Calculus Practice

Calculus Practice

In physics, calculus is a tool – it gets you from place A to place B – kind of like your car. Do you understand in detail how your car engine works? Most drivers don’t. They understand their cars operationally – the accelerator makes you go faster, the steering wheel changes your direction, etc. Treat calculus the same way! It is vast, complex and beautiful — and your math course will take you “under the hood”. In the meantime, let’s get you on the road…

Exercise 1: Derivatives

1.1 What is a derivative? What does it tell you?

1.2 Find the derivative of the following:

A. $y(x) = x^3$

Answer

$\frac{dy}{dx}=3x^2$

B. $y(x) = 4x$

Answer

$\frac{dy}{dx}=4$

C. $y(x) = 0.5x^2$

Answer

$\frac{dy}{dx}=x$

D. $y(x) = 10$

Answer

$\frac{dy}{dx}=0$

E. $x(t) = t^4$

Answer

$\frac{dx}{dt}=4t^3$

F. $x(t) = 4t^3 -6t^2 +2$

Answer

$\frac{dx}{dt}=12t^2 -12t$

G. $v(t) = \frac{4}{t^3}$

Answer

$\frac{dv}{dt}=\frac{-12}{t^4}$

H. $y(x) = 8x^{1/2}$

Answer

$\frac{dy}{dx}=4x^{-1/2}$

I. $y(x) = mx + b$ , where $m, b$ are constants.

Answer

$\frac{dy}{dx}=m$

J. $x(t) = x_0 + v_0t+\frac{1}{2}gt^2$ , where $x_0, v_0, g$ are constants.

Answer

$\frac{dx}{dt}=v_0 + gt$

K. What is the slope of the function: $x(t) = 4t^3 + 3$ at $t=2$ ?

Answer

$\frac{dx}{dt} = 12t^2$
$12t^2$ at $t=2$ , $\frac{dx}{dt} = 48.$

Exercise 2: Anti-derivatives and integrals

2.1 What is an anti-derivative or an integral? What can it tell you?

2.2 Find the original functions:

A. $\frac{dy}{dx} = x^2$

Answer

$y(x) = \frac{1}{3}x^3 + C$

B. $\frac{dy}{dx} = 0.2x^3$

Answer

$0.05x^4 +C$

C. $\frac{dy}{dx} =4x$

Answer

$y(x) = 2x^2+C$

D. $\frac{dy}{dx} = 2$

Answer

$y(x) = 2x+C$

E. $\frac{dy}{dx} = 0$

Answer

$y(x) = C$

F. $\frac{dx}{dt} = 4t^2 -6t +2$

Answer

$x(t) = \frac{4}{3}t^3-3t^2 + 2t + C$

G. $\frac{dv}{dt} = \frac{10}{t^2}$

Answer

$v(t) = -\frac{10}{t}+C$

2.3 Determine the integrals:

H. $\int 2x^3\ dx$

Answer

$= \frac{1}{2}x^4 + C$

I. $\int x\ dx$

Answer

$=\frac{1}{2}x^2 + C$

J. $\int (z^3+5)\ dz$

Answer

$=\frac{1}{4}z^4 +5z + C$

K. $\int dx$

Answer

$=x + C$

L. $\int (at+v_0)\ dt$ , where $a$ and $v_0$ are constants.

Answer

$=\frac{1}{2}at^2 +v_0t + C$

M. $\int_1^3 0.5t^2\ dt$

Answer

$=\frac{1}{6}t^3|^{3}_{1}$
$=13/3$

N. $\int_{-2}^{2}5t\ dt$

Answer

$=\frac{5}{2}t^2|_{-2}^{2}$
$=0$

O. $\int_{1}^{2}\frac{4}{x^3}\ dx$

Answer

$=\frac{-2}{x^2}|_{1}^2$
$=3/2$

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Physics 1C03 Lab Manual Copyright © by Physics 1C03 Team. All Rights Reserved.

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