Assuming that the equation defines x and y implicitly as differentiable functions xequals​f(t), yequals​g(t), find the slope of the curve xequals​f(t), yequals​g(t) at the given value of t.
x cubed plus 2 t squaredequals19​, 2 y cubed minus 2 t squaredequals36​, tequals3

To find the slope of the curve defined by the implicit equations at the given value \( t = 3 \), we need to implicitly differentiate both equations with respect to \( t \) and then calculate the slope as \( \frac{dy}{dx} \). 

The given equations are:

1. \( x^3 + 2t^2 = 19 \)
2. \( 2y^3 - 2t^2 = 36 \)

### Step 1: Differentiate both equations with respect to \( t \).

#### For the first equation \( x^3 + 2t^2 = 19 \):
Differentiate implicitly with respect to \( t \):

\[
\frac{d}{dt}(x^3) + \frac{d}{dt}(2t^2) = \frac{d}{dt}(19)
\]

This gives:

\[
3x^2 \frac{dx}{dt} + 4t = 0
\]

So,

\[
3x^2 \frac{dx}{dt} = -4t
\]

Solving for \( \frac{dx}{dt} \):

\[
\frac{dx}{dt} = \frac{-4t}{3x^2}
\]

#### For the second equation \( 2y^3 - 2t^2 = 36 \):
Differentiate implicitly with respect to \( t \):

\[
2 \cdot 3y^2 \frac{dy}{dt} - \frac{d}{dt}(2t^2) = 0
\]

This simplifies to:

\[
6y^2 \frac{dy}{dt} - 4t = 0
\]

Solving for \( \frac{dy}{dt} \):

\[
6y^2 \frac{dy}{dt} = 4t
\]

\[
\frac{dy}{dt} = \frac{4t}{6y^2} = \frac{2t}{3y^2}
\]

### Step 2: Evaluate at \( t = 3 \).
We need to find \( x \) and \( y \) at \( t = 3 \) by solving the original equations.

#### First equation: \( x^3 + 2t^2 = 19 \)
Substitute \( t = 3 \):

\[
x^3 + 2(3^2) = 19
\]

\[
x^3 + 18 = 19
\]

\[
x^3 = 1
\]

\[
x = 1
\]

#### Second equation: \( 2y^3 - 2t^2 = 36 \)
Substitute \( t = 3 \):

\[
2y^3 - 2(3^2) = 36
\]

\[
2y^3 - 18 = 36
\]

\[
2y^3 = 54
\]

\[
y^3 = 27
\]

\[
y = 3
\]

### Step 3: Calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) at \( t = 3 \).

Using \( x = 1 \) and \( y = 3 \):

\[
\frac{dx}{dt} = \frac{-4(3)}{3(1)^2} = \frac{-12}{3} = -4
\]

\[
\frac{dy}{dt} = \frac{2(3)}{3(3^2)} = \frac{6}{3(9)} = \frac{6}{27} = \frac{2}{9}
\]

### Step 4: Find the slope \( \frac{dy}{dx} \).

The slope of the curve is given by:

\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2}{9}}{-4} = \frac{2}{9} \times \frac{1}{-4} = \frac{-2}{36} = \frac{-1}{18}
\]

### Final Answer:
The slope of the curve at \( t = 3 \) is \( \boxed{-\frac{1}{18}} \).

Would you like any more details on this process or have questions? Here are some related questions to deepen your understanding:

1. What is implicit differentiation, and why is it used here?
2. How can you verify the results of implicit differentiation using numerical methods?
3. How does the concept of parametric differentiation apply to this problem?
4. What are some real-life applications of finding the slope of a curve defined implicitly?
5. How would the problem change if higher-order derivatives were required?

**Tip**: When working with implicit differentiation, always remember to differentiate both sides of the equation with respect to the independent variable, keeping in mind the chain rule.

Assuming that the equation defines x and y implicitly as differentiable functions x = f(t), y = g(t), find the slope of the curve x = f(t), y = g(t) at the given value of t. x^3 + 2t^2 = 19, 2y^3 - 2t^2 = 36, t = 3

Discover how to use implicit differentiation to find the slope of a curve defined by x^3 + 2t^2 = 19 and 2y^3 - 2t^2 = 36 at t = 3. This problem covers key calculus concepts such as the chain rule and parametric differentiation. Suitable for students in Grades 11-12 and university-level calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation