Finding The Gradient Of The Tangent Line To $w=\frac{v^3-7 V+2}{(v+2)(v^2-v+3)}$ At $v=\frac{2}{3}$

by qnaftunila 100 views
Iklan Headers

Finding the gradient of the tangent to a curve at a specific point is a fundamental concept in calculus. It allows us to understand the instantaneous rate of change of a function at that particular point. In this comprehensive guide, we will delve into the process of finding the gradient of the tangent to the curve defined by the equation ${w=\frac{v^3-7 v+2}{(v+2)(v^2-v+3)}}$ at the point $\${v=\frac{2}{3}}$. This involves several key steps, including differentiation, substitution, and simplification. We will break down each step in detail, providing explanations and examples to ensure a clear understanding.

Understanding the Gradient of a Tangent

The gradient of the tangent to a curve at a point represents the slope of the line that touches the curve at that point without crossing it. This slope, also known as the instantaneous rate of change, provides valuable information about the function's behavior at that specific location. For instance, a positive gradient indicates that the function is increasing at that point, while a negative gradient suggests that it is decreasing. A zero gradient signifies a stationary point, where the function momentarily stops increasing or decreasing.

In calculus, the gradient of the tangent is found using the derivative of the function. The derivative, denoted as $\${\frac{dw}{dv}}$ in this case, gives us a general formula for the slope of the tangent at any point on the curve. To find the gradient at a specific point, we simply substitute the corresponding value of the independent variable ($v\$v$ in this case) into the derivative.

Step-by-Step Solution

1. Differentiate the Function

The first step in finding the gradient of the tangent is to differentiate the given function. Our function is:

$\${w=\frac{v^3-7 v+2}{(v+2)(v^2-v+3)}}$

This function is a quotient of two expressions, so we need to apply the quotient rule of differentiation. The quotient rule states that if we have a function of the form $\${w = \frac{u}{v}}$, then its derivative is given by:

$\${\frac{dw}{dv} = \frac{v \frac{du}{dv} - u \frac{dv}{dv}}{v^2}}$

In our case, let:

$\${u = v^3 - 7v + 2} $v=(v+2)(v2−v+3){v = (v+2)(v^2-v+3)}

First, we find the derivatives of $u$ and $v$ separately.

Finding $\${\frac{du}{dv}}$

$\${\frac{du}{dv} = \frac{d}{dv}(v^3 - 7v + 2) = 3v^2 - 7}$

Finding $\${\frac{dv}{dv}}$

To find $\${\frac{dv}{dv}}$, we first need to expand $v$:

$\${v = (v+2)(v^2-v+3) = v^3 - v^2 + 3v + 2v^2 - 2v + 6 = v^3 + v^2 + v + 6}$

Now, we can differentiate $v$:

$\${\frac{dv}{dv} = \frac{d}{dv}(v^3 + v^2 + v + 6) = 3v^2 + 2v + 1}$

Applying the Quotient Rule

Now that we have $\${\frac{du}{dv}}$ and $\${\frac{dv}{dv}}$, we can apply the quotient rule:

$\${\frac{dw}{dv} = \frac{(v^3 + v^2 + v + 6)(3v^2 - 7) - (v^3 - 7v + 2)(3v^2 + 2v + 1)}{(v^3 + v^2 + v + 6)^2}}$

2. Substitute $\${v = \frac{2}{3}}$ into the Derivative

Now that we have the derivative $\${\frac{dw}{dv}}$, we need to substitute $\${v = \frac{2}{3}}$ into the expression to find the gradient of the tangent at that specific point. This step involves careful arithmetic and simplification.

$\${\frac{dw}{dv}|_{v=\frac{2}{3}} = \frac{((\frac{2}{3})^3 + (\frac{2}{3})^2 + \frac{2}{3} + 6)(3(\frac{2}{3})^2 - 7) - ((\frac{2}{3})^3 - 7(\frac{2}{3}) + 2)(3(\frac{2}{3})^2 + 2(\frac{2}{3}) + 1)}{((\frac{2}{3})^3 + (\frac{2}{3})^2 + \frac{2}{3} + 6)^2}}$

Let's break down the calculation step by step:

