a. verify that the given point is on the curve. evaluate 48(x² + y²)² for x = 3 and y = 4. b. determine an…

Explanation:

Step1: Evaluate left - hand side for part a

Substitute (x = 3) and (y = 4) into (48(x^{2}+y^{2})^{2}). First, calculate (x^{2}+y^{2}): [x^{2}+y^{2}=3^{2}+4^{2}=9 + 16=25] Then, ((x^{2}+y^{2})^{2}=25^{2}=625), and (48(x^{2}+y^{2})^{2}=48\times625 = 30000).

Step2: Evaluate right - hand side for part a

Substitute (x = 3) and (y = 4) into (625xy^{2}). We have (xy^{2}=3\times4^{2}=3\times16 = 48), and (625xy^{2}=625\times48=30000). Since the left - hand side equals the right - hand side when (x = 3) and (y = 4), the point ((3,4)) lies on the curve.

Step3: Differentiate the equation (48(x^{2}+y^{2})^{2}=625xy^{2}) implicitly

Let (u=x^{2}+y^{2}), then the left - hand side is (48u^{2}). Using the chain - rule, (\frac{d}{dx}(48u^{2})=96u\frac{du}{dx}=96(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})). The right - hand side: Using the product - rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = 625x) and (v = y^{2}), we get (\frac{d}{dx}(625xy^{2})=625y^{2}+1250xy\frac{dy}{dx}). So, (96(x^{2}+y^{2})(2x + 2y\frac{dy}{dx})=625y^{2}+1250xy\frac{dy}{dx}).

Step4: Substitute (x = 3) and (y = 4) into the differentiated equation

First, (x^{2}+y^{2}=25). The left - hand side of the differentiated equation is (96\times25\times(2\times3 + 2\times4\frac{dy}{dx})=2400\times(6 + 8\frac{dy}{dx})=14400+19200\frac{dy}{dx}). The right - hand side is (625\times4^{2}+1250\times3\times4\frac{dy}{dx}=10000 + 15000\frac{dy}{dx}). Then, (14400+19200\frac{dy}{dx}=10000 + 15000\frac{dy}{dx}). [19200\frac{dy}{dx}-15000\frac{dy}{dx}=10000 - 14400] [4200\frac{dy}{dx}=- 4400] (\frac{dy}{dx}=-\frac{22}{21}).

Step5: Find the equation of the tangent line

Using the point - slope form (y - y_{1}=m(x - x_{1})) where ((x_{1},y_{1})=(3,4)) and (m =-\frac{22}{21}), we have (y - 4=-\frac{22}{21}(x - 3)). [21y-84=-22x + 66] [22x+21y=150]

Answer:

a. Since (48(3^{2}+4^{2})^{2}=48\times25^{2}=48\times625 = 30000) and (625\times3\times4^{2}=625\times48 = 30000), the point ((3,4)) lies on the curve. b. The equation of the tangent line to the curve at the point ((3,4)) is (22x + 21y=150).