- #1
Tori No Otoko
- 1
- 0
Slightly confused at what it wants me to do here?
Derivatives with Quadratic Functions.
- MHB
- Thread starter Tori No Otoko
- Start date
In summary, a derivative is a mathematical concept that measures the rate of change of a function. The derivative of a quadratic function can be found using the power rule, and it plays a critical role in understanding the behavior of quadratic functions. By analyzing derivatives, we can determine the critical points and graph the function.
Slightly confused at what it wants me to do here?
- #2
skeeter
- 1,103
- 1
just an algebra drill ...
$T'(z) = 0 \implies \dfrac{z}{\sqrt{80^2+z^2}} = \dfrac{D-z}{\sqrt{50^2+(z-D)^2}}$
$z\sqrt{50^2+(z-D)^2} = (D-z)\sqrt{80^2+z^2}$
$z^2[50^2+(z-D)^2] = (D-z)^2[80^2+z^2]$
$50^2z^2 + z^2(z-D)^2 = 80^2(D-z)^2 + z^2(D-z)^2$
$0 = 80^2(D-z)^2 - 50^2z^2$
$0 = 80^2(D^2 - 2zD + z^2) - 50^2z^2$
$0 = 80^2D^2 - 80^2 \cdot 2zD + (80^2-50^2)z^2$
$0 = 6400(D^2-2zD) + 3900z^2$
$0 = 64(D^2-2zD) + 39z^2$
$0 = 39z^2 - 128zD + 64D^2$
$T'(z) = 0 \implies \dfrac{z}{\sqrt{80^2+z^2}} = \dfrac{D-z}{\sqrt{50^2+(z-D)^2}}$
$z\sqrt{50^2+(z-D)^2} = (D-z)\sqrt{80^2+z^2}$
$z^2[50^2+(z-D)^2] = (D-z)^2[80^2+z^2]$
$50^2z^2 + z^2(z-D)^2 = 80^2(D-z)^2 + z^2(D-z)^2$
$0 = 80^2(D-z)^2 - 50^2z^2$
$0 = 80^2(D^2 - 2zD + z^2) - 50^2z^2$
$0 = 80^2D^2 - 80^2 \cdot 2zD + (80^2-50^2)z^2$
$0 = 6400(D^2-2zD) + 3900z^2$
$0 = 64(D^2-2zD) + 39z^2$
$0 = 39z^2 - 128zD + 64D^2$
- #3
HOI
- 921
- 2
Notice that the problem gives you T', not T. It is not necessary to differentiate. Just set it equal to 0 and do the algebra to get the final result.
FAQ: Derivatives with Quadratic Functions.
What are derivatives with quadratic functions?
Derivatives with quadratic functions refer to the process of finding the rate of change of a quadratic function at a specific point. This is done by calculating the slope of the tangent line at that point.
How do you find the derivative of a quadratic function?
To find the derivative of a quadratic function, you can use the power rule. This involves multiplying the coefficient of the x term by the exponent, reducing the exponent by 1, and then moving the variable to the front of the expression. For example, the derivative of f(x) = 3x^2 would be f'(x) = 6x.
What is the significance of the derivative in quadratic functions?
The derivative of a quadratic function represents the slope of the tangent line at a specific point. This can be useful in determining the maximum and minimum points of the function, as well as the direction of its concavity.
Can you use derivatives to find the roots of a quadratic function?
Yes, you can use derivatives to find the roots of a quadratic function. The roots of a quadratic function are the points where the function crosses the x-axis, or where f(x) = 0. By setting the derivative of the function equal to 0 and solving for x, you can find the critical points which can then be used to find the roots.
Are there any real-world applications of derivatives with quadratic functions?
Yes, derivatives with quadratic functions have many real-world applications. For example, they can be used in physics to calculate the velocity and acceleration of an object in motion, or in economics to determine the maximum profit or minimum cost of a business. They are also used in engineering and other fields to optimize designs and solve complex problems.