To determine the rate at which the particle is moving vertically when it reaches the point [tex]\((10, 613)\)[/tex] on the curve [tex]\( y = 7x^2 - 8x - 7 \)[/tex], we'll employ the chain rule of calculus. Given that the particle's horizontal velocity is [tex]\( \frac{dx}{dt} = -3 \)[/tex] units per minute, our goal is to find [tex]\(\frac{dy}{dt}\)[/tex].
Here are the steps to find the desired vertical rate of change:
1. Differentiate [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex] to find [tex]\(\frac{dy}{dx}\)[/tex]:
The equation of the curve is:
[tex]\[ y = 7x^2 - 8x - 7 \][/tex]
Differentiating this equation with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(7x^2 - 8x - 7) \][/tex]
[tex]\[ \frac{dy}{dx} = 14x - 8 \][/tex]
2. Evaluate [tex]\(\frac{dy}{dx}\)[/tex] at the point [tex]\((10, 613)\)[/tex]:
Substitute [tex]\( x = 10 \)[/tex] into [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ \frac{dy}{dx}\bigg|_{x=10} = 14(10) - 8 \][/tex]
[tex]\[ \frac{dy}{dx}\bigg|_{x=10} = 140 - 8 \][/tex]
[tex]\[ \frac{dy}{dx}\bigg|_{x=10} = 132 \][/tex]
3. Use the chain rule to find [tex]\(\frac{dy}{dt}\)[/tex]:
The chain rule states that:
[tex]\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \][/tex]
Substitute the values of [tex]\( \frac{dy}{dx} \)[/tex] evaluated at [tex]\( x = 10 \)[/tex] and [tex]\( \frac{dx}{dt} \)[/tex]:
[tex]\[ \frac{dy}{dt} = 132 \cdot (-3) \][/tex]
[tex]\[ \frac{dy}{dt} = -396 \][/tex]
Thus, the rate at which the particle is moving vertically (with respect to the origin) when it is at the point [tex]\((10, 613)\)[/tex] is [tex]\(-396\)[/tex] units per minute.
