg)' = \ln(2) \cos(x) e^{\sin(x) \ln(2)} = \ln(2) \cos(x) 2^{\sin(x)}$$

so

$$\frac{dy}{dx} = \ln(2) \cos(x) 2^{\sin(x)} \cos(b x) + 2^{\sin(x)} \bigg(-b \sin(x)\bigg)$$

or

$$\frac{dy}{dx} = \ln(2) \cos(x) cos(bx)2^{\sin(x)} -b \sin(b x) 2^{\sin(x)}\;.$$

Problem 2. Find the equations of the tangent lines to the graph of

$$2xy+y^2 = 1$$

where $x=1$. Answer. Note first that when $x=1$, we get

$$2y + y^2 = 1$$

which implies $y = - 1 + \sqrt{2}$ or $y = - 1 - \sqrt{2}$. So we have to consider two points. In order to find the equation of the tangent lines we will need the slope given by the derivative of $y$. Since the function is defined implicitly, we will use implicit differentiation. We have

$$2 y + 2x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$$

or

$$(2x+2y) \frac{dy}{dx} = -2 y$$

so

$$\frac{dy}{dx} = - \frac{2y}{2x+2y} = - \frac{y}{x+y}$$

So for the point $(1, -1+\sqrt{2})$, the slope is

$$m = - \frac{-1+\sqrt{2}}{\sqrt{2}}$$

so the equation is

$$y-(-1+\sqrt{2}) = - \frac{-1+\sqrt{2}}{\sqrt{2}} (x-1)\;.$$

And for the point $(1, -1-\sqrt{2})$, the slope is

$$m = - \frac{-1-\sqrt{2}}{-\sqrt{2}} = - \frac{1+\sqrt{2}}{\sqrt{2}}$$

so the equation is

$$y-(-1-\sqrt{2}) = - \frac{1+\sqrt{2}}{\sqrt{2}} (x-1)$$

Problem 3. Use derivatives to identify local maxima, local minima, and points of inflection. Then sketch the graph of the function.

$$\displaystyle f(x) = \frac{x^2}{x^2+1}$$

Answer. In order to find the local minima and maxima, we need to find the critical points of $f(x)$. Since $f(x)$ is differentiable everywhere we only look for the roots of $f'(x)$. So let us first find $f'(x)$. We have

$$f'(x) = \frac{2x(x^2+1)-x^2 2x}{(x^2+1)^2} = \frac{2x}{(x^2+1)^2}\;.$$

So $f'(x)=0$ if and only if $x=0$. Since

$$\left\{\begin{array}{lll} f'(x) < 0 & \mbox{when}& x < 0\;,\\ f'(x) > 0 & \mbox{when}& x > 0\;,\\ \end{array}\right.$$

the first derivative test implies that 0 is a local minima. In order to find inflection points, we need to compute the second derivative. We have

$$f''(x) = \frac{2(x^2+1)^2 - 2x 2 2x(x^2+1)}{(x^2+1)^4} = \frac{2 - 6x^2}{(x^2+1)^3}$$

So the critical of $f'(x)$ are given by $f''(x)=0$. It is easy to see that $f''(x)=0$ if and only if

$$x = \pm \frac{1}{\sqrt{3}}\;.$$

Since

$$\left\{\begin{array}{lll} f''(x) < 0 & \mbox{when}& x < -1/\s... ...(x) < 0 & \mbox{when}& 1/\sqrt{3} < x \;,\\ \end{array}\right.$$

we conclude that both points are inflection points.

Problem 4. Find values of $a$ and $b$ so that the function $f(x) = a\;x\;e^{-b\;x}$ has a local maximum at the point $(2,10)$.

Answer. Note first that $(2,10)$ must be on the graph which implies

$$2 a e^{-2b} = 10\;\;\mbox{or}\;\; a = \frac{5}{e^{-2b}} = 5 e^{2b}\;.$$

In order for $(2,10)$ to be a local maximum, this point must be a critical point of $f(x)$. Since

$$f'(x) = a e^{-b x} - ab xe^{-b x}\;,$$

so

$$0 = a e^{-2b} - ab 2e^{-2b}\;,$$

which implies

$$a (1-2b) = 0\;.$$

Since $a \neq 0$, we get

$$b = \frac{1}{2}\;,$$

so

$$a = 5 e^{2b} = 5 e\;.$$

Problem 5. A rectangle has one side on the x-axis and two corners on the top half of the unit circle (radius 1 centered at the origin). Find the maximum area of such rectangle. What are the coordinates of its vertices?

Answer. Let us introduce some variables. Denote the four vertices of the rectangle by

$$(x,0);\; (x,y);\; (-x,y);\;\; \mbox{and}\;\; (-x,0)\;.$$

We have $x^2+y^2 =1$ or $y = \sqrt{1-x^2}$. The area of the rectangle is

$$A(x) = 2xy = 2x\sqrt{1-x^2}\;.$$

In order to find the maximum of $A(x)$ we will need the derivative

$$A'(x) = 2 \sqrt{1-x^2} + 2x \frac{(-2x)}{2\sqrt{1-x^2}}$$

or

$$A'(x) = 2 \sqrt{1-x^2} - \frac{2x^2}{\sqrt{1-x^2}} = \frac{2-4x^2}{\sqrt{1-x^2}}\;.$$

So $A'(x) = 0$ if and only if

$$x^2 = \frac{1}{2}\;\;\mbox{or}\;\; x = \frac{1}{\sqrt{2}}\;.$$

So

$$y = \frac{1}{\sqrt{2}}\;.$$

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