Math Problem Statement

Olkoon {x(t)=sqrt(t)ln(t)−4 {y(t)=5*sqrt(t),t>1, xy -tason käyrä. i) Määrää käyrää pisteessä (2e−4,5e) sivuavan käyrän tangentti. Anna ensin tangentin kulmakerroin k ko. pisteessä. ii) Määrää se parametrin t arvo, jolle tangentin kulmakerroin on 53 . b) Yhtälö 4(xy^2+y^2)^(1/4)+xy−20=0 määrittelee muuttujan y muuttujan x funktiona (y=f(x) ) pisteen (0,5) ympäristössä. Määrää implisiittisen derivoinnin avulla y′(0) . a) k=__ y=__ t=__ b) y′(0)=__

Solution

Problem Breakdown:

Part (a):

Given: x(t)=tln(t)4,y(t)=5t,t>1x(t) = \sqrt{t} \ln(t) - 4, \quad y(t) = 5\sqrt{t}, \quad t > 1 This describes a curve in the xyxy-plane.

  1. i) Tangent Slope kk at the Point (2e4,5e)(2e^{-4}, 5e):

    • First, find the derivatives x(t)x'(t) and y(t)y'(t) to determine the slope of the tangent line at the point.
    • The slope kk is given by: k=dydx=y(t)x(t)k = \frac{dy}{dx} = \frac{y'(t)}{x'(t)} evaluated at the tt value corresponding to x=2e4x = 2e^{-4} and y=5ey = 5e.
  2. ii) Determine the tt value for which the tangent slope is 5353:

    • Solve the equation k=53k = 53 for tt using the expression found for kk.

Part (b):

The equation 4(xy2+y2)1/4+xy20=04(xy^2 + y^2)^{1/4} + xy - 20 = 0 implicitly defines yy as a function of xx near the point (0,5)(0, 5).

To find y(0)y'(0) using implicit differentiation:

  1. Differentiate the equation with respect to xx.
  2. Substitute x=0x = 0 and y=5y = 5 into the differentiated equation to solve for y(0)y'(0).

I'll proceed with these calculations now.### Part (a) Results

  1. Tangent Slope kk at the Point (2e4,5e)(2e^{-4}, 5e):

    • The parameter tt corresponding to this point is approximately: t=e2W(e4+2)t = e^{2 \cdot W(e^{-4} + 2)} where WW is the Lambert W function.
    • The slope kk of the tangent line at this tt value is: k=5eW(e4+2)2(eW(e4+2)W(e4+2)+eW(e4+2))k = \frac{5 e^{-W(e^{-4} + 2)}}{2 \left( e^{-W(e^{-4} + 2)} W(e^{-4} + 2) + e^{-W(e^{-4} + 2)} \right)}
  2. Parameter tt for which the Tangent Slope is 5353:

    • The parameter tt that gives a tangent slope of 5353 is: t=e10153t = e^{-\frac{101}{53}}

Part (b) Solution Outline:

I will now proceed to Part (b) and determine y(0)y'(0) using implicit differentiation.It appears there was an issue when attempting to solve for y(0)y'(0) in part (b). I’ll revisit the implicit differentiation setup and check for any constraints that might affect the solution, ensuring it’s properly calculated.It seems that the implicit differentiation approach with the current setup isn’t yielding a solution for y(0)y'(0). This could indicate either that there is no solution in the defined region or a more complex behavior in the function around this point.

If you have further constraints or additional details, please let me know. Alternatively, I can explore alternative methods or verify any assumptions to proceed.

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Math Problem Analysis

Mathematical Concepts

Calculus
Parametric Equations
Implicit Differentiation
Tangent Lines

Formulas

Slope of the tangent to a parametric curve: k = dy/dx = y'(t)/x'(t)
Implicit differentiation to find dy/dx from an equation involving x and y

Theorems

Differentiation of Parametric Equations
Implicit Differentiation

Suitable Grade Level

Grades 11-12 or early university