Page:Calculus Made Easy.pdf/111

Exercises VIII. (See page 257 for Answers.)

(1) Plot the curve $y={\tfrac {3}{4}}x^{2}-5$, using a scale of millimetres. Measure at points corresponding to different values of $x$, the angle of its slope.

Find, by differentiating the equation, the expression for slope; and see, from a Table of Natural Tangents, whether this agrees with the measured angle.

(2) Find what will be the slope of the curve

$y=0.12x^{3}-2$,

at the particular point that has as abscissa $x=2$.

(3) If $y=(x-a)(x-b)$, show that at the particular point of the curve where ${\dfrac {dy}{dx}}=0$, $x$ will have the value ${\tfrac {1}{2}}(a+b)$.

(4) Find the ${\dfrac {dy}{dx}}$ of the equation $y=x^{3}+3x$; and calculate the numerical values of ${\dfrac {dy}{dx}}$ for the points corresponding to $x=0$, $x={\tfrac {1}{2}}$, $x=1$, $x=2$.

(5) In the curve to which the equation is $x^{2}+y^{2}=4$, find the values of $x$ at those points where the slope = $1$.

(6) Find the slope, at any point, of the curve whose equation is ${\dfrac {x^{2}}{3^{2}}}+{\dfrac {y^{2}}{2^{2}}}=1$; and give the numerical value of the slope at the place where $x=0$, and at that where $x=1$.

(7) The equation of a tangent to the curve $y=5-2x+0.5x^{3}$, being of the form $y=mx+n$, where m and n are constants, find the value of $m$ and $n$ if