Math  /  Calculus

Question27,28,29,30,31,32,33,34,35\underline{27}, \underline{28}, \underline{29}, \underline{30}, \underline{31}, \underline{32}, \underline{33}, \underline{34}, \underline{35}, and 36\underline{36} Use implicit differentiation to find an equation of the tangent line to the curve at the given point. 3.
31. x2xyy2=1,(2,1)x^{2}-x y-y^{2}=1,(2,1) (hyperbola) Answer 4

Studdy Solution

STEP 1

What is this asking? We need to find the equation of the tangent line to the hyperbola x2xyy2=1x^2 - xy - y^2 = 1 at the point (2,1)(2, 1). Watch out! Remember that *implicit differentiation* is needed because we don't have yy explicitly in terms of xx.
Also, don't forget the product rule when differentiating xyxy!

STEP 2

1. Implicit Differentiation
2. Find the Slope
3. Build the Equation

STEP 3

Alright, let's **kick things off** by differentiating both sides of our equation x2xyy2=1x^2 - xy - y^2 = 1 with respect to xx.
Remember, we're treating yy as a function of xx, so we'll need the chain rule!

STEP 4

ddx(x2)ddx(xy)ddx(y2)=ddx(1) \frac{d}{dx}(x^2) - \frac{d}{dx}(xy) - \frac{d}{dx}(y^2) = \frac{d}{dx}(1) Applying the power rule and product rule gives us: 2x(1y+xdydx)2ydydx=0 2x - (1 \cdot y + x \cdot \frac{dy}{dx}) - 2y\frac{dy}{dx} = 0

STEP 5

Let's **tidy things up** a bit: 2xyxdydx2ydydx=0 2x - y - x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0

STEP 6

Now, we want to **isolate** dydx\frac{dy}{dx}: 2xy=xdydx+2ydydx 2x - y = x\frac{dy}{dx} + 2y\frac{dy}{dx} 2xy=(x+2y)dydx 2x - y = (x + 2y)\frac{dy}{dx} dydx=2xyx+2y \frac{dy}{dx} = \frac{2x - y}{x + 2y}

STEP 7

We're interested in the slope at the point (2,1)(2, 1), so let's **substitute** x=2x = 2 and y=1y = 1 into our expression for dydx\frac{dy}{dx}: dydx=2(2)12+2(1)=412+2=34 \frac{dy}{dx} = \frac{2(2) - 1}{2 + 2(1)} = \frac{4 - 1}{2 + 2} = \frac{3}{4} So, our **slope** is 34\frac{3}{4}!

STEP 8

We have a point (2,1)(2, 1) and a slope 34\frac{3}{4}.
We can use the **point-slope form** of a line: yy1=m(xx1)y - y_1 = m(x - x_1).

STEP 9

Substituting our values, we get: y1=34(x2) y - 1 = \frac{3}{4}(x - 2)

STEP 10

Let's **rewrite** this in slope-intercept form: y1=34x32 y - 1 = \frac{3}{4}x - \frac{3}{2} y=34x32+1 y = \frac{3}{4}x - \frac{3}{2} + 1 y=34x12 y = \frac{3}{4}x - \frac{1}{2}

STEP 11

The equation of the tangent line to the curve x2xyy2=1x^2 - xy - y^2 = 1 at the point (2,1)(2, 1) is y=34x12y = \frac{3}{4}x - \frac{1}{2}.

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