A curve with equation $y = f(x)$ passes through the point (4, 25). Given that $$f'(x) = rac{3}{8}x^2 - 10x + 1, \, x > 0$$ (a) find $f(x)$, simplifying each term. (b) Find an equation of the normal to the curve at the point (4, 25). Give your answer in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are integers to be found.

A-Level Edexcel Maths Pure Differentiation: A curve with equation $y = f(x)$

A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 1

Question 1

A curve with equation $y = f(x)$ passes through the point (4, 25). Given that $$f'(x) = rac{3}{8}x^2 - 10x + 1, \, x > 0$$ (a) find $f(x)$, simplifying each term... show full transcript

Worked Solution & Example Answer:A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 1

Step 1

find $f(x)$, simplifying each term.

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Answer

To find $f(x)$ , we integrate $f'(x)$ :

$f(x) = rac{3}{8} \int x^2 \, dx - 10 \int x \, dx + \int 1 \, dx$

Calculating each integral:

For $\int x^2 \, dx$ , we have: $\frac{x^3}{3} + C_1$ Therefore, $\frac{3}{8} \cdot \frac{x^3}{3} = \frac{1}{8} x^3.$
For $\int x \, dx$ , we have: $\frac{x^2}{2} + C_2$ Therefore, $-10 \cdot \frac{x^2}{2} = -5x^2.$
For $\int 1 \, dx$ , this is simply: $x + C_3.$

Combining all these results, we have:

$f(x) = \frac{1}{8} x^3 - 5x^2 + x + C$

Next, we use the point (4, 25) to find $C$ :

$f(4) = \frac{1}{8}(4^3) - 5(4^2) + 4 + C = 25$

Calculating further:

$f(4) = \frac{1}{8} \cdot 64 - 5 \cdot 16 + 4 + C = 25$

This simplifies to:

$8 - 80 + 4 + C = 25$

Which gives:

$C = 25 + 80 - 8 - 4 = 93.$

Thus, the function is:

$f(x) = \frac{1}{8} x^3 - 5x^2 + x + 93.$

Step 2

Find an equation of the normal to the curve at the point (4, 25).

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Answer

To find the equation of the normal, we first need the slope of the tangent line at (4, 25).

Using $f'(x)$ :

$f'(4) = \frac{3}{8}(4^2) - 10(4) + 1 = \frac{3}{8} \cdot 16 - 40 + 1 = 6 - 40 + 1 = -33.$

The slope of the normal line is the negative reciprocal of the tangent slope:

$\text{slope}_{normal} = -\frac{1}{-33} = \frac{1}{33}.$

Using the point-slope form of a line $y - y_1 = m(x - x_1)$ where $(x_1, y_1) = (4, 25)$ and $m = \frac{1}{33}$ :

$y - 25 = \frac{1}{33}(x - 4).$

Rearranging this gives:

$33(y - 25) = x - 4$

Distributing the left side:

$33y - 825 = x - 4$

Rearranging to standard form yields:

$x - 33y + 821 = 0.$

Thus, the required form is: $ax + by + c = 0$ where $a = 1$ , $b = -33$ , and $c = 821.$

A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 1

Worked Solution & Example Answer:A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 1

find $f(x)$, simplifying each term.

Find an equation of the normal to the curve at the point (4, 25).

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