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WechatIMG190.jpg 7. Lavie blunk Figure 4 Figure 4 shows part of the curve with ...
Dec 12, 2024
WechatIMG190.jpg
Generated Graph
Solution by Steps
step 1
To find the gradient of the tangent to the curve at point P(2,13) P(2, 13) , we first need to differentiate the equation y=2x2+5 y = 2x^2 + 5 . The derivative is given by dydx=4x \frac{dy}{dx} = 4x
step 2
Evaluating the derivative at x=2 x = 2 : dydx∣x=2=4(2)=8 \frac{dy}{dx} \bigg|_{x=2} = 4(2) = 8 . Thus, the gradient of the tangent at point P P is 8 8
step 3
For part (b), we need to find the gradient of the line PQ PQ where Q Q has an x-coordinate of 2+h 2 + h . The y-coordinate of Q Q can be found by substituting x=2+h x = 2 + h into the equation: y=2(2+h)2+5 y = 2(2+h)^2 + 5
step 4
The coordinates of Q Q are (2+h,2(2+h)2+5) (2+h, 2(2+h)^2 + 5) . The gradient of line PQ PQ is given by yQ−yPxQ−xP=(2(2+h)2+5)−13(2+h)−2 \frac{y_Q - y_P}{x_Q - x_P} = \frac{(2(2+h)^2 + 5) - 13}{(2+h) - 2}
step 5
Simplifying the expression for the gradient of line PQ PQ : 2(2+h)2−21h \frac{2(2+h)^2 - 21}{h} . Expanding (2+h)2 (2+h)^2 gives 4+4h+h2 4 + 4h + h^2 , so 2(4+4h+h2)−21=8+8h+2h2−21=2h2+8h−13 2(4 + 4h + h^2) - 21 = 8 + 8h + 2h^2 - 21 = 2h^2 + 8h - 13 . Thus, the gradient is 2h2+8h−13h \frac{2h^2 + 8h - 13}{h}
step 6
The final gradient of line PQ PQ in simplest form is 2h+8−13h 2h + 8 - \frac{13}{h} as h h approaches 0 0
Answer
The gradient of the tangent at point P P is 8 8 , and the gradient of line PQ PQ is 2h+8−13h 2h + 8 - \frac{13}{h} .
Key Concept
The gradient of a curve at a point is found using the derivative, while the gradient of a secant line between two points is calculated using the difference in their coordinates.
Explanation
The gradient of the tangent line gives the instantaneous rate of change at point P P , while the gradient of line PQ PQ represents the average rate of change between points P P and Q Q . As h h approaches 0 0 , the gradient of PQ PQ approaches the gradient of the tangent at P P .
Figure 4 shows part of the curve with equation y = 2x2 + 5 The point P(2,13) lies on the curve. (a) Find the gradient of the tangent to the curve at P. The point q with x coordinate 2 + h also lies on the curve. (b) Find, in terms of h, the gradient of the line PQ. Give your answer in simplest form. ③ (c) Explain briefly the relationship between the answer to (b) and the answer to (a).
Generated Graph
Solution by Steps
step 1
To find the gradient of the tangent to the curve at point P(2,13) P(2, 13) , we first need to differentiate the equation y=2x2+5 y = 2x^2 + 5 . The derivative is given by dydx=4x \frac{dy}{dx} = 4x
step 2
Evaluating the derivative at x=2 x = 2 : dydx∣x=2=4(2)=8 \frac{dy}{dx} \bigg|_{x=2} = 4(2) = 8 . Thus, the gradient of the tangent at point P P is 8 8
step 3
For point Q Q with x x coordinate 2+h 2 + h , the coordinates of Q Q are (2+h,2(2+h)2+5) (2 + h, 2(2 + h)^2 + 5) . The gradient of line PQ PQ is given by yQ−yPxQ−xP=(2(2+h)2+5)−13(2+h)−2 \frac{y_Q - y_P}{x_Q - x_P} = \frac{(2(2 + h)^2 + 5) - 13}{(2 + h) - 2}
step 4
Simplifying the expression for the gradient of line PQ PQ : (2(4+4h+h2)+5−13)h=(8+8h+2h2−8)h=8h+2h2h=8+2h \frac{(2(4 + 4h + h^2) + 5 - 13)}{h} = \frac{(8 + 8h + 2h^2 - 8)}{h} = \frac{8h + 2h^2}{h} = 8 + 2h
step 5
The relationship between the answers to (a) and (b) is that as h h approaches 0 0 , the gradient of line PQ PQ approaches the gradient of the tangent at point P P , which is 8 8
Answer
The gradient of the tangent at point P P is 8 8 , and the gradient of line PQ PQ in terms of h h is 8+2h 8 + 2h .
Key Concept
The gradient of a curve at a point is found using the derivative, while the gradient of a secant line between two points can be expressed in terms of the change in y y over the change in x x .
Explanation
The gradient of the tangent line gives the instantaneous rate of change at a point, while the gradient of line PQ PQ represents the average rate of change between two points on the curve. As h h approaches 0 0 , the average rate of change approaches the instantaneous rate of change.
WechatIMG190.jpg
Solution by Steps
step 1
To find the gradient of the tangent to the curve at point P(2,13) P(2, 13) , we first need to differentiate the equation y=2x2+5 y = 2x^2 + 5 . The derivative is given by dydx=4x \frac{dy}{dx} = 4x
step 2
Evaluating the derivative at x=2 x = 2 : dydx∣x=2=4(2)=8 \frac{dy}{dx} \bigg|_{x=2} = 4(2) = 8 . Thus, the gradient of the tangent at point P P is 8 8
step 3
The point Q Q has an x-coordinate of 2+h 2 + h . The y-coordinate at Q Q can be found by substituting x=2+h x = 2 + h into the equation: y=2(2+h)2+5 y = 2(2 + h)^2 + 5 . Simplifying gives y=2(4+4h+h2)+5=8+8h+2h2+5=13+8h+2h2 y = 2(4 + 4h + h^2) + 5 = 8 + 8h + 2h^2 + 5 = 13 + 8h + 2h^2
step 4
The gradient of the line PQ PQ is given by the formula yQ−yPxQ−xP=(13+8h+2h2)−13(2+h)−2=8h+2h2h=8+2h \frac{y_Q - y_P}{x_Q - x_P} = \frac{(13 + 8h + 2h^2) - 13}{(2 + h) - 2} = \frac{8h + 2h^2}{h} = 8 + 2h
step 5
The relationship between the answers to (a) and (b) is that the gradient of the tangent at point P P (which is 8 8 ) is the limit of the gradient of the line PQ PQ as h h approaches 0 0
Answer
The gradient of the tangent at point P P is 8 8 , and the gradient of the line PQ PQ in terms of h h is 8+2h 8 + 2h .
Key Concept
The gradient of a curve at a point is found using derivatives, while the gradient of a secant line between two points can be expressed in terms of the difference in coordinates.
Explanation
The gradient of the tangent line at point P P represents the instantaneous rate of change of the function at that point, while the gradient of line PQ PQ approximates this rate as h h approaches 0 0 .
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