A curve with equation $y = f(x)$ passes through the point (4, 25). Given that $f'(x) = \frac{3}{8}x^2 - 10x + 1, \; x > 0$ (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 Integration: 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 2 - 2013 - Paper 5

Question 2

A curve with equation $y = f(x)$ passes through the point (4, 25). Given that $f'(x) = \frac{3}{8}x^2 - 10x + 1, \; x > 0$ (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 2 - 2013 - Paper 5

Step 1

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

96%

114 rated

Answer

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

$f(x) = \int f'(x) \, dx$

Substituting for $f'(x)$ gives:

$f(x) = \int \left( \frac{3}{8}x^2 - 10x + 1 \right) dx$

Integrating term by term:

First term: ( \int \frac{3}{8}x^2 , dx = \frac{3}{8} \cdot \frac{x^3}{3} = \frac{1}{8}x^3 )
Second term: ( \int -10x , dx = -5x^2 )
Third term: ( \int 1 , dx = x )

Combining these results, we have:

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

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

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

Calculating:

$25 = \frac{1}{8}(64) - 80 + 4 + C \rightarrow 25 = 8 - 80 + 4 + C \rightarrow 25 = -68 + C \rightarrow C = 93$

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).

99%

104 rated

Answer

To find the equation of the normal, we first need to calculate the derivative $f'(x)$ at $x = 4$ :

From the previous part, we know:

$f'(4) = \frac{3}{8}(4^2) - 10(4) + 1$

Calculating:

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

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

$m_{normal} = -\frac{1}{f'(4)} = -\frac{1}{-33} = \frac{1}{33}$

Now we can use the point-slope form of the line:

$y - y_1 = m(x - x_1)$

Substituting the point (4, 25):

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

Rearranging gives:

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

$33y - 825 = x - 4$

Bringing all terms to one side yields:

$x - 33y + 821 = 0$

This can be rewritten in the required form:

$1x - 33y + 821 = 0$

Thus, the final equation of the normal line is:

$x - 33y + 821 = 0$ 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 2 - 2013 - Paper 5

Worked Solution & Example Answer:A curve with equation $y = f(x)$ passes through the point (4, 25) - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 5

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

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

Join the A-Level students using SimpleStudy...