The Sixth Gate
After walking through the fifth gate, your computer informs you that more than half of the gates have been
crossed. “Only 4 left”, you conclude. “Computer, what is this gate’s riddle?”
Your computer responds:
For the function, , find the value of that would make the slope of the
tangent line -176 at .

Explain, in your own words and with your own work, how you arrived at this result. Be sure to explain
using calculus concepts to best support the work of the game design team.

Ask by Whittaker Burns. in the United States

Mar 30,2025

Question

The Sixth Gate
After walking through the fifth gate, your computer informs you that more than half of the gates have been
crossed. “Only 4 left”, you conclude. “Computer, what is this gate’s riddle?”
Your computer responds:
For the function, , find the value of that would make the slope of the
tangent line -176 at .

Explain, in your own words and with your own work, how you arrived at this result. Be sure to explain
using calculus concepts to best support the work of the game design team.

Ask by Whittaker Burns. in the United States

Mar 30,2025

UpStudy AI · Accepted Answer

We are given the function $$ f(t) = \frac{1}{a}(2t^3 - 15t^2 + 45)^{44} $$ and need to find the value of $$ a $$ such that the slope of the tangent line at $$ t = 2 $$ is -176. The slope of the tangent line at a point is given by the derivative of the function at that point.
To find the slope, we first need to compute the derivative $$ f'(t) $$. We will use the chain rule for differentiation, which states that if $$ y = g(h(t)) $$, then $$ y' = g'(h(t)) \cdot h'(t) $$. Here, let $$ g(u) = \frac{1}{a} u^{44} $$ and $$ h(t) = 2t^3 - 15t^2 + 45 $$. Thus, we need to find $$ g'(u) $$ and $$ h'(t) $$.
Calculating the derivatives:
- $$ g'(u) = \frac{44}{a} u^{43} $$
- $$ h'(t) = 6t^2 - 30t $$

Now, we can apply the chain rule:
$$
f'(t) = g'(h(t)) \cdot h'(t) = \frac{44}{a} (2t^3 - 15t^2 + 45)^{43} \cdot (6t^2 - 30t)
$$
Next, we need to evaluate $$ f'(2) $$:
1. Calculate $$ h(2) = 2(2^3) - 15(2^2) + 45 = 16 - 60 + 45 = 1 $$.
2. Calculate $$ h'(2) = 6(2^2) - 30(2) = 24 - 60 = -36 $$.

Now substituting these values into the derivative:
$$
f'(2) = \frac{44}{a} (1)^{43} \cdot (-36) = \frac{44}{a} \cdot (-36) = -\frac{1584}{a}
$$
We want the slope to be -176, so we set up the equation:
$$
-\frac{1584}{a} = -176
$$
This simplifies to:
$$
\frac{1584}{a} = 176
$$
Now, we can solve for $$ a $$:
$$
1584 = 176a \implies a = \frac{1584}{176}
$$
Calculating this gives:
$$
a = 9
$$
The value of $$ a $$ that makes the slope of the tangent line -176 at $$ t = 2 $$ is $$ a = 9 $$. This conclusion is reached by applying the chain rule to find the derivative of the function and then solving for $$ a $$ based on the desired slope.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Mind Expander

preguntas relacionadas

Latest Calculus Questions