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7 The equation of a curve is 3x2+4xy+3y2=53 x^{2}+4 x y+3 y^{2}=5. (a) Show that $\frac{\m...
Mar 15, 2024
7 The equation of a curve is 3x2+4xy+3y2=53 x^{2}+4 x y+3 y^{2}=5. (a) Show that dy dx=−3x+2y2x+3y\frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{3 x+2 y}{2 x+3 y}. (b) Hence find the exact coordinates of the two points on the curve at which the tangent is parallel to y+2x=0y+2 x=0.
Solution by Steps
step 1
To show that dydx=−3x+2y2x+3y\frac{dy}{dx} = -\frac{3x + 2y}{2x + 3y}, we differentiate the given equation implicitly with respect to x
step 2
Differentiating 3x2+4xy+3y2=53x^2 + 4xy + 3y^2 = 5 gives us 6x+4y+4xdydx+6ydydx=06x + 4y + 4x\frac{dy}{dx} + 6y\frac{dy}{dx} = 0
step 3
Rearrange the terms to isolate dydx\frac{dy}{dx}, resulting in dydx(4x+6y)=−6x−4y\frac{dy}{dx}(4x + 6y) = -6x - 4y
step 4
Divide both sides by 4x+6y4x + 6y to get dydx=−6x+4y4x+6y\frac{dy}{dx} = -\frac{6x + 4y}{4x + 6y}, which simplifies to dydx=−3x+2y2x+3y\frac{dy}{dx} = -\frac{3x + 2y}{2x + 3y}
Answer
dydx=−3x+2y2x+3y\frac{dy}{dx} = -\frac{3x + 2y}{2x + 3y}
Key Concept
Implicit Differentiation
Explanation
To find the derivative of y with respect to x for an equation involving both x and y, we use implicit differentiation and then solve for dydx\frac{dy}{dx}.
step 5
To find the points where the tangent is parallel to y+2x=0y + 2x = 0, we set dydx\frac{dy}{dx} equal to the slope of this line, which is -1
step 6
Substitute y=−2xy = -2x into the original equation to find the x-coordinates of the points
step 7
Substituting gives us 3x2+4x(−2x)+3(−2x)2=53x^2 + 4x(-2x) + 3(-2x)^2 = 5
step 8
Simplifying the equation results in 3x2−8x2+12x2=53x^2 - 8x^2 + 12x^2 = 5, which simplifies to 7x2=57x^2 = 5
step 9
Solving for x gives us x2=57x^2 = \frac{5}{7}, so x=±57x = \pm\sqrt{\frac{5}{7}}
step 10
Using y=−2xy = -2x, we find the y-coordinates: y=−2(±57)y = -2(\pm\sqrt{\frac{5}{7}}), which gives us y=∓207y = \mp\sqrt{\frac{20}{7}}
Answer
The exact coordinates are (57,−257)\left(\sqrt{\frac{5}{7}}, -2\sqrt{\frac{5}{7}}\right) and (−57,257)\left(-\sqrt{\frac{5}{7}}, 2\sqrt{\frac{5}{7}}\right).
Key Concept
Parallel Tangents and Implicit Differentiation
Explanation
To find where the tangent to a curve is parallel to a given line, we set the derivative equal to the slope of the line and solve for the coordinates.
5 The equation of a curve is x2y−ay2=4a3x^{2} y-a y^{2}=4 a^{3}, where aa is a non-zero constant. (a) Show that dy dx=2xy2ay−x2\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 x y}{2 a y-x^{2}}. (b) Hence find the coordinates of the points where the tangent to the curve is parallel to the yy-axis. [4]
Solution by Steps
step 1
Differentiate the equation x2y−ay2=4a3x^{2} y - a y^{2} = 4 a^{3} implicitly with respect to xx
step 2
Applying the product rule to x2yx^{2} y, we get 2xy+x2dydx2xy + x^{2} \frac{dy}{dx}
step 3
Differentiating −ay2-a y^{2} with respect to xx, we get −2aydydx-2ay \frac{dy}{dx}
step 4
The derivative of the constant 4a34 a^{3} with respect to xx is 0
step 5
Combining the derivatives, we have 2xy+x2dydx−2aydydx=02xy + x^{2} \frac{dy}{dx} - 2ay \frac{dy}{dx} = 0
step 6
Solve for dydx\frac{dy}{dx} to get dydx=2xy2ay−x2\frac{dy}{dx} = \frac{2xy}{2ay - x^{2}}
Answer
dydx=2xy2ay−x2\frac{dy}{dx} = \frac{2xy}{2ay - x^{2}}
Key Concept
Implicit Differentiation
Explanation
To find dydx\frac{dy}{dx}, we differentiate both sides of the equation with respect to xx, treating yy as a function of xx and applying the product rule where necessary.
Solution by Steps
step 1
To find where the tangent is parallel to the y-axis, set dydx\frac{dy}{dx} equal to undefined
step 2
From the previous result, dydx=2xy2ay−x2\frac{dy}{dx} = \frac{2xy}{2ay - x^{2}}, the denominator 2ay−x22ay - x^{2} must be 0
step 3
Solve 2ay−x2=02ay - x^{2} = 0 for xx using the equation from asksia-ll calculator
step 4
Substitute x=±2ayx = ±\sqrt{2} \sqrt{a} \sqrt{y} into the original equation x2y−ay2=4a3x^{2} y - a y^{2} = 4 a^{3}
step 5
Simplify to find the coordinates of the points where the tangent is parallel to the y-axis
Answer
The coordinates cannot be determined without further information.
Key Concept
Tangent Parallel to the Y-axis
Explanation
When the tangent to the curve is parallel to the y-axis, the slope of the tangent (dydx\frac{dy}{dx}) is undefined, which occurs when the denominator of the derivative is zero.
4 The parametric equations of a curve are x=cos⁡θ2−sin⁡θ,y=θ+2cos⁡θ. x=\frac{\cos \theta}{2-\sin \theta}, \quad y=\theta+2 \cos \theta . Show that dy dx=(2−sin⁡θ)2\frac{\mathrm{d} y}{\mathrm{~d} x}=(2-\sin \theta)^{2}.
Solution by Steps
step 1
To find the derivative of y y with respect to x x , we first differentiate both x x and y y with respect to θ \theta
step 2
Differentiate x=cos⁡θ2−sin⁡θ x = \frac{\cos \theta}{2 - \sin \theta} with respect to θ \theta to find dxdθ \frac{dx}{d\theta}
step 3
Differentiate y=θ+2cos⁡θ y = \theta + 2\cos \theta with respect to θ \theta to find dydθ \frac{dy}{d\theta}
step 4
To find dydx \frac{dy}{dx} , we divide dydθ \frac{dy}{d\theta} by dxdθ \frac{dx}{d\theta}
step 5
Simplify dydx \frac{dy}{dx} to show that dydx=(2−sin⁡θ)2 \frac{dy}{dx} = (2 - \sin \theta)^2
Answer
dydx=(2−sin⁡θ)2 \frac{dy}{dx} = (2 - \sin \theta)^2
Key Concept
Differentiation of parametric equations
Explanation
To find dydx \frac{dy}{dx} for parametric equations, differentiate both x x and y y with respect to the parameter (in this case, θ \theta ), and then divide dydθ \frac{dy}{d\theta} by dxdθ \frac{dx}{d\theta} .
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