Exit Exam-Part 4: October 11, 2005

Problem 1. Find $$\frac{d^2y}{dx^2}$$ for $x=t^3+t$ and $y=t^2$. Then decide whether the curve is concave up or down at $t=1$.

Answer. We have

$$\frac{dy}{dx} = \frac{\displaystyle \frac{dy}{dt}}{\displaystyle \frac{dx}{dt}}= \frac{2t}{3t^2+1}$$

Since

$$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) =... ...t} \left(\frac{2t}{3t^2+1}\right) \left(\frac{1}{3t^2+1}\right)$$

or

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

When $t=1$, we get

$$\frac{d^2y}{dx^2} = \frac{2}{4^3} = \frac{1}{32}$$

so the curve is concave up at $t=1$.

Problem 2. Find the indefinite integrals

$$\int \left(t \sqrt{t} + \frac{1}{t\sqrt{t}}\right)dt$$

Answer. We have

$$t \sqrt{t} = t t^{1/2} = t^{3/2}$$

so

$$\displaystyle \int \left(t \sqrt{t} + \frac{1}{t\sqrt{t}}\right)dt = \int \left(t^{3/2} + t^{-3/2}\right)dt$$

Since

$$\int t^{r} dt = \frac{t^{r+1}}{r+1} + C$$

whenever $r \neq -1$, we get

$$\int \left(t^{3/2} + t^{-3/2}\right)dt = \frac{t^{3/2 + 1}}{3/2+1} + \frac{t^{-3/2+1}}{-3/2+1} + C$$

or

$$\int \left(t^{3/2} + t^{-3/2}\right)dt = \frac{t^{5/2}}{5/2} + \frac{t^{-1/2}}{-1/2} + C = \frac{2}{5}t^{5/2} - 2t^{-1/2} + C$$

Problem 3. Find the definite integrals

$$\displaystyle \int_0^{\pi/4} \frac{1}{\cos^2(x)}dx$$

Answer. We have

$$\int \frac{1}{\cos^2(x)}dx = \tan(x) + C$$

The fundamental theorem of Calculus gives

$$\int_0^{\pi/4} \frac{1}{\cos^2(x)}dx = \Big[\tan(x)\Big]^{\pi/4}_0 = \tan(\pi/4) - \tan(0) = 1-0 = 1$$

Problem 4. Find the exact area below the curve $y = x^3(1-x)$ and above the x-axis.

Answer. Graphing the function $y = x^3(1-x)$ we see that when $0 \leq x \leq 1$ the function is above the x-axis otherwise it is below. And it crosses the x-axis when $x=0$ and $x=1$. So the area is given by the formula

$$\mbox{Area} = \int_0^1 x^3(1-x) dx$$

Since

$$\int_0^1 x^3(1-x) dx = \int_0^1 (x^3-x^4) dx = \left[\frac{x^4}{4} - \frac{x^5}{5}\right]_0^1$$

so

$$\mbox{Area} = \int_0^1 x^3(1-x) dx = \frac{1}{4} - \frac{1}{5} = \frac{1}{20}$$

Problem 5. Find the derivatives

$$\frac{d}{dt} \left(\int_t^{t^2} \sin(x^2)dx\right)$$

Answer. Recall the formula

$$\frac{d}{dt} \left(\int_{u}^{v} f(x)dx\right) = f(v)\frac{dv}{dt} - f(u)\frac{du}{dt}$$

so

$$\frac{d}{dt} \left(\int_t^{t^2} \sin(x^2)dx\right) = \sin\Big((t^2)^2\Big)2t - \sin(t^2) 1 = 2t \sin(t^4) - \sin(t^2)$$

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