Math Problem Statement

Verify that the given point lies on the curve. Determine an equation of the line tangent to the curve at the given point. x^3 + y^3 = 14xy; (7,7)

Solution

The image you've uploaded presents a question where you are tasked with:

  1. Verifying whether the point (7, 7) lies on the curve defined by the equation x3+y3=14xyx^3 + y^3 = 14xy.
  2. Determining the equation of the tangent line to the curve at the point (7, 7).

Part a: Verify that the point (7, 7) lies on the curve

To verify if the point (7,7)(7, 7) lies on the curve, substitute x=7x = 7 and y=7y = 7 into the given equation x3+y3=14xyx^3 + y^3 = 14xy.

73+73=14(7)(7)7^3 + 7^3 = 14(7)(7) 343+343=14×7×7343 + 343 = 14 \times 7 \times 7 686=686686 = 686

Since both sides are equal, the point (7,7)(7, 7) lies on the curve.

Part b: Find the equation of the tangent line at (7, 7)

To find the tangent line, we first need to differentiate the given equation implicitly to find dydx\frac{dy}{dx}. Differentiating both sides of the equation x3+y3=14xyx^3 + y^3 = 14xy with respect to xx:

ddx(x3)+ddx(y3)=ddx(14xy)\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(14xy) 3x2+3y2dydx=14y+14xdydx3x^2 + 3y^2 \frac{dy}{dx} = 14y + 14x \frac{dy}{dx}

Now, we solve for dydx\frac{dy}{dx} by isolating it:

3y2dydx14xdydx=14y3x23y^2 \frac{dy}{dx} - 14x \frac{dy}{dx} = 14y - 3x^2 (3y214x)dydx=14y3x2\left(3y^2 - 14x\right) \frac{dy}{dx} = 14y - 3x^2 dydx=14y3x23y214x\frac{dy}{dx} = \frac{14y - 3x^2}{3y^2 - 14x}

Now, substitute x=7x = 7 and y=7y = 7 into this equation to find the slope of the tangent line at (7,7)(7, 7):

dydx=14(7)3(72)3(72)14(7)\frac{dy}{dx} = \frac{14(7) - 3(7^2)}{3(7^2) - 14(7)} dydx=9814714798\frac{dy}{dx} = \frac{98 - 147}{147 - 98} dydx=4949=1\frac{dy}{dx} = \frac{-49}{49} = -1

So, the slope of the tangent line at (7,7)(7, 7) is 1-1.

Finally, the equation of the tangent line can be written using the point-slope form:

yy1=m(xx1)y - y_1 = m(x - x_1) Substitute m=1m = -1, x1=7x_1 = 7, and y1=7y_1 = 7:

y7=1(x7)y - 7 = -1(x - 7) y7=x+7y - 7 = -x + 7 y=x+14y = -x + 14

So, the equation of the tangent line at (7,7)(7, 7) is:

y=x+14y = -x + 14

Let me know if you'd like further explanation or have any questions! Here are some related questions:

  1. How would the tangent line change if the point was at (3,5)(3, 5) instead?
  2. How do implicit differentiation and explicit differentiation differ when solving such problems?
  3. What are other methods to find tangent lines besides implicit differentiation?
  4. How does the graph of the equation x3+y3=14xyx^3 + y^3 = 14xy behave in general?
  5. What does the result of dydx=1\frac{dy}{dx} = -1 tell us about the slope of the curve?

Tip: When differentiating implicitly, remember to apply the chain rule carefully, especially when dealing with terms involving yy.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Tangents
Curve Analysis

Formulas

x^3 + y^3 = 14xy
Implicit differentiation formula

Theorems

Chain rule
Point-slope form of a line

Suitable Grade Level

Grades 11-12