MATH SOLVE

4 months ago

Q:
# 1. (1 pt each) The equation of motion of a particle is s = t 3 − 12t 2 + 36t, t ≥ 0, where s is measured in meters and t is in seconds. (a) Find the velocity at time t. (b) What is the velocity after 3 s? (c) When is the particle at rest? (d) When is the particle moving in the positive direction? (e) Find the total distance traveled during the first 8 s. (f) Find the acceleration at time t and after 3 s. 2. (3 pt each) Given p(x) = x n − x, find the intervals over which p(x) is a decreasing function when: (a) n = 2 (b) n = 1 2 (c) n = −1 3. (a) (3 pts) Find equations of both lines through the point (2, −3) that are tangent to the parabola y = x 2 + x. (b) (1 pt) Show that there is no line through the point (2, 7) that is tangent to the parabola. Then sketch a graph to see why. 4. (3 pts) Find a cubic function y = ax3+bx2+cx+d whose graph has horizontal tangents at the points (−2, 6) and (2, 0). 5. (a) (3 pts) Use the Product Rule twice to prove that if f, g, and h are differentiable, then (fgh) 0 = f 0 gh + fg0h + fgh0 . (b) (3 pts) Use part (a) to differentiate y = x sin x cos x 6. (3 pts)

Accepted Solution

A:

