a ) The domain:
x^4 - 16 x² ≥ 0
x² ( x² - 16 ) ≥ 0
x² - 16 ≥ 0
x² ≥ 16
x ∈ ( - ∞, - 4 ] ∪ [ 4 , + ∞ )b ) f ` ( x ) =
=
( 2 x³ - 16 x ) / √(x^4 - 16 x²)c ) The slope of the tangent line at x = 5:
f ` ( 5 ) = ( 2 * 125 - 16 * 5 ) / √ ( 625 - 400 ) = 170 / 15 = 34 / 3
The slope of the line normal to the graph at x = 5:
m = - 3 / 34
Answer:
7,7, and 7
Step-by-step explanation:
If she vase is cubical we know that all 3 dimensions are the same. This means that x^3 = 343. So we know the answer is the cube root of 343, which is 7
The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is
In this case, we have
<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> ) ==> d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )
<em>y(t)</em> = 5 - 2<em>t</em> ==> d<em>y</em>/d<em>t</em> = -2
and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then
Answer:
12
Step-by-step explanation:
Answer:
4
Step-by-step explanation:
