Respuesta :
Answer:
A) w'(8) = -4/5 ;w'(8) is less than 0, so it shows that w(t) function is a decreasing function and thus with the decrease in one function, the other will also decrease.
B) y = 8.8927 ;Thus we have proved that; w(3.5) ≤ 9
C) The function is undefined and thus has a vertical asymptote
Step-by-step explanation:
A) we are to find w'(8)
8 lies in the interval of 6 < t ≤ 10, we'll use;
w(t) = 10 - (1/5)(t - 6)²
Differentiating, we have;
w'(t) = -(2/5)(t - 6)
Thus, w'(8) = -(2/5)(8 - 6)
w'(8) = -4/5
As, w'(8) is less than 0,it shows that w(t) function is a decreasing function.
Thus, there will be a direct relation such that with the decrease in one function, the other will also decrease.
B) To find the line tangent to the graph of W at t=3; 3 falls into the interval of 0 ≤ t < 6,
Thus;
w(t) = 17/2 - (3/2)(cos(πt/6))
Now let's differentiate so we can find w'(3)
So, w'(t) = -(3/2 x -π/6)(sin(πt/6))
w'(t) = (π/4)(sin(πt/6))
So; w'(3) = (π/4)(sin(π*3/6))
w'(3) = (π/4)(sin(π/2))
sin (π/2) in radians = 1
Thus,w'(3) = (π/4)
Now, let's find the y coordinate which is at w(3);
w(3) = 17/2 - (3/2)(cos(π x 3/6))
w(3) = 17/2 - (3/2)(cos(π/2))
(cos(π/2)) in radians = 0
Thus;
w(3) = 17/2
So the tangent line will be;
y - 17/2 = (π/4)(x - 3)
y = (π/4)(x - 3) + 17/2
Now, at t = 3.5, putting 3. 5for x;
y = (π/4)(3.5 - 3) + 17/2
y = 0.5(π/4) + 8.5
y = 0.3927 + 8.5
y = 8.8927
Thus; w(3.5) ≤ 9
C) we want to find the limit as t approaches 2.
Now, 2 falls into the interval of 0 ≤ t < 6,
Thus;
w(t) = 17/2 - (3/2)(cos(πt/6))
So, z = [17/2 - (3/2)(cos(πt/6)) - t³ + ¼]/(t - 2)
At t = 2,we have;
Lim z → 2 = [17/2 - (3/2)(cos(π x 2/6)) - 2³ + ¼]/(2 - 2)
Lim z → 2 = [17/2 - (3/4) - 8 + ¼]/0
The denominator is zero and thus the function is undefined.
So, at t = 2, the function has a vertical asymptote
The depth of a river at a certain point is modeled by the function W. So, the value of W'(8) is -0.68 for t between 0 and 6 and by using the tangent it can be proved that the value of W(3.5) is less than or equal to 9.
Given :
- The depth of a river at a certain point is modeled by the function W
- [tex]\rm W(t) =\left \{ {{\dfrac{17}{2}-\dfrac{3}{2}cos \left(\dfrac{\pi t}{6}\right)\;\;\;\;\;\; for \;\;0\leq t \leq 6} \atop {10-\dfrac{1}{10}(t-6)^2\;\;\;\;\;\;for \;\; 6\leq t \leq 10}} \right.[/tex] ---- (1)
- W(t) is measured in feet and time t is measured in hours.
A) Differentiate W(t) with respect to 't'.
[tex]\rm W'(t) =\left \{ {{0+\dfrac{\pi}{6}\times \dfrac{3}{2} sin \left(\dfrac{\pi t}{6}\right)\;\;\;\;\;\; for \;\;0\leq t \leq 6} \atop {0-\dfrac{1}{5}(t-6)\;\;\;\;\;\;for \;\; 6\leq t \leq 10}} \right.[/tex]
Now, put the value of (t = 8) in order to evaluate W'(8).
[tex]\rm W'(8) =\left \{ {{0+\dfrac{\pi}{6}\times \dfrac{3}{2} sin \left(\dfrac{4\pi}{3}\right)\;\;\;\;\;\; for \;\;0\leq t \leq 6} \atop {0-\dfrac{1}{5}(2)\;\;\;\;\;\;for \;\; 6\leq t \leq 10}} \right.[/tex]
[tex]\rm W'(8) =\left \{ {{-\dfrac{\pi}{6}\times \dfrac{3}{2} \left(\dfrac{\sqrt{3} }{2}\right)\;\;\;\;\;\; for \;\;0\leq t \leq 6} \atop {0-\dfrac{1}{5}(2)\;\;\;\;\;\;for \;\; 6\leq t \leq 10}} \right.[/tex]
[tex]\rm W'(8) =\left \{ {{-0.68 \;\;\;\;\;\; for \;\;0\leq t \leq 6} \atop {-0.4 \;\;\;\;\;\;for \;\; 6\leq t \leq 10}} \right.[/tex]
B) Differentiate W for 3 ≤ t ≤ 3.5, that is:
[tex]\rm W'(t) =\dfrac{\pi}{6}\times \dfrac{3}{2} sin \left(\dfrac{\pi t}{6}\right)\;\;\;\;\;\; for \;\;0\leq t \leq 6[/tex]
Now, put (t = 3) in the above equation.
[tex]\rm W'(3) = \dfrac{\pi}{6}\times \dfrac{3}{2}sin\dfrac{\pi}{2}[/tex]
[tex]\rm W'(3) = \dfrac{\pi}{4}[/tex]
Now, put the value (t=3) in equation (1) only for 3 ≤ t ≤ 3.5.
[tex]\rm W(3) = \dfrac{17}{2}-\dfrac{3}{2}cos\dfrac{\pi}{2}[/tex]
[tex]\rm W(3) = \dfrac{17}{2}[/tex]
W(3) = 8.5
So, the tangent is given by the equation:
[tex]y - \dfrac{17}{2}=\dfrac{\pi}{4}(x-3)[/tex]
Now, put x = 3.5 in above equation.
[tex]y -\dfrac{17}{2}=\dfrac{\pi}{4}\times 0.5[/tex]
y = 8.8927
Therefore, W(3.5) [tex]\leq[/tex] 9
C)
[tex]\lim_{t \to \infty} \dfrac{W(t)-t^3+\frac{1}{4}}{t-2}[/tex]
[tex]\rm A =\dfrac{ \dfrac{17}{2}-\dfrac{3}{2}cos\dfrac{\pi t}{6}-t^3+\dfrac{1}{4}}{t-2}[/tex]
Now, at (t=2) A tends towards 2.
[tex]\rm \lim_{A \to 2} \dfrac{ \dfrac{17}{2}-\dfrac{3}{2}cos\dfrac{\pi t}{6}-t^3+\dfrac{1}{4}}{t-2}[/tex]
= [tex]\rm \dfrac{ \dfrac{17}{2}-\dfrac{3}{2}cos\dfrac{\pi \times 2}{6}-2^3+\dfrac{1}{4}}{2-2}[/tex]
Thus the function is undefined at t = 2.
For more information, refer to the link given below:
https://brainly.com/question/18008819
