Respuesta :
Answer:
[tex]\sf a) \quad Minimum[/tex]
[tex]\textsf{b)} \quad y=\dfrac{x^{4}}{2}-\dfrac{2}{x}+2[/tex]
[tex]\textsf{c)}\quad \textsf{See below for proof.}[/tex]
[tex]\textsf{d)} \quad \dfrac{5}{4}\; \textsf{units per second}[/tex]
Step-by-step explanation:
Part a
Given second derivative:
[tex]\dfrac{\text{d}^2y}{\text{d}x^2}=6x^2-\dfrac{4}{x^3}[/tex]
Given that the curve has a stationary point at (-1, 9/2), we can determine its nature by substituting its x-coordinate into the second derivative.
If the result is positive, the stationary point is a minimum, and if the result is negative, the stationary point is a maximum.
[tex]\begin{aligned}x=-1 \implies \dfrac{\text{d}^2y}{\text{d}x^2}&=6(-1)^2-\dfrac{4}{(-1)^3}\\\\&=6(1)-\dfrac{4}{(-1)}\\\\&=6+4\\\\&=10 > 0 \leftarrow \textsf{minimum}\end{aligned}[/tex]
As the result is positive, this means that the stationary point is a minimum.
[tex]\hrulefill[/tex]
Part b
To find the equation of the curve, we need to integrate twice.
Integrate once to find the equation of the first derivative:
[tex]\dfrac{\text{d}y}{\text{d}x}=\displaystyle \int \left(\dfrac{\text{d}^2y}{\text{d}x^2}\right)\; \text{d}x[/tex]
[tex]\dfrac{\text{d}y}{\text{d}x}=\displaystyle \int \left(6x^2-\dfrac{4}{x^3}\right)\; \text{d}x[/tex]
[tex]\dfrac{\text{d}y}{\text{d}x}=\displaystyle \int \left(6x^2-4x^{-3}\right)\; \text{d}x[/tex]
[tex]\dfrac{\text{d}y}{\text{d}x}=\dfrac{6x^{(2+1)}}{2+1}-\dfrac{4x^{(-3+1)}}{-3+1}+C[/tex]
[tex]\dfrac{\text{d}y}{\text{d}x}=\dfrac{6x^{3}}{3}-\dfrac{4x^{-2}}{-2}+C[/tex]
[tex]\dfrac{\text{d}y}{\text{d}x}=2x^3+\dfrac{2}{x^2}+C[/tex]
To find the constant of integration (C), we can use the fact that at a stationary point, the first derivative is equal to zero.
Therefore, substitute x = -1 into dy/dx, set it to zero, and solve for C.
[tex]\begin{aligned}2(-1)^3+\dfrac{2}{(-1)^2}+C&=0\\\\2(-1)+\dfrac{2}{1}+C&=0\\\\-2+2+C&=0\\\\0+C&=0\\\\C&=0\end{aligned}[/tex]
Therefore, the equation of the first derivative is:
[tex]\dfrac{\text{d}y}{\text{d}x}=2x^3+\dfrac{2}{x^2}[/tex]
Now, integrate again:
[tex]y=\displaystyle \int \dfrac{\text{d}y}{\text{d}x}[/tex]
[tex]y=\displaystyle \int \left(2x^3+\dfrac{2}{x^2}\right)\;\text{d}x[/tex]
[tex]y=\displaystyle \int \left(2x^3+2x^{-2}\right)\;\text{d}x[/tex]
[tex]y=\dfrac{2x^{(3+1)}}{3+1}+\dfrac{2x^{(-2+1)}}{-2+1}+C[/tex]
[tex]y=\dfrac{2x^{4}}{4}+\dfrac{2x^{-1}}{-1}+C[/tex]
[tex]y=\dfrac{x^{4}}{2}-\dfrac{2}{x}+C[/tex]
Given that point (-1, 9/2) is on the curve, substitute this into the equation and solve for C:
[tex]\begin{aligned}\dfrac{(-1)^{4}}{2}-\dfrac{2}{(-1)}+C&=\dfrac{9}{2}\\\\\dfrac{1}{2}-(-2)+C&=\dfrac{9}{2}\\\\\dfrac{1}{2}+2+C&=\dfrac{9}{2}\\\\\dfrac{5}{2}+C&=\dfrac{9}{2}\\\\C&=2\end{aligned}[/tex]
Therefore, the equation of the curve is:
[tex]y=\dfrac{x^{4}}{2}-\dfrac{2}{x}+2[/tex]
[tex]\hrulefill[/tex]
Part c
Stationary points occur when dy/dx = 0.
To show that there is only one stationary point at x = -1, we can set dy/dx to zero and solve for x:
[tex]\begin{aligned}\dfrac{\text{d}y}{\text{d}x}&=0\\\\2x^3+\dfrac{2}{x^2}&=0\\\\2x^5+2&=0\\\\2(x^5+1)&=0\\\\x^5+1&=0\\\\x^5&=-1\\\\x&=\sqrt[5]{-1}\\\\x&=-1\end{aligned}[/tex]
Therefore, this proves that there is only one stationary point at x = -1.
[tex]\hrulefill[/tex]
Part d
To find the rate of increase of the x-coordinate (dx/dt) when the y-coordinate is increasing at a rate of 5 units per second (dy/dt = 5), we can use the chain rule:
[tex]\dfrac{\text{d}y}{\text{d}t} = \dfrac{\text{d}y}{\text{d}x} \times \dfrac{\text{d}x}{\text{d}t}[/tex]
Plug in dy/dx and dy/dt:
[tex]5 = \left(2x^3+\dfrac{2}{x^2}\right) \times \dfrac{\text{d}x}{\text{d}t}[/tex]
Rearrange to isolate dx/dt:
[tex]\dfrac{\text{d}x}{\text{d}t}=\dfrac{5}{2x^3+\dfrac{2}{x^2}}[/tex]
Now, substitute x = 1:
[tex]\dfrac{\text{d}x}{\text{d}t}=\dfrac{5}{2(1)^3+\dfrac{2}{(1)^2}}[/tex]
[tex]\dfrac{\text{d}x}{\text{d}t}=\dfrac{5}{2(1)+\dfrac{2}{1}}[/tex]
[tex]\dfrac{\text{d}x}{\text{d}t}=\dfrac{5}{2+2}[/tex]
[tex]\dfrac{\text{d}x}{\text{d}t}=\dfrac{5}{4}[/tex]
So, the rate of increase of the x-coordinate at the point where x = 1 is 5/4 units per second.