Evaluating the Terms

$\${(\frac{2}{3})^3 = \frac{8}{27}} $(23)2=49{(\frac{2}{3})^2 = \frac{4}{9}} $\${3(\frac{2}{3})^2 = 3(\frac{4}{9}) = \frac{4}{3}}$

Substituting the Values

Now, substitute these values into the expression:

$\${\frac{dw}{dv}|_{v=\frac{2}{3}} = \frac{(\frac{8}{27} + \frac{4}{9} + \frac{2}{3} + 6)(\frac{4}{3} - 7) - (\frac{8}{27} - \frac{14}{3} + 2)(\frac{4}{3} + \frac{4}{3} + 1)}{(\frac{8}{27} + \frac{4}{9} + \frac{2}{3} + 6)^2}}$

3. Simplify the Expression

This is the most challenging part, as it involves simplifying complex fractions. Let's simplify each part separately.

Simplifying the Numerator Terms

First, we simplify the terms inside the parentheses:

$\${\frac{8}{27} + \frac{4}{9} + \frac{2}{3} + 6 = \frac{8 + 12 + 18 + 162}{27} = \frac{200}{27}} $43−7=4−213=−173{\frac{4}{3} - 7 = \frac{4 - 21}{3} = \frac{-17}{3}} $\${\frac{8}{27} - \frac{14}{3} + 2 = \frac{8 - 126 + 54}{27} = \frac{-64}{27}} $43+43+1=83+1=113{\frac{4}{3} + \frac{4}{3} + 1 = \frac{8}{3} + 1 = \frac{11}{3}}

Now, substitute these simplified values back into the expression:

$\${\frac{dw}{dv}|_{v=\frac{2}{3}} = \frac{(\frac{200}{27})(\frac{-17}{3}) - (\frac{-64}{27})(\frac{11}{3})}{(\frac{200}{27})^2}}$

Multiplying the Fractions

$\${(\frac{200}{27})(\frac{-17}{3}) = \frac{-3400}{81}} $(−6427)(113)=−70481{(\frac{-64}{27})(\frac{11}{3}) = \frac{-704}{81}}

Substituting the Multiplied Values

$\${\frac{dw}{dv}|_{v=\frac{2}{3}} = \frac{\frac{-3400}{81} - (\frac{-704}{81})}{(\frac{200}{27})^2}}$

Simplifying the Numerator

$\${\frac{-3400}{81} + \frac{704}{81} = \frac{-2696}{81}}$

Simplifying the Denominator

$\${(\frac{200}{27})^2 = \frac{40000}{729}}$

Final Calculation

Now we have:

$\${\frac{dw}{dv}|_{v=\frac{2}{3}} = \frac{\frac{-2696}{81}}{\frac{40000}{729}}}$

To divide fractions, we multiply by the reciprocal:

$\${\frac{dw}{dv}|_{v=\frac{2}{3}} = \frac{-2696}{81} \times \frac{729}{40000} = \frac{-2696 \times 9}{40000} = \frac{-24264}{40000}}$

Reducing the Fraction

We can simplify this fraction by dividing both numerator and denominator by their greatest common divisor, which is 8:

$\${\frac{-24264}{40000} = \frac{-3033}{5000}}$

4. The Result

Therefore, the gradient of the tangent to the curve at the point $\${v = \frac{2}{3}}$ is:

$\${\frac{dw}{dv}|_{v=\frac{2}{3}} = \frac{-3033}{5000}}$

Conclusion

In conclusion, finding the gradient of the tangent to a curve involves a systematic application of calculus principles. The key steps include differentiating the function, substituting the given point, and simplifying the resulting expression. While the process can be complex, especially when dealing with quotients and fractions, breaking it down into smaller, manageable steps makes it easier to handle. Understanding the quotient rule and practicing algebraic manipulation are crucial skills for success in this type of problem. The gradient of the tangent provides valuable insights into the behavior of a function at a specific point, making it a fundamental concept in calculus and its applications.

This comprehensive guide has walked you through the process of finding the gradient of the tangent to the curve $\${w=\frac{v^3-7 v+2}{(v+2)(v^2-v+3)}}$ at the point $\${v=\frac{2}{3}}$. By understanding each step and practicing similar problems, you can master this essential calculus skill and apply it to various mathematical and real-world scenarios. Remember, the gradient of the tangent is a powerful tool for analyzing the rate of change of functions, and its applications extend far beyond the classroom.