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MATH 131 Fall 2002

  1. a) (10) Find the derivative ${\displaystyle \frac{dy}{dx}}$ of the function $y$, implicitly defined by the equation

(x+2y)^2   =   3x^2y + 6x.

    Solution: Differentiating with respect to $x$ we have

2(x + 2y)(1 + 2y') = 6xy +3x^2y' + 6 \\

    Solving for $y'=\frac{dy}{dx}$, we get

y'(4(x+2y)-3x^2) &=& 6xy + 6 - 2(x+2y) \\
y' &=& \frac{6xy+6 - 2(x+2y)}{4(x+2y)-3x^2}

    b) (5) Find an equation of the tangent line at the point $(1,1)$.

    Solution: The equation is $y-1=m(x-1)$, where $m$ is the value of ${\displaystyle \frac{dy}{dx}}$ when $x=y=1$. From (a),

m = \frac{6\cdot 1 \cdot 1+6 - 2(1+2\cdot 1)}{4(1+2\cdot 1)-3\cdot1^2}
= \frac 2 3 .

  2. (15) A helicopter is hovering in place 300 meters above a highway. A radar on the helicopter measures the rate of change of the distance between the helicopter and approaching cars. What is the measured rate, for a car driving at a speed of 100 kilometers per hour, when it is 500 meters from the helicopter?

    Note: 1 kilometer = 1000 meters.

    Solution: Let $y$ be the distance from the car to the helicopter, and let $x$ be the distance from the car to the point of the road directly below the helicopter. The problem states that $\frac{dx}{dt}=-100 \textrm{km/hour}$, and asks us to find the value of $\frac{dy}{dt}$ at the moment when $y=500 \textrm{m}$. By the Pythagorean Theorem $x=\sqrt{y^2-300^2}$ at any time. Differentiating $x$ and $y$ with respect to time, we get

\frac{dx}{dt}=\frac{2y}{2\sqrt{y^2 -300^2}}\frac{dy}{dt}

    so that

\frac{dy}{dt}=\frac{\sqrt{y^2 -300^2}}{y}\frac{dx}{dt}=\frac{\sqrt{y^2 -300^2}}{y}-100\textrm{km/hour}

    Thus, when $y=500 \textrm{m}$, $\frac{dy}{dt}=\frac{\sqrt{500^2
-300^2}}{500}-100\textrm{km/hour}=\frac 4 5 -100\textrm{km/hour}=-80\textrm{km/hour}$.

  3. (15) a) Find the linearization of the function ${\displaystyle f(x)=\sqrt{x}}$ at the point $a=36$.

    Solution: $\sqrt{x}=f(x) \approx L(x) =

    b) Use the linearization to approximate $\sqrt{35}$.

    Solution: $\sqrt{35}\approx L(35) = 6+\frac{1}{12}(35-36)=\frac{71}{12}$

    c) Is your approximation greater than or less than the actual value? Justify your answer.

    Solution: $f''(x)=\frac{-1}{4}x^{-\frac 3 2}$ is negative for $x>0$, so that the graph of $f(x)=\sqrt{x}$ is concave down on the domain of $f$. This implies that the tangent line at the point (36,6) will lie above the graph of $f(x)$ for $x>0$. Since the graph of $L(x)$ in (a) is this tangent line, our approximation, $L(35)$, will be greater than $\sqrt{35}$.

  4. (15) Let $f(x)=0.25 x^4 + x^3 -5x^2$.

    a) Find the intervals on which $f(x)$ is increasing and those on which it is decreasing.

    Solution: $f'(x)=x^3+3x^2-10x=x(x+5)(x-2)$. The zeros of $f'(x)$ are $x=-5,0,2$. Since $f$ is differentiable for all $x$, these are the only critical points of $f$, so that we only need to test the sign of the derivative on the intervals $(-\infty,-5)$, $(-5,0)$, $(0,2)$, and $(2,\infty)$. We have:

    $(-\infty,-5)$ $f'(x)<0$ decreasing
    $(-5,0)$ $f'(x)>0$ increasing
    $(0,2)$ $f'(x)<0$ decreasing
    $(2,\infty)$ $f'(x)>0$ increasing

    b) Find all the local maxima and local minima of $f$.

    Solution: We have critical points $x=-5,0,2$. Using the chart from (a) we find by the First Derivative Test that $f(-5)=-93.75$ is a local minimum, $f(0)=0$ is a local maximum, and $f(2)=-8$ is a local minimum.

    c) Find the values of the global maximum and global minimum of $f$ over the interval $-6\leq x \leq 4$.

    Solution: We evaluate $f$ at the endpoints of the interval: $f(-6)=-72$ and $f(4)=48$. Comparing these two values with the local minima and maxima we find that $f(-5)=-93.75$ is the global minimum and $f(4)=48$ is the global maximum on this interval.

    d) Graph the function over the interval $-6\leq x \leq 4$, indicating all the local and global extrema.

  5. (10) A particle is said to undergo a simple harmonic motion, if its position $s(t)$ at time $t$ is given by the formula

s(t)   =   A\cos(\omega t + \delta)

    for some constant real numbers $A$, $\omega$, and $\delta$, with $A>0$ and $\omega>0$.

    a) Express the acceleration $a(t)$ in terms of the time variable $t$ and the constants $A$, $\omega$, and $\delta$.

    Solution: $a(t)=s''(t)=-\omega^2 A \cos (\omega t + \delta)$

    b) Assume now, that $s(t)=3\cos(2t)$, (so $A=3$, $\omega=2$ and $\delta=0$). Find the time intervals, where both the position $s(t)$ and velocity $v(t)$ are positive.

    Solution: $s(t)=3\cos(2t)>0$ when $(4k-1)\frac{\pi}{2}<2t<
(4k+1)\frac{\pi}{2}$ for any integer $k$. $v(t)=s'(t)=-6\sin(2t)>0$ when $(4k-2)\frac{\pi}{2}<2t<(4k)\frac{\pi}{2}$ for any integer $k$. Thus both are positive when $2t$ is in the intersection of these families of intervals, i.e., when $(4k-1)\frac{\pi}{2}<2t<
(4k)\frac{\pi}{2}$, or $(4k-1)\frac{\pi}{4}<t<
(4k)\frac{\pi}{4}$ for any intger $k$.

    c) Is the particle in part b) speeding up or slowing down, when both its position and velocity are positive? justify your answer.

    Solution: When $s(t)=3\cos(2t)$, $a(t)=s''(t)=-12\cos(2t)$ which is clearly negative whenever $s(t)$ is positive (they have opposite signs). So in particular, when $s(t)$ and $v(t)$ are both positive, $a(t)$ will be negative. Thus velocity and acceleration will have opposite directions so that the particle will be slowing down.

  6. (15) Consider the function ${\displaystyle f(x) = \ln(x^2+1).}$

    a) Determine the intervals on which the function is concave upward and those on which it is concave downward.

    Solution: $f''(x)=\frac{d}{dx}\frac{2x}{x^2+1}=\frac{2(1-x^2)}{(x^2+1)^2}$ exists for all $x$, so that the only critical points of $f'(x)$ will be where $f''(x)=0$, i.e., where $1-x^2=0$, i.e., $x=\pm 1$. Thus we have to test the sign of the second derivative on the intervals $(-\infty,-1)$, $(-1,1)$, and $(1,\infty)$. We have

    $(-\infty,-1)$ $f''(x)<0$ concave down
    $(-1,1)$ $f''(x)>0$ concave up
    $(1,\infty)$ $f''(x)<0$ concave down

    b) Determine all points of inflection, if any. Justify your answer!

    Solution: Since $f''(x)$ exists for all $x$, our only candidates for inflection points are where $f''(x)=0$, i.e., $x=\pm 1$. From (a), we know that concavity changes at each of these points so that both are inflection points.

  7. (15) Use logarithmic differentiation to find the derivative $\frac{dy}{dx}$ for

y   =   \frac{\sqrt{x}(x^2+1)^7}{\left(x^x\right)}


\ln y &=& \ln \left(\frac{\sqrt{x}(x^2+1)^7}{\left(x^x\right)}...
- \ln x -1 \right)

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Eyal Markman 2002-12-04