QUESTION 1) The particle's equation of motion is [tex]s=t^3-12t^2+36t,t\ge0[/tex] where s is measured in meters and t is in seconds. The velocity at time,[tex]t[/tex] is given by the first derivative of the equation of motion of the particle. [tex]s'(t)=(3t^2-24t+36)ms^{-1}[/tex] Question 1b). To find the velocity after 3 seconds, we put [tex]t=3[/tex] into the velocity function. [tex]\Rightarrow s'(3)=(3(3)^2-24(3)+36)ms^{-1}[/tex] [tex]\Rightarrow s'(3)=(27-72+36)ms^{-1}[/tex] [tex]\Rightarrow s'(3)=-9ms^{-1}[/tex] QUESTION 1C To find the time that the particle is at rest, we equate the velocity (the first derivative formula) to zero and solve for t. [tex]\Rightarrow 3t^2-24t+36=0[/tex] Divide through by 3 to get; [tex]\Rightarrow t^2-8t+12=0[/tex] Factor: [tex]\Rightarrow (t-2)(t-6)=0[/tex] [tex](t-2)=0,(t-6)=0[/tex] [tex]t=2s,t=6s[/tex] The particle is at rest when [tex]t=2s[/tex] and [tex]t=6s[/tex]. QUESTION 1d. The particle is moving in a positive direction when the velocity is greater than zero. [tex]\Rightarrow 3t^2-24t+36\:>\:0[/tex] [tex]\Rightarrow t^2-8t+12\:>\:0[/tex] [tex]\Rightarrow (t-2)(t-6)\:>\:0[/tex] [tex]\Rightarrow t\:<\:2,t\:>\:6[/tex] But [tex]t\ge0[/tex]. This implies that, the particle is moving in a positive direction on the interval, [tex]0\le t\:<\:2,t\:>\:6[/tex] QUESTION 1e. To find the total distance traveled after 8s, we substitute [tex]t=8[/tex] into the equation of motion of the particle. [tex]s(8)=8^3-12(8)^2+36(8)[/tex] [tex]s(8)=512-768+288[/tex] [tex]s(8)=32m[/tex] The particle covered 32m in the first 8 seconds. QUESTION 1f i) The acceleration at time [tex]t[/tex] can be obtained by differentiating the velocity equation. [tex]\Rightarrow s"(t)=6t-24[/tex] ii) To find the acceleration after 3 seconds, we substitute [tex]t=3[/tex] into the equation of acceleration. [tex]\Rightarrow s"(3)=6(3)-24[/tex] [tex]\Rightarrow s"(3)=18-24[/tex] [tex]\Rightarrow s"(3)=-6ms^{-2}[/tex] After 3 seconds, the particle is decelerating at 6 meters per seconds square. QUESTION 2a Given: [tex]p(x)=x^n-x[/tex] [tex]p'(x)=nx^{n-1}-1[/tex] When [tex]n=2[/tex], then [tex]p'(x)=2x^{2-1}-1[/tex] [tex]p'(x)=2x-1[/tex] This implies that, p(x) is decreasing when [tex]p'(x)\:<\:0[/tex] [tex]2x-1\:<\:0[/tex] Therefore the function is decreasing on; [tex]x\:<\:\frac{1}{2}[/tex] QUESTION 2b When [tex]n=\frac{1}{2}[/tex] [tex]p'(x)=\frac{1}{2\sqrt{x}}-1[/tex] To find the interval over which the function is decreasing, we solve the inequality; [tex]\frac{1}{2\sqrt{x}}-1\:<\:0[/tex] Therefore the function is decreasing on the interval; [tex]\Rightarrow x\:>\:\frac{1}{4}[/tex] QUESTION 3a
The given parabola has equation [tex]y=x^2+x...(1)[/tex]
Let the two tangents from the external point; [tex](2,-3)[/tex] have equation;
[tex]y+3=m(x-2)..(2)[/tex]
Put equation (1) into equation (2)
This implies that;
[tex]x^2+x+3=m(x-2)[/tex]
Rewrite to obtain a quadratic equation in [tex]x[/tex].
[tex]x^2+(1-m)x+3+2m=0[/tex]
Since this is the point of intersection of a tangent and a parabola, the discriminant of this quadratic equation must be zero.
[tex]\Rightarrow (1-m)^2-4(3-2m)=0[/tex]
[tex]\Rightarrow m^2-10m-11=0[/tex]
[tex]\Rightarrow m=11\:or\:m=-1[/tex]
We substitute the values of m into equation (2) to obtain the equations of the two tangents to be;
[tex]y=11x-25[/tex] and [tex]y=-x-1[/tex]
QUESTION 3b
Let the equation of the tangents from the external point (2,7) be [tex]y-7=m(x-2)...(1)[/tex]
The given parabola has equation
[tex]y=x^2+x...(2)[/tex]
The discriminant of the intersection of these two equations yields;
[tex]m^2-10m+29=0[/tex]
This quadratic equation has no real roots.
Hence there are no lines through the point (2,7) that are tangents to parabola.
We can see from the graph that this point is lying inside the parabola.
QUESTION 4
The given cubic function is [tex]y=ax^3+bx^2+cx+d[/tex]
[tex]\frac{dy}{dx}=3ax^2+2bx+c[/tex]
The horizontal tangents occurs when [tex]\frac{dy}{dx}=0[/tex].
[tex]\Rightarrow 3ax^2+2bx+c=0[/tex]
This occurs at (2,0).
[tex]\Rightarrow 12a+4b+c=0...(1)[/tex]
and (-2,6) .
[tex]\Rightarrow 12a-4b+c=0...(2)[/tex]
These points also lie on the curve so they must satisfy the equation of the curve;
Substituting (2,0) into the original equation gives;
[tex]8a+4b+2c+d=0...(3)[/tex]
Substituting (-2,6) into the original equation gives;
[tex]-8a+4b-2c+d=0...(4)[/tex]
Solving the four equations simultaneously gives;
[tex]a=\frac{3}{16},b=0,c=-\frac{9}{4},c=3[/tex]
Hence the required cubic function;
[tex]y=\frac{3}{16}x^3-\frac{9}{4}x+3[/tex]
QUESTION 5a
Let [tex]y=fgh[/tex]
where [tex]f,g,[/tex] and [tex]h[/tex] are differentiable.
Using the product rule;
[tex]y'=f'(gh)+f(gh)'[/tex]
Use the product rule again;
[tex]y'=f'(gh)+f(g'h+gh')[/tex]
[tex]y'=f'(gh)+fg'h+fgh'[/tex] as required.