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Mathematics Question on Differential Calculus

The equation of the tangent to the curve x5/2 + y5/2 = 33 at the point(1, 4) is:

A

x + 8y − 33 = 0

B

12x + y − 8 = 0

C

x + 8y − 12 = 0

D

x + 12y − 8 = 0

Answer

x + 8y − 33 = 0

Explanation

Solution

Differentiate the given curve implicitly:

ddx(5x2+5y2)=ddx(33).\frac{d}{dx} \left( \frac{5}{x^2} + \frac{5}{y^2} \right) = \frac{d}{dx}(33).

Using the chain rule:

−10x3−10y3⋅dydx=0.-\frac{10}{x^3} - \frac{10}{y^3} \cdot \frac{dy}{dx} = 0.

Rearrange to find dydx\frac{dy}{dx}:

dydx=−10x3÷−10y3=−y3x3.\frac{dy}{dx} = -\frac{10}{x^3} \div -\frac{10}{y^3} = -\frac{y^3}{x^3}.

At the point (1, 4):

dydx=−4313=−64.\frac{dy}{dx} = -\frac{4^3}{1^3} = -64.

The equation of the tangent is:

y−y1=m(x−x1),y - y_1 = m(x - x_1),

where m=−64m = -64, (x1,y1)=(1,4)(x_1, y_1) = (1, 4). Substituting:

y−4=−64(x−1).y - 4 = -64(x - 1).

Simplify:

y−4=−64x+64⇒x+8y−33=0.y - 4 = -64x + 64 \quad \Rightarrow \quad x + 8y - 33 = 0.