a) Given F(x,y,z) = ln(x^{2}+y^{2}+z^{2})}], x=cos(t)}], y=sin(t)}], and z = 4\sqrt{t} find...

Question:

a) Given {eq}F(x,y,z) = ln(x^{2}+y^{2}+z^{2}){/eq}, x=cos(t)}], {eq}y=sin(t){/eq}, and {eq}z = 4 \sqrt{t}{/eq} find {eq}\left.\begin{matrix} \frac{dF}{dt} \end{matrix}\right|_{t=3}{/eq}

b) Given {eq}f(x,y) = (x^{2} + y^{2})^{-1/2} + ln(xy){/eq} find {eq}\triangledown f{/eq} at {eq}(-1,2) {/eq}

c) Given {eq}f(x,y) = 2xy - 3y^{2}{/eq} find the derivative of f at the point {eq}(5,5){/eq} in the direction of {eq}\vec{v} = (4,-3){/eq}

Application of Derivative:

The derivative of the given function is evaluated by using the differentiation formula {eq}\displaystyle \frac{dF}{dt} {/eq}. The gradient of the function is evaluated by using the partial derivative formula {eq}\displaystyle \nabla f(x, y)=\left \langle f_{x}=\frac{\partial f}{\partial x}, f_{y}=\frac{\partial f}{\partial y} \right \rangle {/eq}. The directional derivative of the function at the point is evaluated by using the formula {eq}\displaystyle D_{\vec{u}} f(P)= \nabla f(P) \cdot \vec{u} {/eq}.

Answer and Explanation: 1

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a) Given function is {eq}\displaystyle F(x,y,z) = \ln(x^{2}+y^{2}+z^{2}) {/eq}, and the parametric equations are {eq}\displaystyle x= \cos (t),...

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Derivatives: The Formal Definition

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Chapter 7 / Lesson 5
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The derivative in calculus is the rate of change of a function. In this lesson, explore this definition in greater depth and learn how to write derivatives.


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